
Exercises1.1.8Exercises

Exercise1
Hint

Draw a rectangle that encompasses the entire shaded area, and one that is encompassed by the shaded area. The shaded area is no more than the area of the bigger rectangle, and no less than the area of the smaller rectangle.

The area is between $1.5$ and $2.5$ square units.

Solution

The diagram on the left shows a rectangle with area $2 \times 1.25=2.5$ square units. Since the blue-shaded region is entirely inside this rectangle, the area of the blue-shaded region is no more than 2.5 square units.

The diagram on the right shows a rectangle with area $2 \times 0.75=1.5$ square units. Since the blue-shaded region contains this entire rectangle, the area of the blue region is no less than 1.5 square units.

So, the area of the blue-shaded region is between 1.5 and 2.5 square units.

Remark: we could also give an obvious range, like “the shaded area is between zero and one million square units.” This would be true, but not very useful or interesting.

Exercise2
Hint

We can improve on the method of Question 1 by using three rectangles that together encompass the shaded region, and three rectangles that together are encompassed by the shaded region.

The shaded area is between 2.75 and 4.25 square units. (Other estimates are possible, but this is a reasonable estimate, using methods from this chapter.)

Solution

• Solution 1: One naive way to solve this is to simply use the same method as Question 1.

The rectangle on the left has area $3 \times 2.25 = 6.75$ square units, and encompasses the entire shaded region. The rectangle on the right has area $3 \times 0.25 = 0.75$ square units, and is entirely contained inside the blue-shaded region. So, the area of the blue-shaded region is between 0.75 and 6.75 square units.

This is a legitimate approximation, but we can easily do much better. The shape of this graph suggests that using the areas of three rectangles would be a natural way to improve our estimate.

• Solution 2: Let's use these rectangles instead:

In the left picture, the red area is $(1 \times 1.25)+(1 \times 2.25)+(1 \times 0.75)=4.25$ square units. In the right picture, the red area is $(1 \times 0.75)+(1 \times 1.75)+(1 \times 0.25)=2.75$ square units. So, the blue shaded area is between 2.75 and 4.25 square units.

Exercise3
Hint

Four rectangles suffice.

The area under the curve is a number in the interval $\left( \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right], \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right]\right)\text{.}$

Solution

Remark: in the solution below, we find the appropriate approximation using trial and error. In Question 46, we take a more systematic approach.

• Try 1: First, we can try by using a single rectangle as an overestimate, and a single rectangle as an underestimate.

The area under the curve is less than the area of the rectangle on the left ($2 \times \frac{1}{2}=1$) and greater than the area of the rectangle on the right ($2 \times \frac{1}{8}=\frac{1}{4}$). So, the area is in the range $\left(\frac{1}{4},1\right)\text{.}$ Unfortunately, this range is too big--we need our range to have length at most 0.2. So, we refine our approximation by using more rectangles.

• Try 2: Let's try using two rectangles each for the upper and lower bounds.

The rectangles in the left picture have area $\left(1 \times \frac{1}{2}\right)+\left(1 \times \frac{1}{4}\right)=\frac{3}{4}\text{,}$ and the rectangles in the right picture have area $\left(1 \times \frac{1}{4}\right)+\left(1 \times \frac{1}{8}\right)=\frac{3}{8}\text{.}$ So, the area under the curve is in the interval $\left(\frac{3}{8},\frac{3}{4}\right)\text{.}$ The length of this interval is $\frac{3}{8}\text{,}$ and $\frac{3}{8} \gt \frac{3}{15}=\frac{1}{5}=0.2\text{.}$ (Indeed, $\frac{3}{8}=0.375 \gt 0.2\text{.}$) Since the length of our interval is still bigger than 0.2, we need even more rectangles.

• Try 3: Let's go ahead and try four rectangles each for the upper and lower estimates.

The area of the rectangles on the left is:

\begin{equation*} \left(\frac{1}{2}\times \frac{1}{2}\right)+ \left(\frac{1}{2}\times \frac{1}{2\sqrt{2}}\right)+ \left(\frac{1}{2}\times \frac{1}{4}\right)+ \left(\frac{1}{2}\times \frac{1}{4\sqrt{2}}\right) = \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right], \end{equation*}

and the area of the rectangles on the right is:

\begin{equation*} \left(\frac{1}{2}\times \frac{1}{2\sqrt{2}}\right)+ \left(\frac{1}{2}\times \frac{1}{4}\right)+ \left(\frac{1}{2}\times \frac{1}{4\sqrt{2}}\right)+ \left(\frac{1}{2}\times \frac{1}{8}\right) = \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right]. \end{equation*}

So, the area under the curve is in the interval $\left( \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right], \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right]\right)\text{.}$ The length of this interval is $\frac{3}{16}\text{,}$ and $\frac{3}{16} \lt \frac{3}{15}=\frac{1}{5}=0.2\text{,}$ as desired. (Indeed, $\frac{3}{16}=0.1875 \lt 0.2\text{.}$)

Note, if we choose any value in the interval $\left( \frac{3}{8}\left[\frac{1}{2}+\frac{1}{\sqrt{2}}\right], \frac{3}{8}\left[1+\frac{1}{\sqrt{2}}\right]\right)$ as an approximation for the area under the curve, our error is no more than 0.2.

Exercise4
Hint

Try drawing a picture.

left

Solution

Since $f(x)$ is decreasing, it is larger on the left endpoint of an interval than on the right endpoint of an interval. So, a left Riemann sum gives a larger approximation. Notice this does not depend on $n\text{.}$

Furthermore, the actual area $\displaystyle\int_0^5f(x)\dee{x}$ is larger than its right Riemann sum, and smaller than its left Riemann sum.

Exercise5
Hint

Try an oscillating function.

Many answers are possible. One example is $f(x)=\sin x\text{,}$ $[a,b]=[0,\pi]\text{,}$ $n=1\text{.}$ Another example is $f(x)=\sin x\text{,}$ $[a,b]=[0,5\pi]\text{,}$ $n=5\text{.}$

Solution

If $f(x)$ is always increasing or always decreasing, then the midpoint Riemann sum will be between the left and right Riemann sums. So, we need a function that goes up and down. Many examples are possible, but let's work with a familiar one: $\sin x\text{.}$

If our intervals have endpoints that are integer multiples of $\pi\text{,}$ then the left and right Riemann sums will be 0, since $\sin(0)=\sin(\pi)=\sin(2\pi)=\cdots=0\text{.}$ The midpoints of these intervals will give $y$-values of 1 and -1. So, for example, we can let $f(x)=\sin x\text{,}$ $[a,b]=[0,\pi]\text{,}$ and $n=1\text{.}$ Then the right and left Riemann sums are 0, while the midpoint Riemann sum is $\pi\text{.}$

We can extend the example of $f(x)=\sin x$ to have more intervals. As long as we have more positive terms than negative, the midpoint approximation will be a positive number, and so it will be larger than both the left and right Riemann sums. So, for example, we can let $f(x)=\sin x\text{,}$ $[a,b]=[0,5\pi]\text{,}$ and $n=5\text{.}$ Then the midpoint Riemann sum is $\pi-\pi+\pi-\pi+\pi=\pi\text{,}$ which is strictly larger than 0 and so it is larger than both the left and right Riemann sums.

Exercise6
Hint

The ordering of the parts is intentional: each sum can be written by changing some small part of the sum before it.

Some of the possible answers are given, but more exist.

1. $\displaystyle\sum_{i=3}^7 i$\quad;\quad $\displaystyle\sum_{i=1}^5 (i+2)$
2. $\displaystyle\sum_{i=3}^7 2i$\quad;\quad $\displaystyle\sum_{i=1}^5 (2i+4)$
3. $\displaystyle\sum_{i=3}^7 (2i+1)$\quad;\quad $\displaystyle\sum_{i=1}^5 (2i+5)$
4. $\displaystyle\sum_{i=1}^8 (2i-1)$\quad;\quad$\displaystyle\sum_{i=0}^7 (2i+1)$
Solution

1. Two possible answers are $\displaystyle\sum_{i=3}^7 i$ and $\displaystyle\sum_{i=1}^5 (i+2)\text{.}$ The first has simpler terms ($i$ versus $i+2$), while the second has simpler indices (we often like to start at $i=1$). Neither is objectively better than the other, but depending on your purposes you might find one more useful.
2. The terms of this sum are each double the terms of the sum from part (a), so two possible answers are $\displaystyle\sum_{i=3}^7 2i$ and $\displaystyle\sum_{i=1}^5 (2i+4)\text{.}$ We often want to write a sum that involves even numbers: it will be useful for you to remember that the term $2i$ (with index $i$) generates evens.
3. The terms of this sum are each one more than the terms of the sum from part (b), so two possible answers are $\displaystyle\sum_{i=3}^7 (2i+1)$ and $\displaystyle\sum_{i=1}^5 (2i+5)\text{.}$ \\ In the last part, we used the expression $2i$ to generate even numbers; $2i+1$ will generate odds. So will the index $2i+5\text{,}$ and indeed, $2i+k$ for any odd number $k\text{.}$ The choice of what you add will depend on the bounds of $i\text{.}$
4. This sum adds up the odd numbers from 1 to 15. From Part (c), we know that the formula $2i+1$ is a simple way of generating odd numbers. Since our first term should be 1 and our last term should be 15, if we use $\sum (2i+1)\text{,}$ then $i$ should run from $0$ to $7\text{.}$ So, one way of expressing our sum in sigma notation is $\displaystyle\sum_{i=0}^7 (2i+1)\text{.}$

Sometimes we like our sum to start at $i=1$ instead of $i=0\text{.}$ If this is our desire, we can use $2i-1$ as our terms, and let $i$ run from 1 to 8. This gives us another way of expressing our sum: $\displaystyle\sum_{i=1}^8 (2i-1)\text{.}$

Exercise7
Hint

If we raise $-1$ to an even power, we get $+1\text{,}$ and if we raise it to an odd power, we get $-1\text{.}$

Some answers are below, but others are possible.

1. $\displaystyle\sum_{i=1}^4 \frac{1}{3^i}$ and $\displaystyle\sum_{i=1}^4 \left(\frac{1}{3}\right)^i$
2. $\displaystyle\sum_{i=1}^4 \frac{2}{3^i}$ and $\displaystyle\sum_{i=1}^4 2\left(\frac{1}{3}\right)^i$
3. $\displaystyle\sum_{i=1}^4(-1)^i \frac{2}{3^i}$ and $\displaystyle\sum_{i=1}^4 \frac{2}{(-3)^i}$
4. $\displaystyle\sum_{i=1}^4(-1)^{i+1} \frac{2}{3^i}$ and $\displaystyle\sum_{i=1}^4 -\frac{2}{(-3)^i}$
Solution

1. The denominators are successive powers of three, so one way of writing this is $\displaystyle\sum_{i=1}^4 \frac{1}{3^i}\text{.}$ Equivalently, the terms we're adding are powers of $1/3\text{,}$ so we can also write $\displaystyle\sum_{i=1}^4 \left(\frac{1}{3}\right)^i\text{.}$
2. This sum is obtained from the sum in (a) by multiplying each term by two, so we can write $\displaystyle\sum_{i=1}^4 \frac{2}{3^i}$ or $\displaystyle\sum_{i=1}^4 2\left(\frac{1}{3}\right)^i\text{.}$
3. The difference between this sum and the previous sum is its alternating sign, minus-plus-minus-plus. This behaviour appears when we raise a negative number to successive powers. We can multiply each term by $(-1)^i\text{,}$ or we can slip a negative into the number that is already raised to the power $i\text{:}$ $\displaystyle\sum_{i=1}^4(-1)^i \frac{2}{3^i}$\,, or $\displaystyle\sum_{i=1}^4 \frac{2}{(-3)^i}\text{.}$
4. This sum is the negative of the sum in part (c), so we can simply multiply each term by negative one: $\displaystyle\sum_{i=1}^4(-1)^{i+1} \frac{2}{3^i}$ or $\displaystyle\sum_{i=1}^4 -\frac{2}{(-3)^i}\text{.}$

Be careful with the second form: a common mistake is to think that $-\dfrac{2}{(-3)^i} = \dfrac{2}{3^i}\text{,}$ but these are not the same.

Exercise8
Hint

Sometimes a little anti-simplification can make the pattern more clear.

1. Re-write as $\frac{1}{3}+\frac{3}{9}+\frac{5}{27}+\frac{7}{81}+\frac{9}{243}\text{.}$
2. Compare to the sum in the hint for (a).
3. Re-write as $1\cdot1000+2\cdot 100+3\cdot10+\frac{4}{1}+\frac{5}{10}+\frac{6}{100}+\frac{7}{1000}\text{.}$

1. $\displaystyle\sum_{i=1}^5 \frac{2i-1}{3^i}$
2. $\displaystyle\sum_{i=1}^5 \frac{1}{3^i+2}$
3. $\displaystyle\sum_{i=1}^7 i\cdot10^{4-i}$\ and $\displaystyle\sum_{i=1}^7 \frac{i}{10^{i-4}}$
Solution

1. If we re-write the second term as $\frac{3}{9}$ instead of $\frac{1}{3}\text{,}$ our sum becomes:

\begin{equation*} \frac{1}{3}+\frac{3}{9}+\frac{5}{27}+\frac{7}{81}+\frac{9}{243} \end{equation*}

The numerators are the first five odd numbers, and the denominators are the first five positive powers of 3. We learned how to generate odd numbers in Question 6, and we learned how to generate powers of three in Question 7. Combining these, we can write our sum as $\displaystyle\sum_{i=1}^5 \frac{2i-1}{3^i}\text{.}$

2. The denominators of these terms differ from the denominators of part (a) by precisely two, while the numerators are simply 1. So, we can modify our previous answer: $\displaystyle\sum_{i=1}^5 \frac{1}{3^i+2}\text{.}$
3. Let's re-write the sum to make the pattern clearer.
\begin{equation*} \begin{array}{cccccccccccccc} &1000&+&200&+&30&+&4&+&\frac{1}{2}&+&\frac{3}{50}&+&\frac{7}{1000}\\ =&1\cdot1000&+&2\cdot 100&+&3\cdot10&+&\frac{4}{1}&+&\frac{5}{10}&+&\frac{6}{100}&+&\frac{7}{1000}\\ =&1\cdot 10^3 &+& 2\cdot 10^2&+&3\cdot 10^1 &+& 4\cdot 10^0 &+& 5 \cdot 10^{-1} &+& 6\cdot 10^{-2} &+& 7 \cdot 10^{-3} \\ =&\textcolor{red}1\cdot 10^{4-\textcolor{red}1} &+& \textcolor{red}2\cdot 10^{4-\textcolor{red}2}&+&\textcolor{red}3\cdot 10^{4-\textcolor{red}3} &+& \textcolor{red}4\cdot 10^{4-\textcolor{red}4} &+& \textcolor{red}5 \cdot 10^{4-\textcolor{red}5} &+& \textcolor{red}6\cdot 10^{4-\textcolor{red}6} &+& \textcolor{red}7 \cdot 10^{4-\textcolor{red}7} \end{array} \end{equation*}

If we let the red numbers be our index $i\text{,}$ this gives us the expression $\displaystyle\sum_{i=1}^7 i\cdot10^{4-i}$\,. Equivalently, we can write $\displaystyle\sum_{i=1}^7 \frac{i}{10^{i-4}}$\,.

Exercise9
Hint

• (a), (b) These are geometric sums.
• (c) You can write this as three separate sums.
• (d) You can write this as two separate sums. Remember that $e$ is a constant. Don't be thrown off by the index being $n$ instead of $i\text{.}$

1. $\dfrac{5}{2}\left[1-\left(\dfrac{3}{5}\right)^{101}\right]$
2. $\dfrac{5}{2}\left(\dfrac{3}{5}\right)^{50}\left[1-\left(\dfrac{3}{5}\right)^{51}\right]$
3. $270$
4. $\dfrac{1-\left(\frac{1}{e}\right)^b}{e-1}+\dfrac{e}{4}\left[b(b+1)\right]^2$
Solution

1. Using Theorem 1.1.6, part (a) with $a=1\text{,}$ $r=\frac{3}{5}$ and $n=100\text{:}$ \begin{equation*} \sum_{i=0}^{100} \left(\dfrac{3}{5}\right)^i = \dfrac{1-\left(\frac{3}{5}\right)^{101}}{1-\frac{3}{5}} = \dfrac{5}{2}\left[1-\left(\frac{3}{5}\right)^{101}\right] \end{equation*}
2. We want to use Theorem 1.1.6, part (a) again, but our sum doesn't start at $\left(\frac{3}{5}\right)^0=1\text{.}$ We have two options: factor out the leading term, or use the difference of two sums that start where we want them to.

• Solution 1: In this solution, we'll make our sum start at 1 by factoring out the leading term. We wrote our work out the long way (expanding the sigma into “dot-dot-dot” notation) for clarity, but it's faster to do the algebra in sigma notation all the way through.

\begin{align*} \displaystyle\sum_{i=50}^{100} \left(\dfrac{3}{5}\right)^i&= \left(\dfrac{3}{5}\right)^{50}+ \left(\dfrac{3}{5}\right)^{51}+ \left(\dfrac{3}{5}\right)^{52}+\cdots+ \left(\dfrac{3}{5}\right)^{100}\\ &= \left(\dfrac{3}{5}\right)^{50}\left[1+ \left(\dfrac{3}{5}\right)+ \left(\dfrac{3}{5}\right)^{2}+\cdots+ \left(\dfrac{3}{5}\right)^{50}\right]\\ &= \left(\dfrac{3}{5}\right)^{50}\dfrac{1-\left(\frac{3}{5}\right)^{51}}{1-\frac{3}{5}}\\ &=\dfrac{5}{2}\left(\dfrac{3}{5}\right)^{50}\left[1-\left(\frac{3}{5}\right)^{51}\right]. \end{align*}
• Solution 2: In this solution, we write our given expression as the difference of two sums, both starting at $i=0\text{.}$

\begin{align*} \displaystyle\sum_{i=50}^{100} \left(\dfrac{3}{5}\right)^i&= \displaystyle\sum_{i=0}^{100} \left(\dfrac{3}{5}\right)^i- \displaystyle\sum_{i=0}^{49} \left(\dfrac{3}{5}\right)^i\\ &=\dfrac{1-\left(\frac{3}{5}\right)^{101}}{1-\frac{3}{5}} - \dfrac{1-\left(\frac{3}{5}\right)^{50}}{1-\frac{3}{5}} \\ &=\dfrac{5}{2}\left[\left(\frac{3}{5}\right)^{50}-\left(\frac{3}{5}\right)^{101}\right]\\ &=\dfrac{5}{2}\left(\dfrac{3}{5}\right)^{50}\left[1-\left(\frac{3}{5}\right)^{51}\right]. \end{align*}
3. Before we can use the equations in Theorem 1.1.6, we'll need to do a little simplification. \begin{align*} \displaystyle\sum_{i=1}^{10} \left(i^2-3i+5\right)&= \displaystyle\sum_{i=1}^{10} i^2 +\displaystyle\sum_{i=1}^{10} -3i +\displaystyle\sum_{i=1}^{10}5\\ &= \displaystyle\sum_{i=1}^{10} i^2 -3\displaystyle\sum_{i=1}^{10} i +5\displaystyle\sum_{i=1}^{10}1\\ &= \frac{1}{6}(10)(11)(21) -3\left(\frac{1}{2}(10\cdot 11)\right) +5\cdot 10\\ &=270 \end{align*}
4. As in part (c), we'll simplify first. The first part (shown here in red) is a geometric sum, but it does not start at $1=\left(\frac{1}{e}\right)^0\text{.}$ \begin{align*} \displaystyle\sum_{n=1}^{b}\left[\textcolor{red}{ \left(\frac{1}{e}\right)^n} +\,\textcolor{blue}{en^3}\right]&= \textcolor{red}{\displaystyle\sum_{n=1}^{b} \left(\frac{1}{e}\right)^n} + \textcolor{blue}{ \displaystyle\sum_{n=1}^{b}en^3}\\ &=\textcolor{red}{\displaystyle\sum_{n=0}^{b} \left(\frac{1}{e}\right)^{n}-1} + \textcolor{blue}{e\displaystyle\sum_{n=1}^{b}n^3}\\ &=\textcolor{red}{\dfrac{1-\left(\frac{1}{e}\right)^{b+1}}{1-\frac{1}{e}}-1} + \textcolor{blue}{e\left[\frac{1}{2}b(b+1)\right]^2}\\ &=\textcolor{red}{\dfrac{\frac{1}{e}-\left(\frac{1}{e}\right)^{b+1}}{1-\frac{1}{e}}} + \textcolor{blue}{e\left[\frac{1}{2}b(b+1)\right]^2}\\ &=\textcolor{red}{\dfrac{1-\left(\frac{1}{e}\right)^b}{e-1}}+\textcolor{blue}{\frac{e}{4}\left[b(b+1)\right]^2} \end{align*}
Exercise10
Hint

1. Write out the terms of the two sums.
2. A change of index is an easier option than expanding the cubic.
3. Which terms cancel?
4. Remember $2n+1$ is odd for every integer $n\text{.}$ The index starts at $n=2\text{,}$ not $n=1\text{.}$

1. $50\cdot 51=2550$
2. $\left[\frac{1}{2}(95)(96)\right]^2-\left[\frac{1}{2}(4)(5)\right]^2$
3. $-1$
4. $-10$
Solution

1. The two pieces are very similar, which we can see by changing the index, or expanding them out: \begin{align*} \displaystyle\sum_{i=50}^{100} (i-50)+\displaystyle\sum_{i=0}^{50} i&= \left(0+1+2+\cdots + 50\right)+\left(0+1+2+\cdots + 50\right)\\ &=\left(1+2+\cdots + 50\right)+\left(1+2+\cdots + 50\right)\\ &=2\left(1+2+\cdots + 50\right)\\ &=2\sum_{i=1}^{50} i\\ &= 2\left(\frac{50\cdot 51}{2}\right)=50\cdot 51=2550 \end{align*}
2. If we expand $(i-5)^3 = i^3-15i^2+75i-225\text{,}$ we can break the sum into four parts, and evaluate each separately. However, it is much simpler to change the index and make the term $(i-5)^3$ into $i^3\text{.}$ \begin{align*} \displaystyle\sum_{i=10}^{100} \left(i-5\right)^3&= 5^3+6^3+7^3+\cdots +95^3\\ \end{align*}

We have a formula to evaluate the sum of cubes if they start at $1\text{,}$ so we turn our expression into the difference of two sums starting at 1:

\begin{align*} &= \left[1^3+2^3+3^3+4^3+5^3+6^3+7^3+\cdots +95^3\right]- \left[1^3+2^3+3^3+4^3\right]\\ &=\displaystyle\sum_{i=1}^{95} i^3 - \displaystyle\sum_{i=1}^4 i^3\\ &=\left[\frac{1}{2}(95)(96)\right]^2-\left[\frac{1}{2}(4)(5)\right]^2\,. \end{align*}
3. Notice every two terms cancel with each other, since the sum is $(-1)+(+1)\text{,}$ etc. Then the terms $n=1$ through $n=10$ cancel, and we're left only with the final term, $(-1)^{11}=-1\text{.}$

Written out more explicitly:

\begin{align*} \displaystyle\sum_{n=1}^{11} (-1)^{n}&=-1+1-1+1-1+1-1+1-1+1-1\\ &=[-1+1]+[-1+1]+[-1+1]+[-1+1]+[-1+1]-1\\ &=0+0+0+0+0-1=-1. \end{align*}
4. For every integer $n\text{,}$ $2n+1$ is odd, so $(-1)^{2n+1}=-1\text{.}$ Then $\displaystyle\sum_{n=2}^{11} (-1)^{2n+1} =\displaystyle\sum_{n=2}^{11} -1 =-10\text{.}$
Exercise11
Hint

Since the sum adds four pieces, there will be four rectangles. However, one might be extremely small.

Solution

The index of the sum runs from 1 to 4: the first, second, third, and fourth rectangles. So, we have four rectangles in our Riemann sum. Let's start by drawing in the intervals along the $x$-axis taken up by these four rectangles. Note each has the same width: $\dfrac{b-a}{4}\text{.}$

Since this is a midpoint Riemann sum, the height of each rectangle is given by the $y$-value of the function in the midpoint of the interval. So, now let's find the height of the function at the midpoints of each of the four intervals.

The left-most interval has a height of about 0, so it gives a “trivial” rectangle with no height and no area. The middle two intervals have rectangles of about the same height, and the right-most interval has the highest rectangle.

Exercise12(*)
Hint

Write out the general formula for the left Riemann sum from Definition 1.1.11 and choose $a\text{,}$ $b$ and $n$ to make it match the given sum.

$n=4\text{,}$ $a=2\text{,}$ and $b=6$

Solution

In general, the {left} Riemann sum for the integral $\int_a^b f(x)\,\,\dee{x}$ is of the form

\begin{gather*} \sum_{k=1}^n f\left(a+(k-1)\frac{b-a}{n}\right)\frac{b-a}{n} \end{gather*}

• To get the limits of summation to match the given sum, we need $n=4\text{.}$
• Then to get the factor multiplying $f$ to match that in the given sum, we need $\frac{b-a}{n}=1\text{,}$ so $b-a=4\text{.}$
• Finally, to get the argument of $f$ to match that in the given sum, we need \begin{gather*} a+(k-1)\frac{b-a}{n}=a-\frac{b-a}{n} +k\frac{b-a}{n}=1+k \end{gather*} Subbing in $n=4$ and $b-a=4$ gives $a-1 +k=1+k\text{,}$ so $a=2$ and $b=6\text{.}$
Exercise13
Hint

Since the sum runs from 1 to 3, there are three intervals. Suppose $2 = \Delta x = \frac{b-a}{n}\text{.}$ You may assume the sum given is a right Riemann sum (as opposed to left or midpoint).

One answer is below, but other interpretations exist.

Solution

The general form of a Riemann sum is $\displaystyle\sum_{i=1}^n \Delta x \cdot f(x_i^*)\text{,}$ where $\Delta x = \frac{b-a}{n}$ is the width of each rectangle, and $f(x_i^*)$ is the height.

There are different ways to interpret the given sum as a Riemann sum. The most obvious is given in Solution 1. You may notice that we make some convenient assumptions in this solution about values for $\Delta x$ and $a\text{,}$ and we assume the sum is a right Riemann sum. Other visualizations of the sum arise from making more exotic choices. Some of these are explored in Solutions 2-4.

All cases have three rectangles, and the three rectangles will have the same areas: 98, 162, and 242 square units, respectively. This is because the terms of the given sum simplify to $98+162+242\text{.}$

• Solution 1:

• Because the index runs from $1$ to $3\text{,}$ there are three intervals: $n=3\text{.}$
• Looking at our sum, it seems reasonable to interpret $\Delta x = 2\text{.}$ Then, since $n=3\text{,}$ we conclude $\frac{b-a}{3}=2\text{,}$ hence $b-a=6\text{.}$
• If $\Delta x = 2\text{,}$ then $f(x_i^*)=\left(5+2i\right)^2\text{.}$ Recall that $x_i^*$ is the $x$-coordinate we use to decide the height of the $i$th rectangle. In a right Riemann sum, $x_i^* = a+i\cdot\Delta x\text{.}$ So, using $2=\Delta x\text{,}$ we can let $f(x_i^*)=f(a+2i)=\left(5+2i\right)^2\text{.}$ This fits with the function $f(x)=x^2\text{,}$ and $a=5\text{.}$
• Since $b-a=6\text{,}$ and $a=5\text{,}$ this tells us $b=11$

To sum up, we can interpret the Riemann sum as a right Riemann sum, with three intervals, of the function $f(x)=x^2$ from $x=5$ to $x=11\text{.}$

• Solution 2: We could have chosen a different value for $\Delta x\text{.}$

• The index of the sum runs from 1 to 3, so we have $n=3\text{.}$
• We didn't have to interpret $\Delta x$ as 2--that was just the path of least resistance. We could have chosen it to be any other number--for the sake of argument, let's say $\Delta x=10\text{.}$ (Positive numbers are easiest to interpret, but negatives are technically allowed as well.)
• Then $10=\frac{b-a}{n}=\frac{b-a}{3}\text{,}$ so $b-a=30\text{.}$
• Let's use the paradigm of a right Riemann sum, and match up the terms of the sum given in the problem to the terms in the definition: \begin{align*} \Delta x \cdot f\left(a+i\cdot \Delta x\right)&= 2\cdot\left(5+2i\right)^2\\ 10 \cdot f(a+10i)&= 2\cdot\left(5+2i\right)^2\\ f(a+10i)&=\frac{1}{5}\cdot\left(5+2i\right)^2\\ f(a+10i)&=\frac{1}{5}\cdot\left(5+\frac{1}{5}\cdot 10i\right)^2 \end{align*}
• The easiest value of $a$ in this case is $a=0\text{.}$ Then $f(\textcolor{red}{10i}) = \frac{1}{5}\cdot\left(5+\frac{1}{5}\cdot \textcolor{red}{10i}\right)^2\text{,}$ so $f(\textcolor{red}{x})= \frac{1}{5}\cdot\left(5+\frac{1}{5}\cdot \textcolor{red}{x}\right)^2\text{.}$
• If $a=0$ and $b-a=30\text{,}$ then $b=30\text{.}$
• To sum up: $n=3\text{,}$ $a=0\text{,}$ $b=30\text{,}$ $\Delta x = 10\text{,}$ and $f(x)= \frac{1}{5}\cdot\left(5+\frac{x}{5} \right)^2\text{.}$

By changing $\Delta x\text{,}$ we changed the widths of the rectangles. The rectangles in this picture are wider and shorter than the rectangles in Solution 1. Their areas are the same: 98, 162, and 242.

• Solution 3: We could have chosen a different value of $a\text{.}$

• Suppose $\Delta x = 2\text{,}$ and we interpret our sum as a right Riemann sum, but we didn't assume $a=5\text{.}$ We could have chosen $a$ to be any number--say, $a=1\text{.}$
• Let's match up what we're given in the problem to what we're given as a definition: \begin{align*} \Delta x \cdot f\left(a+i\cdot\Delta x\right)&=2\cdot\left(5+2i\right)^2\\ 2 \cdot f\left(1+2i\right)&=2\cdot\left(5+2i\right)^2\\ f\left(1+2i\right)&=\left(5+2i\right)^2\\ f\left(1+2i\right)&=\left(4+1+2i\right)^2 \end{align*}
• Since $f(\textcolor{red}{1+2i})=\left(4+\textcolor{red}{1+2i}\right)^2\text{,}$ we have $f(\textcolor{red}{x})=\left(4+\textcolor{red}{x}\right)^2$
• Since $a=1$ and $\frac{b-a}{3}=2\text{,}$ in this case $b=7\text{.}$
• To sum up: $n=3\text{,}$ $a=1\text{,}$ $b=7\text{,}$ $\Delta x=2\text{,}$ and $f(x)=(4+x)^2\text{.}$

This picture is a lot like the picture in Solution 1, but shifted to the left. By changing $a\text{,}$ we changed the left endpoint of our region.

• Solution 4: We could have chosen a different kind of Riemann sum.

• We didn't have to assume that we were dealing with a right Riemann sum. Suppose $\Delta x =2\text{,}$ and we have a midpoint Riemann sum.
• Let's match up what we're given in the problem with what we're given in the definition: \begin{align*} \Delta x \cdot f\left(a+\left(i-\tfrac{1}{2}\right)\Delta x\right)&=2\cdot\left(5+2i\right)^2\\ 2 \cdot f\left(a+\left(i-\tfrac{1}{2}\right)2\right)&=2\cdot\left(5+2i\right)^2\\ f\left(a+\left(i-\tfrac{1}{2}\right)2\right)&=\left(5+2i\right)^2\\ f\left(a+2i-1\right)&=\left(5+2i\right)^2\\ f\left((a-1)+2i\right)&=\left(5+2i\right)^2 \end{align*}
• It is now convenient to set $a-1=5\text{,}$ hence $a=6\text{.}$
• Then $f(\textcolor{red}{5+2i})=(\textcolor{red}{5+2i})^2\text{,}$ so $f(\textcolor{red}{x})=\textcolor{red}{x}^2$
• Since $2=\frac{b-a}{3}$ and $a=6\text{,}$ we see $b=12\text{.}$
• To sum up: $n=3\text{,}$ $a=6\text{,}$ $b=12\text{,}$ $\Delta x = 2\text{,}$ and $f(x)=x^2\text{.}$

By choosing to interpret our sum as a midpoint Riemann sum instead of a right Riemann sum, we changed where our rectangles intersect the graph $y=f(x)\text{:}$ instead of the graph hitting the right corner of the rectangle, it hits in the middle.

Exercise14
Hint

Let $\Delta x = \dfrac{\pi}{20}\text{.}$ Then what is $b-a\text{?}$

Many interpretations are possible--see the solution to Question 13 for a more thorough discussion--but the most obvious is given below.

Solution

Many interpretations are possible--see the solution to Question 13 for a more thorough discussion--but the most obvious is given below. Recall the definition of a left Riemann sum:

\begin{equation*} \sum_{i=1}^n \Delta x \cdot f\left(a+(i-1)\Delta x\right) \end{equation*}

We chose a left Riemann sum instead of right or midpoint because our given sum has $(i-1)$ in it, rather than $(i-\frac{1}{2})$ or simply $i\text{.}$

• Since the sum has five terms ($i$ runs from 1 to 5), there are 5 rectangles. That is, $n=5\text{.}$
• In the definition of the Riemann sum, note that the term $\Delta x$ appears twice: once multiplied by the entire term, and once multiplied by $i-1\text{.}$ So, a convenient choice for $\Delta x$ is $\frac{\pi}{20}\text{,}$ because this is the constant that is both multiplied at the start of the term, and multiplied by $i-1\text{.}$
• Since $\dfrac{\pi}{20}=\Delta x = \dfrac{b-a}{n} = \dfrac{b-a}{5}\text{,}$ we see $b-a=\dfrac{5\pi}{20}=\dfrac{\pi}{4}\text{.}$
• We match the terms in the definition with the terms in the problem: \begin{align*} f(a+(i-1)\Delta x) & = \tan\left(\frac{\pi (i-1)}{20}\right)\\ f\left(a+(i-1)\frac{\pi}{20}\right) & = \tan\left((i-1)\frac{\pi }{20}\right) \end{align*} So, we choose $a=0$ and $f(x) = \tan x\text{.}$
• Since $a=0$ and $b-a=\frac{\pi}{4}\text{,}$ we see $b=\frac{\pi}{4}\text{.}$

We note that the first rectangle of the five is a “trivial” rectangle, with height (and area) 0.

Exercise15(*)
Hint

Notice that the index starts at $k=0\text{,}$ instead of $k=1\text{.}$ Write out the given sum explicitly without using summation notation, and sketch where the rectangles would fall on a graph of $y=f(x)\text{.}$

Then try to identify $b-a\text{,}$ and $n\text{,}$ followed by “right”, “left”, or “midpoint”, and finally $a\text{.}$

Three answers are possible. It is a midpoint Riemann sum for $f$ on the interval $[1,5]$ with $n =4\text{.}$ It is also a left Riemann sum for $f$ on the interval $[1.5,5.5]$ with $n =4\text{.}$ It is also a right Riemann sum for $f$ on the interval $[0.5,4.5]$ with $n =4\text{.}$

Solution

Since there are four terms in the sum, $n=4\text{.}$ (Note the sum starts at $k=0\text{,}$ instead of $k=1\text{.}$) Since the function is multiplied by 1, $1=\Delta x=\dfrac{b-a}{n}=\dfrac{b-a}{4}\text{,}$ hence $b-a=4\text{.}$

We can choose to view the given sum as a left, right, or midpoint Riemann sum. The choice we make determines the interval. Note that the heights of the rectangles are determined when $x = 1.5,\, 2.5,\, 3.5,$ and $4.5\text{.}$

• Option 1: right Riemann sum. If our sum is a right Riemann sum, then we take the heights of the rectangles from the right endpoint of each interval.

Then $a=0.5$ and $b=4.5\text{.}$ Therefore: $\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1$ is a right Riemann sum on the interval $[0.5,4.5]$ with $n=4\text{.}$

• Option 2: left Riemann sum. If our sum is a left Riemann sum, then we take the heights of the rectangles from the left endpoint of each interval.

Then $a=1.5$ and $b=5.5\text{.}$ Therefore: $\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1$ is a left Riemann sum on the interval $[1.5,5.5]$ with $n=4\text{.}$

• Option 3: midpoint Riemann sum. If our sum is a midpoint Riemann sum, then we take the heights of the rectangles from the midpoint of each interval.

Then $a=1$ and $b=5\text{.}$ Therefore: $\sum\limits_{k=0}^3 f (1.5 + k) \cdot 1$ is a midpoint Riemann sum on the interval $[1,5]$ with $n=4\text{.}$

Exercise16
Hint

The area is a triangle.

$\dfrac{25}{2}$

Solution

The area in question is a triangle with base 5 and height 5, so its area is $\dfrac{25}{2}\text{.}$

Exercise17
Hint

There is one triangle of positive area, and one of negative area.

$\dfrac{21}{2}$

Solution

There is a positive and a negative portion of this area. The positive area is a triangle with base 5 and height 5, so area $\dfrac{25}{2}$ square units. The negative area is a triangle with base $2$ and height $2\text{,}$ so negative area $\dfrac{4}{2}=2$ square units. So, the net area is $\dfrac{25}{2}-\dfrac{4}{2}=\dfrac{21}{2}$ square units.

Exercise18(*)
Hint

Review Definition 1.1.11.

$\sum\limits_{i=1}^{50} \Big(5+\big(i-\frac{1}{2}\big)\frac{1}{5}\Big)^8 \ \frac{1}{5}$

Solution

In general, the midpoint Riemann sum is given by

\begin{gather*} \sum_{i=1}^n f\Big(a+\big(i-\frac{1}{2}\big)\De x\Big)\ \De x \, , \qquad \mbox{where } \De x = \frac{b-a}{n}. \end{gather*}

In this problem we are told that $f(x)=x^8\text{,}$ $a=5\text{,}$ $b=15$ and $n=50\text{,}$ so that $\De x = \frac{b-a}{n} = \frac{1}{5}$ and the desired Riemann sum is:

\begin{gather*} \sum_{i=1}^{50} \Big(5+\big(i-\frac{1}{2}\big)\frac{1}{5}\Big)^8 \ \frac{1}{5} \end{gather*}
Exercise19(*)

$54$

Solution

The given integral has interval of integration going from $a=-1$ to $b=5\text{.}$ So when we use three approximating rectangles, all of the same width, the common width is $\Delta x=\frac{b-a}{n} = 2\text{.}$ The first rectangle has left endpoint $x_0=a=-1\text{,}$ the second has left hand endpoint $x_1=a+\Delta x=1\text{,}$ and the third has left hand end point $x_2=a+2\Delta x=3\text{.}$ So

\begin{gather*} \int_{-1}^5 x^3\,\,\dee{x} \approx \big[f(x_0)+f(x_1)+f(x_2)\big]\Delta x =\big[(-1)^3+1^3+3^3\big]\times2 =54 \end{gather*}
Exercise20(*)
Hint

You'll want the limit as $n$ goes to infinity of a sum with $n$ terms. If you're having a hard time coming up with the sum in terms of $n\text{,}$ try writing a sum with a finite number of terms of your choosing. Then, think about how that sum would change if it had $n$ terms.

$\displaystyle\int_{-1}^{7}f(x)\,\,\dee{x}=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^{n} f\left(-1+\frac{8i}{n}\right)\frac{8}{n}$

Solution

In the given integral, the domain of integration runs from $a=-1$ to $b=7\text{.}$ So, we have $\Delta x = \frac{(b-a)}{n}= \frac{(7-(-1))}{n} = \frac{8}{n}\text{.}$ The left-hand end of the first subinterval is at $x_0=a=-1\text{.}$ So, the right-hand end of the $i^{\rm th}$ interval is at $x_i^* = -1+\frac{8i}{n}\text{.}$ So:

\begin{gather*} \int_{-1}^{7}f(x)\,\,\dee{x} =\lim_{n\to\infty}\sum_{i=1}^{n} f\left(-1+\frac{8i}{n}\right)\frac{8}{n} \end{gather*}
Exercise21(*)
Hint

The main step is to express the given sum as the right Riemann sum,

\begin{equation*} \sum_{i=1}^{n} f(a+i\De x)\Delta x. \end{equation*}

Don't be afraid to guess $\De x$ and $f(x$) (review Definition 1.1.11). Then write out explicitly $\sum\limits_{i=1}^{n} f(a+i\De x)\Delta x$ with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

$f(x) = \sin^2 (2 + x)$ and $b=4$

Solution

We identify the given sum as the right Riemann sum $\sum\limits_{i=1}^{n} f(a+i\De x)\Delta x\text{,}$ with $a=0$ (that's specified in the statement of the question). Since $\frac{4}{n}$ is multiplied in every term, and is also multiplied by $i\text{,}$ we let $\Delta x = \frac{4}{n}\text{.}$ Then $x_i^* = a+i\De x=\frac{4i}{n}$ and $f(x) = \sin^2 (2 + x)\text{.}$ So, $b=a+n\De x=0+n\cdot\frac{4}{n}=4\text{.}$

Exercise22(*)
Hint

The main step is to express the given sum as the right Riemann sum $\sum\limits_{k=1}^{n} f(a+k\De x)\Delta x\text{.}$ Don't be afraid to guess $\De x$ and $f(x$) (review Definition 1.1.11). Then write out explicitly $\sum\limits_{k=1}^{n} f(a+k\De x)\Delta x$ with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

$f(x)=x\sqrt{1-x^2}$

Solution

The given sum is of the form

\begin{gather*} \lim_{n\rightarrow\infty}\sum_{k=1}^n \frac{k}{n^2}\sqrt{1-\frac{k^2}{n^2}} =\lim_{n\rightarrow\infty}\sum_{k=1}^n \left(\frac{1}{n}\right) \frac{k}{n}\sqrt{1-\left(\frac{k}{n}\right)^2} =\lim_{n\rightarrow\infty}\sum_{k=1}^n \De x f(x_k^*) \end{gather*}

with $\De x=\frac{1}{n}\text{,}$ $a=0\text{,}$ $x_k^*=\frac{k}{n}=a+k\De x$ and $f(x)=x\sqrt{1-x^2}\text{.}$ Since $x_0^*=0$ and $x_n^*=1\text{,}$ the right hand side is the definition (using the right Riemann sum) of $\int_0^1 f(x)\,\,\dee{x} \text{.}$

Exercise23(*)
Hint

The main step is to express the given sum in the form $\sum_{i=1}^{n} f(x_i^*)\Delta x\text{.}$ Don't be afraid to guess $\De x\text{,}$ $x_i^*$ (for either a left or a right or a midpoint sum — review Definition 1.1.11) and $f(x$). Then write out explicitly $\sum_{i=1}^{n} f(x_i^*)\Delta x$ with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

$\int_0^3 e^{-x/3}\cos(x)\,\,\dee{x}$

Solution

As $i$ ranges from $1$ to $n\text{,}$ $3i/n$ range from $3/n$ to $3$ with jumps of $\De x=3/n\text{,}$ so this is

\begin{gather*} \lim_{n\to\infty}\displaystyle\sum_{i=1}^{n} \frac{3}{n} e^{-i/n} \cos(3i/n) =\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x =\int_a^b f(x)\,\,\dee{x} \end{gather*}

where $x_i^* = 3i/n\text{,}$ $f(x) = e^{-x/3}\cos(x)\text{,}$ $a=x_0=0$ and $b=x_n=3\text{.}$ Thus

\begin{gather*} \lim_{n\to\infty}\displaystyle\sum_{i=1}^{n} \frac{3}{n} e^{-i/n} \cos(3i/n) =\int_0^3 e^{-x/3}\cos(x)\,\,\dee{x} \end{gather*}
Exercise24(*)
Hint

The main step is to express the given sum in the form $\sum\limits_{i=1}^{n} f(x_i^*)\Delta x\text{.}$ Don't be afraid to guess $\De x\text{,}$ $x_i^*$ (probably, based on the symbol $R_n\text{,}$ assuming we have a right Riemann sum — review Definition 1.1.11) and $f(x$). Then write out explicitly $\sum\limits_{i=1}^{n} f(x_i^*)\Delta x$ with your guess substituted in, and compare the result with the given sum. Adjust your guess if they don't match.

$\displaystyle\int_0^1 x e^{x}\,\,\dee{x}$

Solution

As $i$ ranges from $1$ to $n\text{,}$ the exponent $\frac{i}{n}$ ranges from $\frac{1}{n}$ to $1$ with jumps of $\De x=\frac{1}{n}\text{.}$ So let's try $x_i^* =\frac{ i}{n}\text{,}$ $\Delta x=\frac{1}{n}\text{.}$ Then:

\begin{gather*} R_n= \sum_{i=1}^{n} \frac{i e^{i/n}}{n^2} = \sum_{i=1}^{n} \frac{i}{n} e^{i/n} \frac{1}{n} = \sum_{i=1}^{n} x_i^* e^{x_i^*} \Delta x = \sum_{i=1}^{n} f(x_i^*) \Delta x \end{gather*}

with $f(x)= x e^x\text{,}$ and the limit

\begin{gather*} \lim_{n\to\infty} R_n =\lim_{n\to\infty} \sum_{i=1}^{n} f(x_i^*)\Delta x =\int_a^b f(x)\,\,\dee{x} \end{gather*}

Since we chose $x_i^* = \frac{i}{n} = 0+i\De x\text{,}$ we let $a=0\text{.}$ Then $\frac{1}{n}=\De x = \frac{b-a}{n}=\frac{b}{n}$ tells us $b=1\text{.}$ Thus,

\begin{gather*} \lim_{n\to\infty}R_n =\int_0^1 xe^{x}\,\,\dee{x}\,. \end{gather*}
Exercise25(*)
Hint

Try several different choices of $\De x$ and $x_i^*\text{.}$

Possible answers include: $\displaystyle\int\limits_0^2 e^{-1-x}\ \,\dee{x}\text{,}$\quad $\displaystyle\int\limits_1^3 e^{-x}\ \,\dee{x}\text{,}$ \quad $2\displaystyle\int_{1/2}^{3/2} e^{-2x}\ \,\dee{x}\text{,}$ \quad and \quad $2\displaystyle\int\limits_0^1 e^{-1-2x}\ \,\dee{x}\text{.}$

Solution

• Choice #1: If we set $\De x = \frac{2}{n}$ and $x_i^*= \frac{2i}{n}\text{,}$ i.e. $x_i^* = a + i\De x$ with $a=0\text{,}$ then \begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-x_i^*}\De x \bigg) \hidewidth \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with $f(x) = e^{-1-x}$} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with $a=x_0=0$ and $b=x_n=2$} \\ &=\int_0^2 e^{-1-x}\ \,\dee{x} \end{align*}
• Choice #2: If we set $\De x = \frac{2}{n}$ and $x_i^*= 1+\frac{2i}{n}\text{,}$ i.e. $x_i^* = a + i\De x$ with $a=1\text{,}$ then \begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-x_i^*}\De x \bigg) \hidewidth \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with $f(x) = e^{-x}$} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with $a=x_0=1$ and $b=x_n=3$} \\ &=\int_1^3 e^{-x}\ \,\dee{x} \end{align*}
• Choice #3: If we set $\De x = \frac{1}{n}$ and $x_i^*= \frac{i}{n}\text{,}$ i.e. $x_i^* = a + i\De x$ with $a=0\text{,}$ then \begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2x_i^*}\ 2\De x \bigg) \hidewidth \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with $f(x) = 2e^{-1-2x}$} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with $a=x_0=0$ and $b=x_n=1$} \\ &=2\int_0^1 e^{-1-2x}\ \,\dee{x} \end{align*}
• Choice #4: If we set $\De x = \frac{1}{n}$ and $x_i^*= \frac{1}{2}+\frac{i}{n}\text{,}$ i.e. $x_i = a + i\De x$ with $a=\frac{1}{2}\text{,}$ then \begin{align*} \lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-1-2i/n}\cdot \frac{2}{n} \bigg) &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n e^{-2x_i}\ 2\De x \bigg) \hidewidth \\ &=\lim_{n\rightarrow\infty} \bigg( \sum_{i=1}^n f(x_i^*)\De x \bigg) &&\text{with $f(x) = 2e^{-2x}$} \\ &=\int_a^b f(x)\,\,\dee{x}&&\text{with $a=x_0=\frac{1}{2}$ and $b=x_n=\frac{3}{2}$} \\ &=2\int_{1/2}^{3/2} e^{-2x}\ \,\dee{x} \end{align*}
Exercise26
Hint

Let $x=r^3\text{,}$ and re--write the sum in terms of $x\text{.}$

$\dfrac{r^{3n+3}-1}{r-1}$

Solution

This is similar to the familiar form of a geometric sum, but the powers go up by threes. So, we make a subsitution. If $x=r^3\text{,}$ then:

\begin{equation*} 1+r^3+r^6+r^9+\cdots+r^{3n}=1+x+x^2+x^3+\cdots+x^n \end{equation*}

Now, using Equation 1.1.3,

\begin{equation*} 1+x+x^2+x^3+\cdots+x^n = \frac{x^{n+1}-1}{x-1} \end{equation*}

Substituting back in $x=r^3\text{,}$ we find our sum is equal to $\dfrac{(r^3)^{n+1}-1}{r^3-1}\text{,}$ or $\dfrac{r^{3n+3}-1}{r^3-1}\text{.}$

Exercise27
Hint

Note the sum does not start at $r^0=1\text{.}$

$r^5\left(\dfrac{r^{96}-1}{r-1}\right)$

Solution

The sum does not start at $1\text{,}$ so we need to do some algebra. We can either factor out the first term, or subtract off the initial terms that are missing.

• Solution 1: If we factor out $r^5\text{,}$ then what's left fits the form of Equation 1.1.3: \begin{equation*} r^5+r^6+r^7+\cdots+r^{100}=r^5\left[1+r+r^2+\cdots + r^{95}\right]= r^5\left(\frac{r^{96}-1}{r-1}\right)\ . \end{equation*}
• Solution 2: We know how to evaluate sums of this form if they start at 1, so we re-write our sum as follows: \begin{align*} r^5+r^6+r^7+\cdots+r^{100}&=\left(1+r+r^2+r^3+r^4+r^5+\cdots+r^{100}\right) - \left(1+r+r^2+r^3+r^4\right) \\ &=\frac{r^{101}-1}{r-1} - \frac{r^5-1}{r-1}\\ &=\frac{r^{101}-1-r^5+1}{r-1}=\frac{r^{101}-r^5}{r-1}=r^5\left(\frac{r^{96}-1}{r-1}\right)\ . \end{align*}
Exercise28(*)
Hint

Draw a picture. See Example 1.1.15.

$5$

Solution

Recall that

\begin{align*} |x|=\begin{cases} -x &\text{if $x\le 0$} \\ x &\text{if $x\ge 0$} \end{cases} \end{align*}

so that

\begin{align*} |2x|=\begin{cases} -2x &\text{if $x\le 0$}\\ 2x &\text{if $x\ge 0$} \end{cases} \end{align*}

To picture the geometric figure whose area the integral represents observe that

• at the left hand end of the domain of integration $x=-1$ and the integrand $|2x|=|-2|=2$ and
• as $x$ increases from $-1$ towards $0\text{,}$ the integrand $|2x|=-2x$ decreases linearly, until
• when $x$ hits $0$ the integrand hits $|2x|=|0|=0$ and then
• as $x$ increases from $0\text{,}$ the integrand $|2x|=2x$ increases linearly, until
• when $x$ hits $+2\text{,}$ the right hand end of the domain of integration, the integrand hits $|2x|=|4|=4\text{.}$

So the integral $\int_{-1}^2 |2x|\ \,\dee{x}$ is the area of the union of the two shaded triangles (one of base $1$ and of height $2$ and the other of base $2$ and height $4$) in the figure on the right below and

\begin{equation*} \int_{-1}^2 |2x|\ \,\dee{x} = \frac{1}{2}\times 1\times 2 + \frac{1}{2}\times 2\times 4 = 5 \end{equation*}
Exercise29
Hint

Draw a picture. Remember $|x| = \left\{\begin{array}{rc}x&x\ge 0\\-x&x \lt 0\end{array}\right.$\ .

16

Solution

The area we want is two triangles, both above the $x$-axis. Each triangle has base $4$ and height $4\text{,}$ so the total area is $2\cdot\left(\dfrac{4\cdot 4}{2}\right)=16\text{.}$

If you had a hard time sketching the function, recall that the absolute value of a number leaves it unchanged if it is positive or zero, and flips the sign if it is negative. So, when $t-1 \ge 0$ (that is, when $t \ge 1$), our function is simply $f(t)=|t-1|=t-1\text{.}$ On the other hand, when $t=1$ is negative (that is, when $t \lt 1$), the absolute value changes the sign, so $f(t) = |t-1|=-(t-1)=-t+1\text{.}$

Exercise30
Hint

Draw a picture: the area we want is a trapezoid. If you don't remember a formula for the area of a trapezoid, think of it as the difference of two triangles.

$\dfrac{b^2-a^2}{2}$

Solution

The area we want is a trapezoid with base $(b-a)$ and heights $a$ and $b\text{,}$ so its area is $\dfrac{(b-a)(b+a)}{2}=\dfrac{b^2-a^2}{2}\text{.}$

Instead of using a formula for the area of a trapezoid, you can find the blue area as the area of a triangle with base and height $b\text{,}$ minus the area of a triangle with base and height $a\text{.}$

Exercise31
Hint

You can draw a very similar picture to Question 30, but remember the areas are negative.

$\dfrac{b^2-a^2}{2}$

Solution

The area is negative. The shape is a trapezoid with base length $(b - a)$ and heights $0-a=-a$ and $0-b=-b$ (note: those are nonnegative numbers), so its area is $\dfrac{(b-a)(-b-a)}{2}=\dfrac{-b^2+a^2}{2}\text{.}$ Since the shape is below the $x$-axis, we change its sign. Thus, the integral evaluates to\ $\dfrac{b^2-a^2}{2}\text{.}$

The signs can be a little hard to keep track of. The base of our trapezoid is $|a-b|\text{;}$ since $b \gt a\text{,}$ this is $b-a\text{.}$ The heights of the trapezoid are $|a|$ and $|b|\text{;}$ since these are both negative, $|a|=-a$ and $|b|=-b\text{.}$

We note that this is the same result as in Question 30.

Exercise32
Hint

If $y=\sqrt{16-x^2}\text{,}$ then $y$ is nonnegative, and $y^2+x^2=16\text{.}$

$4\pi$

Solution

If $y=\sqrt{16-x^2}\text{,}$ then $y$ is nonnegative, and $y^2+x^2=16\text{.}$ So, the graph $y=\sqrt{16-x^2}$ is the upper half of a circle of radius 4. Since $x$ only runs from 0 to 4, we have a quarter of a circle of radius 4. Then the area under the curve is $\dfrac{1}{4}\left[\pi\cdot 4^2\right]=4\pi\text{.}$

Exercise33(*)
Hint

Sketch the graph of $f(x)\text{.}$

$\displaystyle\int_0^3 f(x)\,\,\dee{x} = 2.5$

Solution

Here is a sketch the graph of $f(x)\text{.}$

There is a linear increase from $x=0$ to $x=1\text{,}$ followed by a constant. Using the interpretation of $\int_0^3 f(x)\,\,\dee{x}$ as the area between $y=f(x)$ and the $x$--axis with $x$ between $0$ and $3\text{,}$ we can break this area into:

• $\int_0^1 f(x)\,\,\dee{x}\text{:}$ a right-angled triangle of height $1$ and base $1$ and hence area $0.5\text{.}$
• $\int_1^3 f(x)\,\,\dee{x}\text{:}$ a rectangle of height $1$ and base $2$ and hence area $2\text{.}$

Summing up: $\int_0^3 f(x)\,\,\dee{x} = 2.5\text{.}$

Exercise34(*)
Hint

At which time in the interval, for example, $0\le t\le 0.5\text{,}$ is the car moving the fastest?

53 m

Solution

The car's speed increases with time. So its highest speed on any time interval occurs at the right hand end of the interval and the best possible upper estimate for the distance traveled is given by the right Riemann sum with $\Delta x =0.5\text{,}$ which is

\begin{equation*} \big[v(0.5)+v(1.0)+v(1.5)+v(2.0)\big]\times 0.5 =\big[14+22+30+40\big]\times 0.5= 53\text{ m} \end{equation*}
Exercise35
Hint

What are the possible speeds the car could have reached at time $t=0.25\text{?}$

true

Solution

There is a key detail in the statement of Question 34: namely, that the car is continuously accelerating. So, although we don't know exactly what's going on in between our brief snippets of information, we know that the car is not going any faster during an interval than at the end of that interval. Therefore, the car certainly travelled no farther than our estimation.

We ask this question in order to point out an important detail. If we did not have the information that the car was continuously accelerating, we would not be able to give a certain upper bound on its distance travelled. It would be possible that, when the car is not being observed (for example, when $t=0.25$), it is going much faster than when it is being observed.

Exercise36
Hint

You need to know the speed of the plane at the midpoints of your intervals, so (for example) noon to 1pm is not one of your intervals.

3200 km

Solution

First, note that the distance travelled by the plane is equal to the area under the curve of its speed.

We need to know the speed of the plane at the midpoints of our intervals. So (for example) noon to 1pm is not one of your intervals--we don't know the speed at 12:30. (A common idea is to average the two end values, 700 and 800. This is a fine approximation, but it is not a Riemann sum.) So, we use the two intervals 12:00 to 2:00, and 2:00 to 4:00. Then our intervals have length 2 hours, and at the midpoints of the intervals the speed of the plane is 700 kph and 900 kph, respectively. So, our midpoint Riemann sum gives us:

\begin{equation*} 700(2)+900(2) = 3200 \end{equation*}

an approximation of 3200 km travelled by the plane from noon to 4:00 pm.

Remark: if we had been asked to approximate the distance travelled from 11:30 am to 4:30 pm, then we could have used the midpoint rule with five intervals and made use of every entry in the data table. With the question as stated, however, we ignore three out of five entries in the table because they are not the midpoints of our intervals.

Exercise37(*)
Hint

Sure looks like a Riemann sum.

(a) There are many possible answers. Two are $\int_{-2}^0 \sqrt{4-x^2}\,\,\dee{x}$ and $\int_0^2 \sqrt{4-(-2+x)^2}\,\,\dee{x}\text{.}$

(b) $\pi$

Solution

• Solution #1: Set $x_i^*=-2+\frac{2i}{n}\text{.}$ Then $a=x_0=-2$ and $b=x_n=0$ and $\Delta x=\frac{2}{n}\text{.}$ So

\begin{align*} \lim_{n\rightarrow\infty} \sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2} &= \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*)\Delta x\qquad \text{ with $f(x) = \sqrt{4-x^2}$ and $\Delta x = \frac{2}{n}$ } \\ &=\int_{-2}^0 \sqrt{4-x^2}\,\,\dee{x} \end{align*}

For the integral $\int_{-2}^0 \sqrt{4-x^2}\,\,\dee{x}\text{,}$ $y=\sqrt{4-x^2}$ is equivalent to $x^2+y^2=4\text{,}$ $y\ge 0\text{.}$ So the integral represents the area between the upper half of the circle $x^2+y^2=4$ (which has radius $2$) and the $x$-axis with $-2\le x\le 0\text{,}$ which is a quarter circle with area $\frac{1}{4}\cdot \pi\, 2^2 = \pi\text{.}$

• Solution #2: Set $x_i^*=\frac{2i}{n}\text{.}$ Then $a=x_0=0$ and $b=x_n=2$ and $\Delta x=\frac{2}{n}\text{.}$ So

\begin{align*} \lim_{n\rightarrow\infty} \sum_{i=1}^n\frac{2}{n}\sqrt{4-\left(-2+\frac{2i}{n}\right)^2} &= \lim_{n\rightarrow\infty} \sum_{i=1}^n f(x_i^*)\Delta x\quad \text{ with $f(x) = \sqrt{4-(-2+x)^2}$, $\Delta x = \frac{2}{n}$ } \\ &=\int_0^2 \sqrt{4-(-2+x)^2}\,\,\dee{x} \end{align*}

For the integral $\int_{0}^2 \sqrt{4-(-2+x)^2}\,\,\dee{x}$\ , $y=\sqrt{4-(x-2)^2}$ is equivalent to $(x-2)^2+y^2=4\text{,}$ $y\ge 0\text{.}$ So the integral represents the area between the upper half of the circle $(x-2)^2+y^2=4$ (which is centered at $(2,0)$ and has radius $2$) and the $x$-axis with $0\le x\le 2\text{,}$ which is a quarter circle with area $\frac{1}{4}\cdot \pi\, 2^2 = \pi\text{.}$

Exercise38(*)
Hint

For part (b): don't panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given identity. The third step is to evaluate the limit $n\rightarrow\infty\text{.}$

(a) $30$ \qquad (b) $41 \frac{1}{4}$

Solution

(a) The left Riemann sum is defined as

\begin{gather*} L_n = \sum_{i=1}^{n} f(x_{i-1})\Delta x \qquad\text{with }x_i=a+i\De x \end{gather*}

We subdivide into $n=3$ intervals, so that $\Delta x = \frac{b-a}{n} =\frac{3-0}{3}=1\text{,}$ $x_0=0\text{,}$ $x_1=1$ and $x_2=2\text{.}$ The function $f(x) = 7 + x^3$ has the values $f(x_0) = 7+0^3=7\text{,}$ $f(x_1) = 7+1^3=8\text{,}$ and $f(x_2) = 7+2^3=15\text{,}$ from which we evaluate

\begin{gather*} L_3 = \big[f(x_0)+f(x_1)+f(x_2)\big]\De x = \big[7+8+15\big]\times 1= 30 \end{gather*}

(b) We divide into $n$ intervals so that $\Delta x = \frac{b-a}{n}=\frac{3}{n}$ and $x_i = a+i\De x= \frac{3i}{n}\text{.}$ The right Riemann sum is therefore:

\begin{gather*} R_n = \sum_{i=1}^{n} f(x_i)\Delta x = \sum_{i=1}^{n} \left[ 7 + \frac{(3i)^3}{n^3} \right] \frac{3}{n} = \sum_{i=1}^{n} \left[ \frac{21}{n} + \frac{81\,i^3}{n^4} \right] \end{gather*}

To calculate the sum:

\begin{align*} R_n &= \left( \frac{21}{n} \sum_{i=1}^{n} 1 \right) + \left( \frac{81}{n^4} \sum_{i=1}^{n} i^3 \right)\\ &=\left( \frac{21}{n}\times n \right)+\left( \frac{81}{n^4} \times \frac{n^4 +2n^3 + n^2}{4} \right)\\ &= 21 + \frac{81}{4}(1 +2/n + 1/n^2) \end{align*}

To evaluate the limit exactly, we take $n \to \infty\text{.}$ The expressions involving $1/n$ vanish leaving:

\begin{gather*} \int_0^3 (7 + x^3) \,\,\dee{x} = \lim_{n \to \infty} R_n = 21 + \frac{81}{4} = 41 \frac{1}{4} \end{gather*}
Exercise39(*)
Hint

The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit as $n\rightarrow\infty\text{.}$

$\dfrac{56}{3}$

Solution

In general, the right--endpoint Riemann sum approximation to the integral $\int_a^b f(x)\,\,\dee{x}$ using $n$ rectangles is

\begin{gather*} \sum_{i=1}^n f(a+i\De x) \De x \end{gather*}

where $\De x=\frac{b-a}{n}\text{.}$ In this problem, $a=2\text{,}$ $b=4\text{,}$ and $f(x)=x^2\text{,}$ so that $\De x=\frac{2}{n}$ and the right--endpoint Riemann sum approximation becomes

\begin{align*} \sum_{i=1}^n f\Big(2+\frac{2i}{n}\Big) \frac{2}{n}&= \sum_{i=1}^n \Big(2+\frac{2i}{n}\Big)^2 \frac{2}{n}\\ &=\sum_{i=1}^n \left(4+\frac{8i}{n}+\frac{4i^2}{n^2}\right)\frac{2}{n}\\ &=\sum_{i=1}^n \left(\frac{8}{n}+\frac{16i}{n^2}+\frac{8i^2}{n^3}\right)\\ &=\sum_{i=1}^n \frac{8}{n}+\sum_{i=1}^n \frac{16i}{n^2} +\sum_{i=1}^n \frac{8i^2}{n^3} \cr &=\frac{8}{n}\sum_{i=1}^n 1+\frac{16}{n^2}\sum_{i=1}^n i +\frac{8}{n^3}\sum_{i=1}^n i^2\cr &=\frac{8}{n}n + \frac{16}{n^2}\cdot\frac{n(n+1)}{2} +\frac{8}{n^3}\cdot\frac{n(n + 1)(2n + 1)}{6} \cr &=8 + 8\Big(1+\frac{1}{n}\Big) +\frac{4}{3}\Big(1+\frac{1}{n}\Big)\Big(2+\frac{1}{n}\Big) \cr \end{align*}

So

\begin{gather*} \int_2^4 x^2\ \,\dee{x}=\lim_{n\rightarrow\infty} \Big[8 + 8\Big(1+\frac{1}{n}\Big) +\frac{4}{3}\Big(1+\frac{1}{n}\Big) \big(2+\frac{1}{n}\Big)\Big] =8+8+\frac{4}{3}\times 2 =\frac{56}{3} \end{gather*}
Exercise40(*)
Hint

The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formulas. The third step is to evaluate the limit $n\rightarrow\infty\text{.}$

$6$

Solution

We'll use right Riemann sums with $a=0$ and $b=2\text{.}$ When there are $n$ rectangles, $\Delta x = \frac{b-a}{n}=\frac{2}{n}$ and $x_i = a+i\De x=2i/n\text{.}$ So we need to evaluate

\begin{align*} \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i)\De x &=\lim_{n\to\infty} \sum_{i=1}^{n} \left( (x_i)^3 + x_i\right) \De x\\ &=\lim_{n\to\infty} \sum_{i=1}^{n} \left( \left(\frac{2i}{n}\right)^3 + \frac{2i}{n}\right) \frac{2}{n} \\ & = \lim_{n\to\infty} \frac{2}{n} \sum_{i=1}^{n} \left(\frac{8i^3}{n^3} + \frac{2i}{n}\right) \\ &= \lim_{n\to\infty} \left( \frac{16}{n^4} \sum_{i=1}^{n} i^3 + \frac{4}{n^2} \sum_{i=1}^{n} i \right) \\ & = \lim_{n\to\infty} \left( \frac{16(n^4+2n^3+n^2)}{n^4 \cdot 4} + \frac{4(n^2+n)}{n^2 \cdot 2} \right) \\ &= \lim_{n\to\infty} \left( \frac{16}{4}\left(1+\frac{2}{n} + \frac1{n^2} \right) + \frac{4}{2}\left(1+\frac{1}{n}\right)\right) \\ & = \frac{16}{4} + \frac{4}{2} = 6. \end{align*}
Exercise41(*)
Hint

You've probably seen this hint before. It is worth repeating. Don't panic! Just take it one step at a time. The first step is to write down the Riemann sum. The second step is to evaluate the sum, using the given formula. The third step is to evaluate the limit $n\rightarrow\infty\text{.}$

$12$

Solution

We'll use right Riemann sums with $a=1\text{,}$ $b=4$ and $f(x) =2x-1\text{.}$ When there are $n$ rectangles, $\Delta x = \frac{b-a}{n}=\frac{3}{n}$ and $x_i = a+i\De x=1 + 3i/n\text{.}$ So we need to evaluate

\begin{align*} \lim_{n\to\infty} \sum_{i=1}^{n} f(x_i) \De x & = \lim_{n\to\infty} \sum_{i=1}^{n} \left( 2x_i - 1\right) \De x\\ & =\lim_{n\to\infty} \sum_{i=1}^{n} \left(2+ \frac{6i}{n}-1\right)\frac{3}{n} \\ & = \lim_{n\to\infty} \frac{3}{n} \sum_{i=1}^{n} \left(\frac{6i}{n}+1\right) \\ &= \lim_{n\to\infty} \left( \frac{18}{n^2} \sum_{i=1}^{n} i + \frac{3}{n} \sum_{i=1}^{n} 1 \right) \\ & = \lim_{n\to\infty} \left(\frac{18\cdot n(n+1)}{n^2\cdot 2} + \frac{3}{n}n\right)\\ &= \lim_{n\to\infty} \left(9\left(1+\frac{1}{n}\right) + 3\right) \\ & = 9 + 3 = 12. \end{align*}
Exercise42
Hint

Using the definition of a right Riemann sum, we can come up with an expression for $f(-5+10i)\text{.}$ In order to find $f(x)\text{,}$ set $x=-5+10i\text{.}$

$f(x)=\dfrac{3}{10}\left(\dfrac{x}{5}+8\right)^2\sin\left(\dfrac{2x}{5}+2\right)$

Solution

Using the definition of a right Riemann sum,

\begin{align*} \displaystyle\sum_{i=1}^{10} 3(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} \Delta x f(a+i\Delta x)\\ \end{align*}

Since $\Delta x = 10$ and $a=-5\text{,}$

\begin{align*} \displaystyle\sum_{i=1}^{10} 3(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} 10 f(-5+10i)\\ \end{align*}

Dividing both expressions by 10,

\begin{align*} \displaystyle\sum_{i=1}^{10} \frac{3}{10}(7+2i)^2\sin(4i) &= \displaystyle\sum_{i=1}^{10} f(-5+10i)\\ \end{align*}

So, we have an expression for $f(-5+10i)\text{:}$

\begin{align*} f(-5+10i) &= \frac{3}{10}(7+2i)^2\sin(4i)\\ \end{align*}

In order to find $f(x)\text{,}$ let $x=-5+10i\text{.}$ Then $i=\frac{x}{10}+\frac{1}{2}\text{.}$

\begin{align*} f(x) &= \frac{3}{10}\left(7+2\left(\frac{x}{10}+\frac{1}{2}\right)\right)^2\sin\left(4\left(\frac{x}{10}+\frac{1}{2}\right)\right)\\ &=\frac{3}{10}\left(\frac{x}{5}+8\right)^2\sin\left(\frac{2x}{5}+2\right)\ . \end{align*}
Exercise43
Hint

Recall that for a positive constant $a\text{,}$ $\diff{}{x}\left\{a^x\right\} = a^x \log a\text{,}$ where $\log a$ is the natural logarithm (base $e$) of $a\text{.}$

$\dfrac{1}{\log 2}$

Solution

As in the text, we'll set up a Riemann sum for the given integral. Right Riemann sums have the simplest form, so we use a right Riemann sum, but we could equally well use left or midpoint.

\begin{align*} \int_0^1 2^x\dee{x}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x f({a+i\Delta x})\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{1}{n} f\left(\frac{i}{n}\right)\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{1}{n}\cdot 2^{i/n}\\ &=\lim_{n \to \infty}\frac{1}{n}\left(2^{1/n}+2^{2/n}+2^{3/n}+\cdots + 2^{n/n}\right)\\ &=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left(1+2^{1/n}+2^{2/n}+\cdots + 2^{\frac{n-1}{n}}\right)\\ &=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left(1+2^{1/n}+\left(2^{1/n}\right)^2+\cdots + \left(2^{1/n}\right)^{n-1}\right)\\ \end{align*}

The sum in parenthesis has the form of a geometric sum, with $r=2^{1/n}\text{:}$

\begin{align*} &=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left( \frac{\left(2^{1/n}\right)^n-1}{2^{1/n}-1} \right)\\ &=\lim_{n \to \infty}\frac{2^{1/n}}{n}\left( \frac{2-1}{2^{1/n}-1} \right)\\ &=\lim_{n \to \infty} \frac{2^{1/n}}{n(2^{1/n}-1)}\\ \end{align*}

Note as $n \to \infty\text{,}$ $1/n \to 0\text{,}$ so the numerator has limit 1, while the denominator has indeterminate form $\infty\cdot 0\text{.}$ So, we'll do a little algebra to get this into a l'Hôpital-style indeterminate form:

\begin{align*} &=\lim_{n \to \infty} \frac{\frac{1}{n}\cdot2^{1/n}}{2^{1/n}-1}\\ &=\lim_{n \to \infty} \underbrace{\frac{\frac{1}{n}}{1-2^{-1/n}}}_{\atp{\mathrm{num}\to 0}{\mathrm{den}\to 0}}\\ \end{align*}

Now we can use l'Hôpital's rule. Recall $\diff{}{x}\left\{2^x\right\}=2^x\log x\text{,}$ where $\log x$ is the natural logarithm of $x\text{,}$ also sometimes written $\ln x\text{.}$ We'll need to use the chain rule when we differentiate the denominator.

\begin{align*} &=\lim_{n \to \infty} \frac{\frac{-1}{n^2}}{-2^{-1/n}\log 2 \cdot \frac{1}{n^2}}\\ &=\lim_{n \to \infty} \frac{2^{1/n}}{\log 2}\\ &=\frac{1}{\log 2} \end{align*}

Using a calculator, we see this is about 1.44 square units.

Exercise44
Hint

Part (a) follows the same pattern as Question 43--there's just a little more algebra involved, since our lower limit of integration is not 0.

(a) $\dfrac{1}{\log 10}\left(10^b-10^a\right)$\\ (b) $\dfrac{1}{\log c}\left(c^b-c^a\right)\text{;}$ yes, it agrees.

Solution

As in the text, we'll set up a Riemann sum for the given integral. Right Riemann sums have the simplest form:

\begin{align*} \int_a^b 10^x\dee{x}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x f({a+i\Delta x})\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{b-a}{n} f\left(a+i\frac{b-a}{n}\right)\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{b-a}{n}\cdot 10^{a+i\frac{b-a}{n}}\\ &=\lim_{n \to \infty}\sum_{i=1}^n \frac{b-a}{n}\cdot 10^a\cdot \left(10^{\frac{b-a}{n}}\right)^{i}\\ &=\lim_{n \to \infty} \frac{b-a}{n}\cdot 10^a\left(\left(10^{\frac{b-a}{n}}\right)^1+ \left(10^{\frac{b-a}{n}}\right)^2 +\left(10^{\frac{b-a}{n}}\right)^3 +\cdots + \left(10^{\frac{b-a}{n}}\right)^n \right)\\ &=\lim_{n \to \infty} \frac{b-a}{n} \cdot10^{a}\cdot10^{\frac{b-a}{n}} \left(1+\left(10^{\frac{b-a}{n}}\right)+ \left(10^{\frac{b-a}{n}}\right)^2 +\cdots + \left(10^{\frac{b-a}{n}}\right)^{n-1} \right)\\ \end{align*}

Now the sum in parentheses has the form of a geometric sum, with $r=10^{\frac{b-a}{n}}\text{:}$

\begin{align*} &=\lim_{n \to \infty} \frac{b-a}{n}\cdot 10^{a}\cdot 10^{\frac{b-a}{n}}\left( \frac{\left(10^{\frac{b-a}{n}}\right)^n-1}{10^{\frac{b-a}{n}}-1} \right)\\ &=\lim_{n \to \infty} \frac{\textcolor{blue}{b-a}}{n}\cdot \textcolor{red}{10^{a}}\cdot 10^{\frac{b-a}{n}}\left( \frac{\textcolor{purple}{10^{b-a}-1}}{10^{\frac{b-a}{n}}-1} \right)\\ \end{align*}

The coloured parts do not depend on $n\text{,}$ so for simplicity we can move them outside the limit.

\begin{align*} &=\textcolor{blue}{(b-a)}\cdot\textcolor{red}{10^a}\left(\textcolor{purple}{10^{b-a}-1}\right)\lim_{n \to \infty} \frac{1}{n}\cdot \left( \frac{ 10^{\frac{b-a}{n}} }{10^{\frac{b-a}{n}}-1} \right)\\ &={(b-a)}\cdot\left({10^{b}-10^a}\right)\lim_{n \to \infty} \underbrace{\left( \frac{ 1/n}{1-10^{-\frac{b-a}{n}}} \right)}_{\atp{\mathrm{num}\to 0}{\mathrm{den}\to 0}}\\ \end{align*}

Now we can use l'Hôpital's rule. Recall $\diff{}{x}\left\{10^x\right\}=10^x\log x\text{,}$ where $\log x$ is the natural logarithm of $x\text{,}$ also sometimes written $\ln x\text{.}$ For the denominator, we will have to use the chain rule.

\begin{align*} &={(b-a)}\cdot\left({10^{b}-10^a}\right)\lim_{n \to \infty} \left( \frac{ -1/n^2}{-10^{-\frac{b-a}{n}}\cdot \log 10 \cdot \frac{b-a}{n^2}} \right)\\ &={(b-a)}\cdot\left({10^{b}-10^a}\right)\lim_{n \to \infty} \left( \frac{ 1}{10^{-\frac{b-a}{n}}\cdot \log 10 \cdot (b-a)} \right)\\ &={(b-a)}\cdot\left({10^{b}-10^a}\right) \left( \frac{ 1}{ \log 10 \cdot (b-a)} \right)\\ &=\frac{1}{\log 10}\left(10^b-10^a\right) \end{align*}

For part (b), we can guess that if 10 were changed to $c\text{,}$ our answer would be

\begin{equation*} \int_a^b c^x \dee{x}=\frac{1}{\log c}\left(c^b-c^a\right) \end{equation*}

In Question 43, we had $a=0\text{,}$ $b=1\text{,}$ and $c=2\text{.}$ In this case, the formula we guessed above gives

\begin{equation*} \int_0^1 2^x \dee{x}=\frac{1}{\log 2}\left(2^1-2^0\right)=\frac{1}{\log2} \end{equation*}

This does indeed match the answer we calculated.

(In fact, we can directly show $\displaystyle\int_a^b c^x \dee{x}=\dfrac{1}{\log c}\left(c^b-c^a\right)$ using the method of this problem.)

Exercise45
Hint

Your area can be divided into a section of a circle and a triangle. Then you can use geometry to find the area of each piece.

$\frac{\pi}{4} -\frac{1}{2} \arccos(a) + \frac{1}{2}a\sqrt{1-a^2}$

Solution

First, we note $y=\sqrt{1-x^2}$ is the upper half of a circle of radius 1, centred at the origin. We're taking the area under the curve from 0 to $a\text{,}$ so the area in question is as shown in the picture below.

In order to use geometry to find this area, we break it up into two pieces: a sector of a circle, and a triangle, shown below.

• Area of sector: The sector is a portion of a circle with radius 1, with inner angle $\theta\text{.}$ So, its area is $\frac{\theta}{2\pi}\left(\mbox{area of circle}\right) = \frac{\theta}{2\pi}\left(\pi\right) = \frac{\theta}{2}\text{.}$

Our job now is to find $\theta$ in terms of $a\text{.}$ Note $\frac{\pi}{2}-\theta$ is the inner angle of the red triangle, which lies in the unit circle. So, $\cos\left(\frac{\pi}{2}-\theta\right)=a\text{.}$ Then $\frac{\pi}{2}-\theta= \arccos(a)\text{,}$ and so $\theta = \frac{\pi}{2} - \arccos(a)\text{.}$

Then the area of the sector is $\frac{\pi}{4} - \frac{1}{2}\arccos(a)$ square units.

• Area of triangle: The triangle has base $a\text{.}$ Its height is the $y$-value of the function when $x=a\text{,}$ so its height is $\sqrt{1-a^2}\text{.}$ Then the area of the triangle is $\frac{1}{2}a\sqrt{1-a^2}\text{.}$

We conclude $\displaystyle\int_0^a \sqrt{1-x^2}\dee{x} = \frac{\pi}{4} -\frac{1}{2} \arccos(a) + \frac{1}{2}a\sqrt{1-a^2}\text{.}$

Exercise46
Hint

1. The difference between the upper and lower bounds is the area that is outside of the smaller rectangles but inside the larger rectangles. Drawing both sets of rectangles on one picture might make things clearer. Look for an easy way to compute the area you want.
2. Use your answer from Part (a). Your answer will depend on $f\text{,}$ $a\text{,}$ and $b\text{.}$

1. $\left[f(b)-f(a)\right]\cdot\dfrac{b-a}{n}$
2. Choose $n$ to be an integer that is greater than or equal to $100\left[f(b)-f(a)\right](b-a)\text{.}$
Solution

1. The difference between our upper and lower bounds is the difference in areas between the larger set of rectangles and the smaller set of rectangles. Drawing them on a single picture makes this a little clearer.

Each of the rectangles has width $\frac{b-a}{n}\text{,}$ since we took a segment of the $x$-axis with length $b-a$ and chopped it into $n$ pieces. We could calculate the height of each rectangle, but it would be a little complicated, since it differs for each of them. An easier method is to notice that the area we want to calculate can be imagined as a single rectangle:

The rectangle has base $\frac{b-a}{n}\text{.}$ Its highest coordinate is $f(a)\text{,}$ and its lowest is $f(b)\text{,}$ so its height is $f(b)-f(a)\text{.}$ Therefore, the difference in area between our lower bound and our upper bound is:

\begin{equation*} \left[f(b)-f(a)\right]\cdot\frac{b-a}{n} \end{equation*}
2. We want to give a range with length at most 0.01, and guarantee that the area under the curve $y=f(x)$ is inside that range. In the previous part, we figured out that when we use $n$ rectangles, the length of our range is $\left[f(b)-f(a)\right]\cdot\frac{b-a}{n}\text{.}$ So, all we have to do is set this to be less than or equal to 0.01, and solve for $n\text{:}$

\begin{align*} \left[f(b)-f(a)\right]\cdot\frac{b-a}{n}&\leq 0.01 \\ 100\left[f(b)-f(a)\right]\cdot(b-a)&\leq n \end{align*}

We can choose $n$ to be an integer that is greater than or equal to $100\left[f(b)-f(a)\right]\cdot(b-a)\text{.}$ Using that many rectangles, we find an upper and lower bound for the area under the curve. If we choose any number between our upper and lower bound as an approximation for the area under the curve, our error is no more than 0.01.

Remark: this question depends on the fact that $f$ is decreasing and positive from $a$ to $b\text{.}$ In general, bounding errors on approximations like this is not so straightforward.

Exercise47
Hint

Since $f(x)$ is linear, there exist real numbers $m$ and $c$ such that $f(x)=mx+c\text{.}$ It's a little easier to first look at a single triangle from each sum, rather than the sums in their entirety.

true (but note, for a non-linear function, it is possible that the midpoint Riemann sum is not the average of the other two)

Solution

Since $f(x)$ is linear, there exist real numbers $m$ and $c$ such that $f(x)=mx+c\text{.}$ Now we can do some calculations. Suppose we have a rectangle in our Riemann sum that takes up the interval $[x,x+w]\text{.}$

• If we are using a left Riemann sum, our rectangle has height $f(x)=mx+c\text{.}$ Then it has area $w(mx+c)\text{.}$
• If we are using a right Riemann sum, our rectangle has height $f(x+w)=m(x+w)+c=mx+c+mw\text{.}$ Then it has area $w(mx+c+mw)\text{.}$
• If we are using a midpoint Riemann sum, our rectangle has height $f(x+\frac{1}{2}w)=m(x+\frac{1}{2}w)+c=mx+c+\frac{1}{2}mw\text{.}$ Then it has area $w\left(mx+c+\frac{1}{2}w\right)\text{.}$

So, for each rectangle in our sums, the midpoint rectangle has the same area as the average of the left and right rectangles:

\begin{equation*} w\left(mx+c+\frac{1}{2}mw\right) = \dfrac{\textcolor{blue}{w(mx+c)}+\textcolor{red}{w(mx+c+mw)}}{2} \end{equation*}

It follows that the midpoint Riemann sum has a value equal to the average of the values of the left and right Riemann sums. To see this, let the rectangles in the midpoint Riemann sum have areas $M_1,M_2,\ldots,M_n\text{,}$ let the rectangles in the left Riemann sum have areas $\textcolor{blue}{L_1,L_2,\ldots,L_n}\text{,}$ and let the rectangles in the right Riemann sum have areas $\textcolor{red}{R_1,R_2,\ldots,R_n}\text{.}$ Then the midpoint Riemann sum evaluates to $M_1+M_2+\cdots+M_n\text{,}$ and:

\begin{align*} \dfrac{\textcolor{blue}{[L_1+L_2+\ldots+L_n]}+ \textcolor{red}{[R_1+R_2+\ldots+R_n]} }{2} &= \dfrac{\textcolor{blue}{L_1}+ \textcolor{red}{R_1} }{2}+ \dfrac{\textcolor{blue}{L_2}+ \textcolor{red}{R_2} }{2}+ \cdots + \dfrac{\textcolor{blue}{L_n}+ \textcolor{red}{R_n} }{2} \\ &=M_1+M_2+\cdots+M_n \end{align*}

So, the statement is true.

(Note, however, it is false for many non-linear functions $f(x)\text{.}$)

Exercises1.2.3Exercises

Exercise1
Hint

1. What is the length of this figure?
2. Think about cutting the area into two pieces vertically.
3. Think about cutting the area into two pieces another way.

Possible drawings:

Solution

1. $\displaystyle\int_a^a f(x)\,\dee{x}=0$

The area under the curve is zero, because it's a region with no width.

2. $\displaystyle\int_a^b f(x)\,\dee{x}=\textcolor{blue}{ \displaystyle\int_a^c f(x)\,\dee{x}} +\textcolor{red}{ \int_c^b f(x)\dee{x} }$

If we assume $a \leq c \leq b\text{,}$ then this identity simply tells us that if we add up the area under the curve from $a$ to $c\text{,}$ and from $c$ to $b\text{,}$ then we get the whole area under the curve from $a$ to $b\text{.}$

(The situation is slightly more complicated when $c$ is not between $a$ and $b\text{,}$ but it still works out.)

3. $\displaystyle\int_a^b \left( f(x) + g(x) \right)\,\dee{x} =\textcolor{blue}{ \displaystyle\int_a^b f(x)\,\dee{x}} +\textcolor{red}{ \displaystyle\int_a^b g(x)\,\dee{x}}$

The blue-shaded area in the picture above is $\displaystyle\int_a^b f(x)\dee{x}\text{.}$ The area under the curve $f(x)+g(x)$ but above the curve $f(x)$ (shown in red) is $\displaystyle\int_a^b g(x)\dee{x}\text{.}$

Exercise2
Hint

Use the identity $\int\limits_a^b f(x)\dee{x} = \int\limits_a^c f(x)\dee{x}+ \int\limits_c^b f(x)\dee{x}\text{.}$

$\sin b-\sin a$

Solution

Using the identity

\begin{align*} \int\limits_a^b f(x)\dee{x} &= \int\limits_a^c f(x)\dee{x}+ \int\limits_c^b f(x)\dee{x}\ ,\\ \end{align*}

we see

\begin{align*} \int\limits_a^b\cos x\dee{x} &= \int\limits_a^0 \cos x\dee{x}+ \int\limits_0^b \cos x\dee{x}\\ &= -\int\limits_0^a \cos x\dee{x}+ \int\limits_0^b \cos x\dee{x}\\ &=-\sin a + \sin b\\ &=\sin b - \sin a \end{align*}
Exercise3(*)

(a) False. For example, the function

\begin{align*} f(x) = \begin{cases} 0 & \text{for } x \lt 0 \\ 1 & \text{for } x \ge 0 \end{cases} \end{align*}

provides a counterexample.

(b) False. For example, the function $f(x)=x$ provides a counterexample.

(c) False. For example, the functions

\begin{align*} f(x) = \begin{cases} 0 & \text{for } x \lt \frac{1}{2} \\ 1 & \text{for } x\ge\frac{1}{2} \\ \end{cases} &&\mbox{and}&&g(x) = \begin{cases} 0 & \text{for } x\ge \frac{1}{2}\\ 1 & \text{for } x \lt \frac{1}{2} \end{cases} \end{align*}

provide a counterexample.

Solution

(a) False. For example if

\begin{align*} f(x) = \begin{cases} 0 & \text{for } x \lt 0 \\ 1 & \text{for } x\ge0 \end{cases} \end{align*}

then $\textcolor{blue}{\int_{-3}^{-2} f(x) \dee{x}=0}$ and $\textcolor{red}{-\int_{3}^{2} f(x) \dee{x}=-1}\text{.}$

(b) False. For example, if $f(x)=x\text{,}$ then $\textcolor{blue}{\int_{-3}^{-2} f(x)\,\dee{x} }$ is negative while $\textcolor{red}{\int_2^3 f(x)\,\dee{x} }$ is positive, so they cannot be the same.

(c) False. For example, consider the functions

\begin{align*} f(x) = \begin{cases} 0 & \text{for } x \lt \frac{1}{2}\\ 1 & \text{for } x\ge\frac{1}{2} \end{cases} &&\mbox{and}&&g(x) = \begin{cases} 0 & \text{for } x\ge \frac{1}{2} \\ 1 & \text{for } x \lt \frac{1}{2} \end{cases} \end{align*}

Then $f(x)\cdot g(x)=0$ for all $x\text{,}$ so $\int_0^1 f(x)\cdot g(x) \dee{x}=0\text{.}$ However, $\textcolor{blue}{\int_0^1 f(x) \dee{x}= \frac{1}{2}}$ and $\textcolor{red}{\int_0^1 g(x) \dee{x}= \frac{1}{2}}\text{,}$ so $\int_0^1 f(x)\dee{x} \cdot \int_0^1 g(x) \dee{x}= \frac{1}{4}\text{.}$

Exercise4
Hint

Note that the limits of the integral given are in the opposite order from what we might expect: the smaller number is the top limit of integration.

Recall $\De x = \frac{b-a}{n}\text{.}$

(a) $-\dfrac{1}{20}\text{,}$ (b) positive, (c) negative, (d) positive.

Solution

1. $\Delta x = \dfrac{b-a}{n}=\dfrac{0-5}{100} = -\dfrac{1}{20}$

Note: if we were to use the Riemann-sum definition of a definite integral, this is how we would justify the identity $\int\limits_a^b f(x)\dee{x}=-\int\limits_b^a f(x)\dee{x}\text{.}$

2. The heights of the rectangles are given by $f(x_i)\text{,}$ where $x_i = a+i\Delta x = 5 - \frac{i}{20}\text{.}$ Since $f(x)$ only gives positive values, $f(x_i) \gt 0\text{,}$ so the heights of the rectangles are positive.
3. Our Riemann sum is the sum of the signed areas of individual rectangles. Each rectangle has a negative base ($\Delta x$) and a positive height ($f(x_i)$). So, each term of our sum is negative. If we add up negative numbers, the sum is negative. So, the Riemann sum is negative.
4. Since $f(x)$ is always above the $x$-axis, $\int\limits_0^5 f(x)\dee{x}$ is positive.
Exercise5(*)
Hint

Split the “target integral” up into pieces that can be evaluated using the given integrals.

$-21$

Solution

The operation of integration is linear (that's part (d) of the “arithmetic of integration” Theorem 1.2.1), so that:

\begin{align*} \int_2^3 [6 f(x) -3 g(x)]\,\dee{x} &= \int_2^3 6 f(x)\,\dee{x} - \int_2^3 3 g(x)\,\dee{x} \\ &= 6 \int_2^3 f(x)\,\dee{x} - 3\int_2^3 g(x)\,\dee{x} = (6 \times (-1)) - (3 \times 5) = -21 \end{align*}
Exercise6(*)
Hint

Split the “target integral” up into pieces that can be evaluated using the given integrals.

$-6$

Solution

The operation of integration is linear (that's part (d) of the “arithmetic of integration” Theorem 1.2.1), so that:

\begin{align*} \int_0^2 [2 f(x) +3 g(x)]\,\dee{x} &= \int_0^2 2 f(x)\,\dee{x} + \int_0^2 3 g(x)\,\dee{x} \\ &= 2 \int_0^2 f(x)\,\dee{x} + 3\int_0^2 g(x)\,\dee{x} = (2 \times 3) + (3 \times (-4)) = -6 \end{align*}
Exercise7(*)
Hint

Split the “target integral” up into pieces that can be evaluated using the given integrals.

20

Solution

Using part (d) of the “arithmetic of integration” Theorem 1.2.1, followed by parts (c) and (b) of the “arithmetic for the domain of integration” Theorem 1.2.3,

\begin{align*} \int_{-1}^2 \big[3g(x)-f(x)\big]\,\dee{x} &=3\int_{-1}^2 g(x)\,\dee{x}-\int_{-1}^2 f(x)\,\dee{x} \\ &=3\int_{-1}^0 g(x)\,\dee{x}+3\int_0^2 g(x)\,\dee{x} -\int_{-1}^0 f(x)\,\dee{x}-\int_0^2 f(x)\,\dee{x} \\ &=3\int^0_{-1} g(x)\,\dee{x}+3\int_0^2 g(x)\,\dee{x} +\int_0^{-1} f(x)\,\dee{x}-\int_0^2 f(x)\,\dee{x} \\ &=3\times 3+3\times 4 + 1 - 2 = 20 \end{align*}
Exercise8
Hint

For part (a), use the symmetry of the integrand. For part (b), the area $\int \limits_{0}^1 \sqrt{1-x^2}\dee{x}$ is easy to find--how is this useful to you?

1. $\frac{\pi}{4} - \frac{1}{2}\arccos(-a)-\frac{1}{2}a\sqrt{1-a^2}$
2. $\frac{1}{2}\arccos(a)-\frac{1}{2}a\sqrt{1-a^2}$
Solution

1. Since $\sqrt{1-x^2}$ is an even function, \begin{align*} \displaystyle\int_{a}^0 \sqrt{1-x^2}\dee{x} &=\displaystyle\int_{0}^{|a|} \sqrt{1-x^2}\dee{x} = \frac{\pi}{4} - \frac{1}{2}\arccos(|a|)+\frac{1}{2}|a|\sqrt{1-|a|^2}\\ &=\frac{\pi}{4} - \frac{1}{2}\arccos(-a)-\frac{1}{2}a\sqrt{1-a^2} \end{align*}
2. Note $\displaystyle\int_{0}^1 \sqrt{1-x^2}\dee{x}=\frac{\pi}{4}\text{,}$ since the area under the curve represents one-quarter of the unit circle. Then, \begin{align*} \displaystyle\int_{a}^1 \sqrt{1-x^2}\dee{x}&= \displaystyle\int_{0}^1 \sqrt{1-x^2}\dee{x}- \displaystyle\int_{0}^a \sqrt{1-x^2}\dee{x}\\ &=\frac{\pi}{4}-\left(\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}\right)\\ &=\frac{1}{2}\arccos(a)-\frac{1}{2}a\sqrt{1-a^2} \end{align*}
Exercise9(*)
Hint

The evaluation of this integral was also the subject of Question 1.2.3.9 in Section 1.1. This time try using the method of Example 1.2.7.

$5$

Solution

Recall that

\begin{align*} |x|=\begin{cases} -x &\text{if } x\le 0 \\ x &\text{if } x\ge 0 \end{cases} \end{align*}

so that

\begin{align*} |2x|=\begin{cases} -2x &\text{if } x\le 0 \\ 2x &\text{if } x\ge 0 \end{cases} \end{align*}

Also recall, from Example 1.2.6, that

\begin{align*} \int_a^b x\dee{x}&=\frac{b^2-a^2}{2} \end{align*}

So

\begin{align*} \int_{-1}^2 |2x|\dee{x} &=\int_{-1}^0 |2x|\dee{x}+\int_0^2 |2x|\dee{x} =\int_{-1}^0 (-2x)\dee{x}+\int_0^2 2x\dee{x} \\ &= -2\int_{-1}^0 x\dee{x}+2\int_0^2 x\dee{x} =-2\cdot\frac{0^2-(-1)^2}{2} +2\cdot\frac{2^2-0^2}{2} \\ &=1+4=5 \end{align*}
Exercise10
Hint

Use symmetry.

0

Solution

We note that the integrand $f(x)=x|x|$ is an odd function, because $f(-x)=-x|-x|=-x|x|=-f(x)\text{.}$ Then by Theorem 1.2.12 part (b), $\displaystyle\int_{-5}^5 x|x| \dee{x}=0\text{.}$

Exercise11
Hint

Check Theorem 1.2.12.

5

Solution

Using Theorem 1.2.12 part (a),

\begin{align*} 10&=\int_{-2}^2 f(x)\dee{x}=2\int_{0}^2f(x)\dee{x}\\ 5&=\int_{0}^2f(x)\dee{x}\\ \end{align*}

Also,

\begin{align*} \int_{-2}^2 f(x)\dee{x}&=\int_{-2}^0 f(x)\dee{x}+\int_{0}^2 f(x)\dee{x}\\ \end{align*}

So,

\begin{align*} \int_{-2}^0 f(x)\dee{x}&=\int_{-2}^2 f(x)\dee{x}-\int_{0}^2 f(x)\dee{x}\\ &=10-5=5 \end{align*}

Indeed, for any even function $f(x)\text{,}$ $\int\limits_{-a}^0 f(x)\dee{x} = \int\limits_{0}^a f(x)\dee{x}\text{.}$

Exercise12(*)
Hint

Split the integral into a sum of two integrals. Interpret each geometrically.

$20 +2\pi$

Solution

\begin{gather*} \int_{-2}^{2} \left(5+\sqrt{4-x^2}\right)\dee{x} = \textcolor{blue}{ \int_{-2}^{2} 5\,\dee{x} }+ \textcolor{red}{\int_{-2}^{2} \sqrt{4-x^2}\,\dee{x}} \end{gather*}

The first integral represents the area of a rectangle of height 5 and width 4 and so equals $20\text{.}$ The second integral represents the area above the $x$--axis and below the curve $y=\sqrt{4-x^2}$ or $x^2+y^2=4\text{.}$ That is a semicircle of radius 2, which has area $\frac{1}{2}\pi 2^2\text{.}$ So

\begin{gather*} \int_{-2}^{2} \left(5+\sqrt{4-x^2}\right)\dee{x} = \textcolor{blue}{20} +\textcolor{red}{2\pi} \end{gather*}
Exercise13(*)
Hint

Hmmmm. Looks like a complicated integral. It's probably a trick question. Check for symmetries.

$0$

Solution

Note that the integrand $f(x) = \frac{\sin x}{\log(3+x^2)}$ is an odd function, because:

\begin{equation*} f(-x) = \frac{\sin(-x)}{\log(3+(-x)^2)}=\frac{-\sin x}{\log(3+x^2)} =- f(x) \end{equation*}

The domain of integration $-2012 \le x \le 2012$ is symmetric about $x=0\text{.}$ So, by Theorem 1.2.12,

\begin{equation*} \int_{-2012}^{+2012} \frac{\sin x}{\log(3+x^2)}\dee{x} = 0 \end{equation*}
Exercise14(*)
Hint

Check for symmetries again.

$0$

Solution

Note that the integrand $f(x) = x^{1/3}\cos x$ is an odd function, because:

\begin{equation*} f(-x) = (-x)^{1/3}\cos(-x)= - x^{1/3}\cos x =- f(x) \end{equation*}

The domain of integration $-2012 \le x \le 2012$ is symmetric about $x=0\text{.}$ So, by Theorem 1.2.12,

\begin{equation*} \int_{-2012}^{+2012}x^{1/3}\cos x\,\dee{x} = 0 \end{equation*}
Exercise15
Hint

What does the integrand look like to the left and right of $x=3\text{?}$

0

Solution

Our integrand $f(x)=(x-3)^3$ is neither even nor odd. However, it does have a similar symmetry. Namely, $f(3+x)=-f(3-x)\text{.}$ So, $f$ is “negatively symmetric” across the line $x=3\text{.}$ This suggests that the integral should be 0: the positive area to the right of $x=3$ will be the same as the negative area to the left of $x=3\text{.}$

Another way to see this is to notice that the graph of $f(x)=(x-3)^3$ is equivalent to the graph of $g(x)=x^3$ shifted three units to the right, and $g(x)$ is an odd function. So,

\begin{equation*} \textcolor{red}{\int_{0}^6 (x-3)^3\,\dee{x}} = \textcolor{blue}{\int_{-3}^3 x^3\,\dee{x}}=0 \end{equation*}
Exercise16
Hint

In part (b), you'll have to factor a constant out through a square root. Remember the upper half of a circle looks like $\sqrt{r^2-x^2}\text{.}$

(a) $y = \dfrac{1}{b}\sqrt{1-(ax)^2}$

(b) $\displaystyle\frac{a}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{\frac{1}{a^2}-x^2}\dee{x}$

(c) $\dfrac{\pi}{ab}$

Solution

1. \begin{align*} (ax)^2+(by)^2&=1\\ by&=\sqrt{1-(ax)^2}\\ y &= \frac{1}{b}\sqrt{1-(ax)^2} \end{align*}
2. The values of $x$ in the domain of the function above are those that satisfy $1-(ax)^2 \geq 0\text{.}$ That is, $-\frac{1}{a}\leq x \leq \frac{1}{a}\text{.}$ Therefore, the upper half of the ellipse has area \begin{align*} \displaystyle\frac{1}{b}&\displaystyle\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{1-(ax)^2}\dee{x}\\ \end{align*}

The upper half of a circle has equation $y=\sqrt{r^2-x^2}\text{.}$

\begin{align*} &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{a^2\left(\frac{1}{a^2}-x^2\right)}\dee{x}\\ &=\frac{1}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}a\sqrt{\frac{1}{a^2}-x^2}\dee{x}\\ &=\frac{a}{b}\int_{-\frac{1}{a}}^{\frac{1}{a}}\sqrt{\frac{1}{a^2}-x^2}\dee{x} \end{align*}
3. The function $y=\sqrt{\dfrac{1}{a^2}-x^2}$ is the upper-half of the circle centred at the origin with radius $\dfrac{1}{a}\text{.}$ So, the expression from (b) evaluates to $\left(\dfrac{a}{b}\right)\dfrac{\pi}{2a^2} = \dfrac{\pi}{2ab}\text{.}$

The expression from (b) was half of the ellipse, so the area of the ellipse is $\dfrac{\pi}{ab}\text{.}$

Remark: this was a slightly long-winded way of getting the result. The reasoning is basically this:

• The area of the unit circle $x^2+y^2=1$ is $\pi$\ .
• The ellipse $(ax)^2+y^2=1$ is obtained by shrinking the unit circle horizontally by a factor of $a\text{.}$ So, its area is $\dfrac{\pi}{a}$\ .
• Further, the ellipse $(ax)^2+(by)^2=1$ is obtained from the previous ellipse by shrinking it vertically by a factor of $b\text{.}$ So, its area is $\dfrac{\pi}{ab}$\ .
Exercise17
Hint

For two functions $f(x)$ and $g(x)\text{,}$ define $h(x)=f(x)\cdot g(x)\text{.}$ If $h(-x)=h(x)\text{,}$ then the product is even; if $h(-x)=-h(x)\text{,}$ then the product is odd.

The table will not be the same as if we were multiplying even and odd numbers.

 $\times$ even odd even even odd odd odd even
Solution

Let's recall the definitions of even and odd functions: $f(x)$ is even if $f(-x)=f(x)$ for every $x$ in its domain, and $f(x)$ is odd if $f(-x)=-f(x)$ for every $x$ in its domain.

Let $h(x)=f(x)\cdot g(x)\text{.}$

• even $\times$ even: If $f$ and $g$ are both even, then $h(-x)=f(-x)\cdot g(-x) = f(x)\cdot g(x)=h(x)\text{,}$ so their product is even.
• odd $\times$ odd: If $f$ and $g$ are both odd, then $h(-x)=f(-x)\cdot g(-x) =[- f(x)]\cdot [-g(x)]=f(x)\cdot g(x)=h(x)\text{,}$ so their product is even.
• even $\times$ odd: If $f$ is even and $g$ is odd, then $h(-x)=f(-x)\cdot g(-x) = f(x)\cdot[- g(x)]=-[f(x)\cdot g(x)]=-h(x)\text{,}$ so their product is odd. Because multiplication is commutative, the order we multiply the functions in doesn't matter.

We note that the table would be the same as if we were adding (not multiplying) even and odd numbers (not functions).

Exercise18
Hint

Note $f(0)=f(-0)\text{.}$

$f(0)=0\text{;}$ $g(0)$ can be any real number

Solution

Since $f(x)$ is odd, $f(0)=-f(-0)=-f(0)\text{.}$ So, $f(0)=0\text{.}$

However, this restriction does not apply to $g(x)\text{.}$ For example, for any constant $c\text{,}$ let $g(x)=c\text{.}$ Then $g(x)$ is even and $g(0)=c\text{.}$ So, $g(0)$ can be any real number.

Exercise19
Hint

If $f(x)$ is even and odd, then $f(x)=-f(x)$ for every $x\text{.}$

$f(x)=0$ for every $x$

Solution

Let $x$ be any real number.

• $f(x)=f(-x)$ (since $f(x)$ is even), and
• $f(x)=-f(-x)$ (since $f(x)$ is odd).
• So, $f(x)=-f(x)\text{.}$
• Then (adding $f(x)$ to both sides) we see $2f(x)=0\text{,}$ so $f(x)=0\text{.}$

So, $f(x)=0$ for every $x\text{.}$

Exercise20
Hint

Think about mirroring a function across an axis. What does this do to the slope?

The derivative of an even function is odd, and the derivative of an odd function is even.

Solution

• Solution 1: Suppose $f(x)$ is an odd function. We investigate $f'(x)$ using the chain rule:

\begin{align*} f(-x)&=-f(x)& \mbox{(odd function)}\\ \diff{}{x}\{f(-x)\}&=\diff{}{x}\{-f(x)\}\\ -f'(-x)&=-f'(x) & \mbox{(chain rule)}\\ f'(-x)&=f'(x) \end{align*}

So, when $f(x)$ is odd, $f'(x)$ is even.

Similarly, suppose $f(x)$ is even.

\begin{align*} f(-x)&=f(x)& \mbox{(even function)}\\ \diff{}{x}\{f(-x)\}&=\diff{}{x}\{f(x)\}\\ -f'(-x)&=f'(x) & \mbox{(chain rule)}\\ f'(-x)&=-f'(x) \end{align*}

So, when $f(x)$ is even, $f'(x)$ is odd.

• Solution 2: Another way to think about this problem is to notice that “mirroring” a function changes the sign of its derivative. Then since an even function is “mirrored once” (across the $y$-axis), it should have $f'(x)=-f'(-x)\text{,}$ and so the derivative of an even function should be an odd function. Since an odd function is “mirrored twice” (across the $y$-axis and across the $x$-axis), it should have $f'(x)=-(-f'(-x))=f'(-x)\text{.}$ So the derivative of an odd function should be even. These ideas are presented in more detail below.

First, we consider the case where $f(x)$ is even, and investigate $f'(x)\text{.}$

The whole function has a mirror-like symmetry across the $y$-axis. So, at $x$ and $-x\text{,}$ the function will have the same “steepness,” but if one is increasing then the other is decreasing. That is, $f'(-x)=-f'(x)\text{.}$ (In the picture above, compare the slope at some point $a_i$ with its corresponding point $-a_i\text{.}$) So, $f'(x)$ is odd when $f(x)$ is even.

Second, let's consider the case where $f(x)$ is odd, and investigate $f'(x)\text{.}$ Suppose the blue graph below is $y=f(x)\text{.}$ If $f(x)$ were even, then to the left of the $y$-axis, it would look like the orange graph, which we'll call $y=g(x)\text{.}$

From our work above, we know that, for every $x \gt 0\text{,}$ $-f'(x)=g'(-x)\text{.}$ When $x \lt 0\text{,}$ $f(x)=-g(x)\text{.}$ So, if $x \gt 0\text{,}$ then $-f'(x)=g'(-x)=-f'(-x)\text{.}$ In other words, $f'(x)=f'(-x)\text{.}$ Similarly, if $x \lt 0\text{,}$ then $f'(x)=-g'(x)=f'(-x)\text{.}$ Therefore $f'(x)$ is even. (In the graph below, you can anecdotally verify that $f'(a_i)=f'(-a_i)\text{.}$)

Exercises1.3.2Exercises

Exercise1(*)

$e^2-e^{-2}$

Solution

The Fundamental Theorem of Calculus Part 2 (Theorem 1.3.1) tells us that

\begin{align*} \int_1^{\sqrt5} f(x)\,\dee{x} &= F(\sqrt5) - F(1) \\ &= \big( e^{(\sqrt5^2-3)} + 1 \big) - \big( e^{(1^2-3)} + 1 \big) \\ &= e^{5-3} - e^{1-3} = e^2-e^{-2} \end{align*}
Exercise2(*)
Hint

First find the general antiderivative by guessing and checking.

$F(x) = \dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+\dfrac{1}{2}\text{.}$

Solution

First, let's find a general antiderivative of $x^3-\sin(2x)\text{.}$

• One function with derivative $x^3$ is $\dfrac{x^4}{4}\text{.}$
• To find an antiderivative of $\sin(2x)\text{,}$ we might first guess $\cos(2x)\text{;}$ checking, we see $\diff{}{x}\{\cos(2x)\}=-2\sin(2x)\text{.}$ So, we only need to multiply by $-\dfrac{1}{2}\text{:}$ $\displaystyle\diff{}{x}\left\{-\dfrac{1}{2}\cos 2x\right\}=\sin(2x)\text{.}$

So, the general antiderivative of $f(x)$ is $\dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+C\text{.}$ To satisfy $F(0)=1\text{,}$ we need  9 The symbol $\iff$ is read “if and only if”. This is used in mathematics to express the logical equivalence of two statements. To be more precise, the statement $P \iff Q$ tells us that $P$ is true whenever $Q$ is true and $Q$ is true whenever $P$ is true.

\begin{gather*} \Big[\frac{x^4}{4}+\frac{1}{2}\cos 2x+C\Big]_{x=0}=1 \iff \frac{1}{2} + C = 1 \iff C=\frac{1}{2} \end{gather*}

So $F(x) = \dfrac{x^4}{4}+\dfrac{1}{2}\cos 2x+\dfrac{1}{2}\text{.}$

Exercise3(*)
Hint

Be careful. Two of these make no sense at all.

(a) True \qquad (b) False \qquad (c) False, unless $\int_a^b f(x)\,\dee{x}=\int_a^b xf(x)\,\dee{x} = 0\text{.}$

Solution

(a) This is true, by part 2 of the Fundamental Theorem of Calculus, Thereom 1.3.1, with $G(x)=f(x)$ and $f(x)$ replaced by $f'(x)\text{.}$

(b) This is not only false, but it makes no sense at all. The integrand is strictly positive so the integral has to be strictly positive. In fact it's $+\infty\text{.}$ The Fundamental Theorem of Calculus does not apply because the integrand has an infinite discontinuity at $x=0\text{.}$

(c) This is not only false, but it makes no sense at all, unless $\int_a^b f(x)\,\dee{x}=\int_a^b xf(x)\,\dee{x} = 0\text{.}$ The left hand side is a number. The right hand side is a number times $x\text{.}$

For example, if $a=0\text{,}$ $b=1$ and $f(x) = 1\text{,}$ then the left hand side is $\int_0^1 x\,\dee{x} = \frac{1}{2}$ and the right hand side is $x\int_0^1 \dee{x}=x\text{.}$

Exercise4
Hint

Check by differentiating.

false

Solution

This is a tempting thought:

\begin{align*} \int \frac{1}{x}\dee{x}&=\log|x|+C\\ \end{align*}

so perhaps similarly

\begin{align*} \int \frac{1}{x^2}\dee{x}&\stackrel{?}{=}\log|x^2|+C=\log(x^2)+C\\ \end{align*}

We check by differentiating:

\begin{align*} \diff{}{x}\{\log(x^2)\} &= \diff{}{x}\{2\log x\}=\frac{2}{x} \neq \frac{1}{x^2} \end{align*}

So, it wasn't so easy: false.

When we're guessing antiderivatives, we often need to adjust our original guesses a little. Changing constants works well; changing functions usually does not.

Exercise5
Hint

Check by differentiating.

false

Solution

This is tempting:

\begin{align*} \diff{}{x}\{\sin(e^x)\} &= e^x\cos(e^x)\\ \end{align*}

so perhaps

\begin{align*} \diff{}{x}\left\{\frac{\sin(e^x)}{e^x}\right\} &\stackrel{?}{=} \cos(e^x)\\ \end{align*}

We check by differentiating:

\begin{align*} \diff{}{x}\left\{\frac{\sin(e^x)}{e^x}\right\} &=\frac{e^x\left(\cos(e^x)\cdot e^x\right)-\sin(e^x)e^x}{e^{2x}} &\mbox{(quotient rule)}\\ & = \cos (e^x) - \frac{\sin(e^x)}{e^x}\\ &\neq \cos(e^x) \end{align*}

So, the statement is false.

When we're guessing antiderivatives, we often need to adjust our original guesses a little. Dividing by constants works well; dividing by functions usually does not.

Exercise6
Hint

Use the Fundamental Theorem of Calculus Part 1.

$\sin(x^2)$

Solution

“The instantaneous rate of change of $F(x)$ with respect to $x$” is another way of saying “$F'(x)$”. From the Fundamental Theorem of Calculus Part 1, we know this is $\sin (x^2)\text{.}$

Exercise7
Hint

Use the Fundamental Theorem of Calculus, Part 1.

$\sqrt[3]{e}$

Solution

The slope of the tangent line to $y=F(x)$ when $x=3$ is exactly $F'(3)\text{.}$ By the Fundamental Theorem of Calculus Part 1, $F'(x) = e^{1/x}\text{.}$ Then $F'(3) = e^{1/3} = \sqrt[3]{e}\text{.}$

Exercise8
Hint

You already know that $F(x)$ is an antiderivative of $f(x)\text{.}$

For any constant $C\text{,}$ $F(x)+C$ is an antiderivative of $f(x)\text{.}$ So, for example, $F(x)$ and $F(x)+1$ are both antiderivatives of $f(x)\text{.}$

Solution

For any constant $C\text{,}$ $F(x)+C$ is an antiderivative of $f(x)\text{,}$ because $\diff{}{x}\{F(x)+C\} = \diff{}{x}\{F(x)\} = f(x)\text{.}$ So, for example, $F(x)$ and $F(x)+1$ are both antiderivatives of $f(x)\text{.}$

Exercise9
Hint

(a) Recall $\diff{}{x}\{\arccos x\} = \frac{-1}{\sqrt{1-x^2}}\text{.}$\\ (b) All antiderivatives of $\sqrt{1-x^2}$ differ from one another by a constant. You already know one antiderivative.

1. We differentiate with respect to $a\text{.}$ Recall $\diff{}{x}\{\arccos x\} = \frac{-1}{\sqrt{1-x^2}}\text{.}$ To differentiate $\frac{1}{2}a\sqrt{1-a^2}\text{,}$ we use the product and chain rules. \begin{align*} \diff{}{a}\left\{\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}\right\} &= 0-\frac{1}{2}\cdot\frac{-1}{\sqrt{1-a^2}} + \left(\frac{1}{2}a\right)\cdot\frac{-2a}{2\sqrt{1-a^2}} + \frac{1}{2}\sqrt{1-a^2}\\ &=\frac{1}{2\sqrt{1-a^2}}- \frac{a^2}{2\sqrt{1-a^2}}+\frac{1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{1-a^2+1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{2(1-a^2)}{2\sqrt{1-a^2}}\\ &=\sqrt{1-a^2} \end{align*}
2. $F(x) = \dfrac{5\pi}{4}-\dfrac{1}{2}\arccos(x)+\dfrac{1}{2}x\sqrt{1-x^2}$
Solution

1. We differentiate with respect to $a\text{.}$ Recall $\diff{}{x}\{\arccos x\} = \frac{-1}{\sqrt{1-x^2}}\text{.}$ To differentiate $\frac{1}{2}a\sqrt{1-a^2}\text{,}$ we use the product and chain rules. \begin{align*} \diff{}{a}\left\{\frac{\pi}{4} - \frac{1}{2}\arccos(a)+\frac{1}{2}a\sqrt{1-a^2}\right\} &= 0-\frac{1}{2}\cdot\frac{-1}{\sqrt{1-a^2}} + \left(\frac{1}{2}a\right)\cdot\frac{-2a}{2\sqrt{1-a^2}} + \frac{1}{2}\sqrt{1-a^2}\\ &=\frac{1}{2\sqrt{1-a^2}}- \frac{a^2}{2\sqrt{1-a^2}}+\frac{1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{1-a^2+1-a^2}{2\sqrt{1-a^2}}\\ &=\frac{2(1-a^2)}{2\sqrt{1-a^2}}\\ &=\sqrt{1-a^2} \end{align*}
2. Let $G(x) = \frac{\pi}{4}-\frac{1}{2}\arccos(x)+\frac{1}{2}x\sqrt{1-x^2}\text{.}$ We showed in part (a) that $G(x)$ is an antiderivative of $\sqrt{1-x^2}\text{.}$ Since $F(x)$ is also an antiderivative of $\sqrt{1-x^2}\text{,}$ $F(x) = G(x)+C$ for some constant $C$ (this is Lemma 1.3.8).

Note $G(0)=\displaystyle\int_0^0\sqrt{1-x^2}\dee{x} =0\text{,}$ so if $F(0)=\pi\text{,}$ then $F(x)=G(x)+\pi\text{.}$ That is,

\begin{equation*} F(x) = \frac{5\pi}{4}-\frac{1}{2}\arccos(x)+\frac{1}{2}x\sqrt{1-x^2}\ . \end{equation*}
Exercise10
Hint

In order to apply the Fundamental Theorem of Calculus Part 2, the integrand must be continuous over the interval of integration.

(a) 0 \qquad (b),(c) The FTC does not apply, because the integrand is not continuous over the interval of integration.

Solution

1. The antiderivative of $\cos x$ is $\sin x\text{,}$ and $\cos x$ is continuous everywhere, so $\displaystyle\int_{-\pi}^\pi \cos x \dee{x} = \sin(\pi)-\sin(-\pi) = 0\text{.}$
2. Since $\sec^2 x$ is discontinuous at $x=\pm\frac{\pi}{2}\text{,}$ the Fundamental Theorem of Calculus Part 2 does not apply to $\displaystyle\int_{-\pi}^\pi \sec^2 x \dee{x}\text{.}$
3. Since $\frac{1}{x+1}$ is discontinuous at $x=-1\text{,}$ the Fundamental Theorem of Calculus Part 2 does not apply to $\displaystyle\int_{-2}^0 \frac{1}{x+1}\dee{x}\text{.}$
Exercise11
Hint

Use the definition of $F(x)$ as an area.

Solution

Using the definition of $F\text{,}$ $\textcolor{red}{F(x)}$ is the area under the curve from $a$ to $x\text{,}$ and $\textcolor{blue}{F(x+h)}$ is the area under the curve from $a$ to $x+h\text{.}$ These are shown on the same diagram, below.

Then the area represented by $\textcolor{blue}{F(x+h)}-\textcolor{red}{F(x)}$ is the area that is outside the red, but inside the blue. Equivalently, it is $\int\limits_{x}^{x+h} f(t)\dee{t}\text{.}$

Exercise12
Hint

$F(x)$ represents net signed area.

(a) zero \qquad (b) increasing when $0 \lt x \lt 1$ and $3 \lt x \lt 4\text{;}$ decreasing when $1 \lt x \lt 3$

Solution

We evaluate $F(0)$ using the definition: $F(0) = \int_0^0 f(t)\dee{t}=0\text{.}$ Although $f(0) \gt 0\text{,}$ the area from $t=0$ to $t=0$ is zero.\\ As $x$ moves along, $F(x)$ adds bits of signed area. If it's adding positive area, it's increasing, and if it's adding negative area, it's decreasing. So, $F(x)$ is increasing when $0 \lt x \lt 1$ and $3 \lt x \lt 4\text{,}$ and $F(x)$ is decreasing when $1 \lt x \lt 3\text{.}$

Exercise13
Hint

Note $G(x)=-F(x)\text{,}$ when $F(x)$ is defined as in Question 12.

(a) zero \qquad (b) $G(x)$ is increasing when $1 \lt x \lt 3\text{,}$ and it is decreasing when $0 \lt x \lt 1$ and when $3 \lt x \lt 4\text{.}$

Solution

This question is nearly identical to Question 12, with

\begin{equation*} G(x) = \displaystyle\int_x^0 f(t)\dee{t} = -\displaystyle\int_0^x f(t)\dee{t}=-F(x). \end{equation*}

So, $G(x)$ increases when $F(x)$ decreases, and vice-versa. Therefore: $G(0)=0\text{,}$ $G(x)$ is increasing when $1 \lt x \lt 3\text{,}$ and $G(x)$ is decreasing when $0 \lt x \lt 1$ and when $3 \lt x \lt 4\text{.}$

Exercise14
Hint

Using the definition of the derivative, $F'(x) = \displaystyle\lim_{h \to 0}\dfrac{F(x+h)-F(x)}{h}\text{.}$

The area of a trapezoid with base $b$ and heights $h_1$ and $h_2$ is $\frac{1}{2}b(h_1+h_2)\text{.}$

Using the definition of the derivative,

\begin{align*} F'(x) & = \displaystyle\lim_{h \to 0}\dfrac{F(x+h)-F(x)}{h}\\ &=\lim_{h \to 0}\dfrac{\int_a^{x+h} t\dee{t}-\int_a^x t\dee{t}}{h}\\ &=\lim_{h \to 0}\dfrac{\int_x^{x+h} t\dee{t}}{h}\\ \end{align*}

The numerator describes the area of a trapezoid with base $h$ and heights $x$ and $x+h\text{.}$

\begin{align*} &=\lim_{h \to 0}\dfrac{\frac{1}{2}h(x+x+h)}{h}\\ &=\lim_{h \to 0}\left(x+\frac{1}{2}h\right)\\ &=x \end{align*}

So, $F'(x)=x\text{.}$

Solution

Using the definition of the derivative,

\begin{align*} F'(x) & = \displaystyle\lim_{h \to 0}\dfrac{F(x+h)-F(x)}{h}\\ &=\lim_{h \to 0}\dfrac{\int_a^{x+h} t\dee{t}-\int_a^x t\dee{t}}{h}\\ &=\lim_{h \to 0}\dfrac{\int_x^{x+h} t\dee{t}}{h}\\ \end{align*}

The numerator describes the area of a trapezoid with base $h$ and heights $x$ and $x+h\text{.}$

\begin{align*} &=\lim_{h \to 0}\dfrac{\frac{1}{2}h(x+x+h)}{h}\\ &=\lim_{h \to 0}\left(x+\frac{1}{2}h\right)\\ &=x \end{align*}

So, $F'(x)=x\text{.}$

Exercise15
Hint

There is only one!

$f(t)=0$

Solution

If $F(x)$ is constant, then $F'(x)=0\text{.}$ By the Fundamental Theorem of Calculus Part 1, $F'(x)=f(x)\text{.}$ So, the only possible continuous function fitting the question is $f(x)=0\text{.}$

This makes intuitive sense: if moving $x$ doesn't add or subtract area under the curve, then there must not be any area under the curve--the curve should be the same as the $x$-axis.

As an aside, we mention that there are other, non-continuous functions $f(t)$ such that $\int_0^x f(t)\dee{t} = 0$ for all $x\text{.}$ For example, $f(t) = \left\{\begin{array}{cc} 0 & x \neq 0\\ 1 & x=0 \end{array}\right.\text{.}$ These kinds of removable discontinuities will not factor heavily in our discussion of integrals.

Exercise16
Hint

If $\diff{}{x}\{F(x)\}=f(x)\text{,}$ that tells us $\int f(x)\dee{x} = F(x)+C\text{.}$

$\int \log(ax)\dee{x}= x\log(ax)-x+C\text{,}$ where $a$ is a given constant, and $C$ is any constant.

Solution

\begin{align*} \diff{}{x}\{x\log(ax)-x\}&=x\left(\frac{a}{ax}\right)+\log(ax)-1 &\mbox{(product rule, chain rule)}\\ &=\log(ax)\\ \end{align*}

So, we know

\begin{align*} \int \log(ax)\dee{x}&= x\log(ax)-x+C&\mbox{where $a$ is a given constant, and $C$ is any constant.} \end{align*}

Remark: $\int \log(ax)\dee{x}$ can be calculated using the method of Integration by Parts, which you will learn in Section 1.7.

Exercise17
Hint

When you're differentiating, you can leave the $e^x$ factored out.

$\int x^3e^x\dee{x}=e^x\left(x^3-3x^2+6x-6\right)+C$

Solution

\begin{align*} \diff{}{x}\left\{e^x\left(x^3-3x^2+6x-6\right)\right\}&=e^x\left(3x^2-6x+6\right)+e^x\left(x^3-3x^2+6x-6\right) &\mbox{(product rule)}\\ &=e^x\left(3x^2-6x+6+x^3-3x^2+6x-6\right)\\ &=x^3e^x\\ \end{align*}

So,

\begin{align*} \int x^3e^x\dee{x}&=e^x\left(x^3-3x^2+6x-6\right)+C \end{align*}

Remark: $\int x^3e^x\dee{x}$ can be calculated using the method of Integration by Parts, which you will learn in Section 1.7.

Exercise18
Hint

After differentiation, you can simplify pretty far. Keep at it!

$\displaystyle\int \dfrac{1}{\sqrt{x^2+a^2}}\dee{x} = \log\left|x+\sqrt{x^2+a^2}\right|+C$ when $a$ is a given constant. As usual, $C$ is an arbitrary constant.

Solution

\begin{align*} \diff{}{x}\left\{\log\left|x+\sqrt{x^2+a^2}\right|\right\}&= \frac{1}{x+\sqrt{x^2+a^2}}\cdot \left(1+\frac{1}{2\sqrt{x^2+a^2}}\cdot 2x\right) &\mbox{(chain rule)}\\ &=\frac{1+\frac{x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}} =\frac{\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}}}{x+\sqrt{x^2+a^2}}\\ &=\frac{1}{\sqrt{x^2+a^2}}\\ \end{align*}

So,

\begin{align*} \int \frac{1}{\sqrt{x^2+a^2}}\dee{x} &= \log\left|x+\sqrt{x^2+a^2}\right|+C \end{align*}

Remark: $\int \frac{1}{\sqrt{x^2+a^2}}\dee{x}$ can be calculated using the method of Trigonometric Substitution, which you will learn in Section 1.9.

Exercise19
Hint

This derivative also simplifies considerably. You might need to add fractions by finding a common denominator.

$\displaystyle\int \dfrac{x}{\sqrt{x(a+x)}}\dee{x}=\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)+C$

Solution

Using the chain rule:

\begin{align*} \diff{}{x}&\left\{\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)\right\}\\ &= \frac{x+(a+x)}{2\sqrt{x(a+x)}}-a\left(\frac{1}{\sqrt{x}+\sqrt{a+x}}\cdot\left(\frac{1}{2\sqrt{x}}+\frac{1}{2\sqrt{a+x}}\right)\right)\\ &= \frac{2x+a}{2\sqrt{x(a+x)}}- a\left( \frac{1}{\sqrt{x}+\sqrt{a+x}}\cdot\left( \frac{\sqrt{a+x}+\sqrt{x}}{2\sqrt{x(a+x)}}\right)\right)\\ &= \frac{2x+a}{2\sqrt{x(a+x)}}- a\left(\frac{1}{2\sqrt{x(a+x)}}\right)\\ &=\frac{2x}{2\sqrt{x(a+x)}} = \frac{x}{\sqrt{x(a+x)}}\\ \end{align*}

So,

\begin{align*} \int \frac{x}{\sqrt{x(a+x)}}\dee{x}&=\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)+C \end{align*}

Remark: $\int \frac{x}{\sqrt{x(a+x)}}\dee{x}$ can be calculated using the method of Trigonometric Substitution, which you will learn in Section 1.9.

Exercise20(*)
Hint

Guess a function whose derivative is the integrand, then use the Fundamental Theorem of Calculus Part 2.

$5-\cos 2$

Solution

By the Fundamental Theorem of Calculus,

\begin{align*} \int_0^2 \big(x^3+\sin x)\,\dee{x} &=\left[\frac{x^4}{4}-\cos x\right]_0^2\\ &=\left(\frac{2^4}{4}-\cos 2\right)-\left(0 -\cos 0\right) \\ &= 4-\cos2+1=5-\cos2. \end{align*}
Exercise21(*)
Hint

Split the given integral up into two integrals.

$2$

Solution

By part (d) of our “Arithmetic of Integration” theorem, Theorem 1.2.1,

\begin{equation*} \int_1^2 \frac{x^2+2}{x^2}\,\dee{x} =\int_1^2 \Big[1+\frac{2}{x^2}\Big]\,\dee{x} =\int_1^2 \dee{x} + 2\int_1^2 \frac{1}{x^2}\,\dee{x} \end{equation*}

Then by the Fundamental Theorem of Calculus Part 2,

\begin{gather*} \int_1^2 \dee{x} + 2\int_1^2 \frac{1}{x^2}\,\dee{x} =\Big[x\Big]_1^2 + 2\Big[-\frac{1}{x}\Big]_1^2 =\big[2-1\big] + 2\Big[-\frac{1}{2}+1\Big] =2 \end{gather*}
Exercise22
Hint

The integrand is similar to $\dfrac{1}{1+x^2}\text{,}$ so something with arctangent seems in order.

$\dfrac{1}{5}\arctan(5x)+C$

Solution

The integrand is similar to $\dfrac{1}{1+x^2}\text{,}$ which is the derivative of arctangent. Indeed, we have

\begin{align*} \displaystyle\int \dfrac{1}{1+25x^2}\dee{x} &= \int \dfrac{1}{1+(5x)^2}\dee{x}.\\ \end{align*}

So, a reasonable first guess for the antiderivative might be

\begin{align*} \textcolor{red}{F(x)} &\textcolor{red}{\stackrel{?}{=} \arctan(5x)}.\\ \end{align*}

However, because of the chain rule,

\begin{align*} \color{red}{F'(x)} &\color{red}{= \dfrac{5}{1+(5x)^2}}.\\ \end{align*}

In order to “fix” the numerator, we make a second guess:

\begin{align*} \color{blue}{F(x)} &\color{blue}{= \frac{1}{5}\arctan(5x)}\\ \color{blue}{F'(x)} &\color{blue}{= \dfrac{1}{5}\left(\dfrac{5}{1+(5x)^2}\right) = \dfrac{1}{1+25x^2}}\\ \mbox{So,}\qquad \displaystyle\int \dfrac{1}{1+25x^2}\dee{x}&=\frac{1}{5}\arctan(5x)+C. \end{align*}
Exercise23
Hint

The integrand is similar to $\dfrac{1}{\sqrt{1-x^2}}\text{,}$ so factoring out $\sqrt{2}$ from the denominator will make it look like some flavour of arcsine.

$\arcsin\left(\dfrac{x}{\sqrt{2}}\right)+C$

Solution

The integrand is similar to $\dfrac{1}{\sqrt{1-x^2}}\text{.}$ In order to formulate a guess for the antiderivative, let's factor out $\sqrt{2}$ from the denominator:

\begin{align*} \displaystyle\int \dfrac{1}{\sqrt{2-x^2}}\dee{x}&= \displaystyle\int \dfrac{1}{\sqrt{2\left(1-\frac{x^2}{2}\right)}}\dee{x}\\ &=\displaystyle\int \dfrac{1}{ \sqrt{2}\sqrt{1-\frac{x^2}{2}}}\dee{x}\\ &= \int\frac{1}{\sqrt{2}}\cdot \dfrac{1}{ \sqrt{1-\left(\dfrac{x}{\sqrt2}\right)^2}}\dee{x}\\ \end{align*}

At this point, we might guess that our antiderivative is something like $F(x) = \arcsin\left(\dfrac{x}{\sqrt{2}}\right)\text{.}$ To explore this possibility, we can differentiate, and see what we get.

\begin{align*} \diff{}{x}\left\{\arcsin\left(\dfrac{x}{\sqrt{2}}\right)\right\}&=\frac{1}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{1-\left(\dfrac{x}{\sqrt{2}}\right)^2}}\\ \end{align*}

This is exactly what we want! So,

\begin{align*} \int \dfrac{1}{\sqrt{2-x^2}}\dee{x}&=\arcsin\left(\dfrac{x}{\sqrt2}\right)+C \end{align*}
Exercise24
Hint

We know how to antidifferentiate $\sec^2 x\text{,}$ and there is an identity linking $\sec^2 x$ with $\tan^2 x\text{.}$

$\tan x - x +C$

Solution

We know that $\int \sec^2 x\dee{x} = \tan x +C\text{,}$ and $\sec^2x = \tan^2 x + 1\text{,}$ so

\begin{align*} \int \tan^2 x\dee{x} &= \int \sec^2x - 1 \dee{x}\\ &=\int \sec^2 x \dee{x} - \int 1 \dee{x}\\ &=\tan x - x + C \end{align*}
Exercise25
Hint

Recall $2\sin x \cos x = \sin(2x)\text{.}$

$-\dfrac{3}{4}\cos(2x)+C\text{,}$ or equivalently, $\dfrac{3}{2}\sin^2 x+C$

Solution

• Solution 1: This might not obviously look like the derivative of anything familiar, but it does look like half of a familiar trig identity: $2\sin x \cos x = \sin(2x)\text{.}$ \begin{align*} \int 3 \sin x \cos x \dee{x}&=\int \frac{3}{2}\cdot 2\sin x \cos x \dee{x}\\ &=\int \frac{3}{2} \sin(2x)\dee{x}\\ \end{align*}

So, we might guess that the antiderivative is something like $-\cos(2x)\text{.}$ We only need to figure out the constants.

• Solution 2: You might notice that the integrand looks like it came from the chain rule, since $\cos x$ is the derivative of $\sin x\text{.}$ Using this observation, we can work out the antideriative: \begin{align*} \diff{}{x}\left\{\sin^2 x\right\}&=2\sin x \cos x\\ \diff{}{x}\left\{\frac{3}{2}\sin^2 x\right\}&=3\sin x \cos x\\ \mbox{So,}\qquad \int 3\sin x \cos x \dee{x}&=\frac{3}{2}\sin^2 x+C \end{align*}

These two answers look different. Using the identity $\cos(2x)=1 - 2 \sin^2(x)\text{,}$ we reconcile them:

\begin{align*} -\frac{3}{4}\cos(2x)+C&= -\frac{3}{4}\left(1-2\sin^2x\right)+C\\ &=\frac{3}{2}\sin^2 x + \left(C-\frac{3}{4}\right) \end{align*}

The $\frac{3}{4}$ here is not significant. Remember that $C$ is used to designate a constant that can take any value between $-\infty$ and $+\infty\text{.}$ So $C-\frac{3}{4}$ is also just a constant that can take any value between $-\infty$ and $+\infty\text{.}$ As the two answers we found differ by a constant, they are equivalent.

Exercise26
Hint

$\cos^2 x = \dfrac{1+\cos(2x)}{2}$

$\dfrac{1}{2}x+\dfrac{1}{4}\sin(2x)+C$

Solution

It's not immediately obvious which function has $\cos^2 x$ as its derivative, but we can make the situation a little clearer by using the identity $\cos^2 x = \dfrac{1+\cos(2x)}{2}\text{:}$

\begin{align*} \int \cos^2 x \dee{x}&=\int \frac{1}{2}\cdot \left(1+\cos(2x)\right)\dee{x}\\ &=\int \frac{1}{2}\dee{x} + \int \frac{1}{2}\cos(2x)\dee{x}\\ &=\frac{1}{2}x + C+ \int \frac{1}{2}\cos(2x)\dee{x}\\ \end{align*}

For the remaining integral, we might guess something like $F(x) = \sin(2x)\text{.}$ Let's figure out the appropriate constant:

\begin{align*} \diff{}{x}\left\{ \sin(2x) \right\}&=2\cos(2x)\\ \diff{}{x}\left\{ \frac{1}{4}\sin(2x) \right\}&=\frac{1}{2}\cos(2x)\\ \mbox{So,}\qquad \int \frac{1}{2}\cos(2x)\dee{x} &=\frac{1}{4}\sin(2x)+C\\ \mbox{Therefore,}\qquad \int \cos^2 x \dee{x}&=\frac{1}{2}x+\frac{1}{4}\sin(2x)+C \end{align*}
Exercise27(*)

$F'\left(\frac{\pi}{2}\right)=\log(3)$\qquad $G'\left(\frac{\pi}{2}\right)=-\log(3)$

Solution

By the Fundamental Theorem of Calculus Part 1, \begin{alignat*}{3} F'(x)&=\diff{}{x} \int_0^x \log(2+\sin t)\,\dee{t} &&=\log(2+\sin x) \\ G'(y)&=\diff{}{y}\bigg[- \int_0^y \log(2+\sin t)\,\dee{t}\bigg] &&=-\log(2+\sin y) \\ \end{alignat*} So,

Exercise28(*)
Hint

There is a good way to test where a function is increasing, decreasing, or constant, that also has something to do with topic of this section.

$f(x)$ is increasing when $-\infty \lt x \lt 1$ and when $2 \lt x \lt \infty\text{.}$

Solution

By the Fundamental Theorem of Calculus Part 1,

\begin{equation*} f'(x) = 100(x^2-3x+2)e^{-x^2} = 100(x-1)(x-2)e^{-x^2} \end{equation*}

As $f(x)$ is increasing whenever $f'(x) \gt 0$ and $100 e^{-x^2}$ is always strictly bigger than $0\text{,}$ we have $f(x)$ increasing if and only if $(x-1)(x-2) \gt 0\text{,}$ which is the case if and only if $(x-1)$ and $(x-2)$ are of the same sign. Both are positive when $x \gt 2$ and both are negative when $x \lt 1\text{.}$ So $f(x)$ is increasing when $-\infty \lt x \lt 1$ and when $2 \lt x \lt \infty\text{.}$

Remark: even without the Fundamental Theorem of Calculus, since $f(x)$ is the area under a curve from 1 to $x\text{,}$ $f(x)$ is increasing when the curve is above the $x$-axis (because we're adding positive area), and it's decreasing when the curve is below the $x$-axis (because we're adding negative area).

Exercise29(*)
Hint

See Example 1.3.5.

$F'(x)=-\dfrac{\sin x}{\cos^3x+6}$

Solution

Write $G(x)={\displaystyle\int_0^x \frac{1}{t^3+6}\,\dee{t}}\text{.}$ By the Fundamental Theorem of Calculus Part 1, $G'(x)=\dfrac{1}{x^3+6}\text{.}$ Since $F(x)=G(\cos x)\text{,}$ the chain rule gives us

\begin{gather*} F'(x)=G'(\cos x)\cdot(-\sin x)=-\frac{\sin x}{\cos^3x+6} \end{gather*}
Exercise30(*)
Hint

See Example 1.3.5.

$4x^3e^{(1+x^4)^2}$

Solution

Define $g(x)= \displaystyle\int_0^x e^{t^2}\dee{t}\text{.}$ By the Fundamental Theorem of Calculus Part 1, $g'(x)=e^{x^2}\text{.}$ As $f(x)=g(1+x^4)$ the chain rule gives us

\begin{gather*} f'(x)=4x^3g'(1+x^4)=4x^3e^{(1+x^4)^2} \end{gather*}
Exercise31(*)
Hint

See Example 1.3.5.

$\big(\sin^6 x+8)\cos x$

Solution

Define $g(x)=\int_0^x (t^6+8)\dee{t}\text{.}$ By the fundamental theorem of calculus, $g'(x)=x^6+8\text{.}$ We are to compute the derivative of $f(x)=g(\sin x)\text{.}$ The chain rule gives

\begin{gather*} \diff{}{x}\left\{\int_0^{\sin x}(t^6+8)\dee{t}\right\} =g'(\sin x)\cdot \cos x=\big(\sin^6 x+8\big)\cos x \end{gather*}
Exercise32(*)
Hint

See Example 1.3.5.

$F'(1)=3e^{-1}$

Solution

Let $G(x)= \displaystyle\int_0^{x}e^{-t}\sin\left(\frac{\pi t}{2}\right)\,\dee{t}\text{.}$ By the Fundamental Theorem of Calculus Part 1, $G'(x)=e^{-x}\sin\big(\frac{\pi x}{2}\big)$ and, since $F(x)=G(x^3)\text{,}$ $F'(x)=3x^2G'(x^3)=3x^2e^{-x^3}\sin\big(\frac{\pi x^3}{2}\big)\text{.}$ Then $F'(1)=3e^{-1}\sin\big(\frac{\pi }{2}\big) =3e^{-1}\text{.}$

Exercise33(*)
Hint

See Example 1.3.6.

$\displaystyle{}\frac{\sin{u}}{1+\cos^3 u}$

Solution

Define $\displaystyle G(x) = \int_x^0 \frac{\dee{t}}{1+t^3} = - \int_0^x \frac1{1+t^3}\,\dee{t}\text{,}$ so that $\displaystyle G'(x) = - \frac1{1+x^3}$ by the Fundamental Theorem of Calculus Part 1. Then by the chain rule,

\begin{gather*} \diff{}{u} \left\{ \int_{\cos u}^0 \frac{\dee{t}}{1+t^3} \right\} = \diff{}{u} G(\cos u) = G'(\cos u) \cdot \diff{}{u}\cos u = {-}\frac{1}{1+\cos^3 u} \cdot (-\sin{u}). \end{gather*}
Exercise34(*)
Hint

Apply $\diff{}{x}$ to both sides.

$f(x)=2x$

Solution

Applying $\diff{}{x}$ to both sides of $x^2=1+\int_1^x f(t)\dee{t}$ gives, by the Fundamental Theorem of Calculus Part 1, $2x=f(x)\text{.}$

Exercise35(*)
Hint

What is the title of this section?

$f(4)=4\pi$

Solution

Apply $\diff{}{x}$ to both sides of $x \sin(\pi x) = \int_0^x f(t)\, \dee{t}\text{.}$ Then, by the Fundamental Theorem of Calculus Part 1, \begin{alignat*}{3} &&f(x) =\diff{}{x} \int_0^x f(t)\, \dee{t}&=\diff{}{x}\big\{x \sin(\pi x)\big\} \\ \implies&& f(x) &= \diff{}{x}\big\{x \sin(\pi x)\big\} =\sin(\pi x)+\pi x\cos(\pi x) \\ \implies&& f(4)&= \sin(4\pi)+4\pi\cos(4\pi)=4\pi \end{alignat*}

Exercise36(*)
Hint

See Example 1.3.6.

(a) $(2x+1)e^{-x^2}$ \qquad (b) $x=-1/2$

Solution

(a) Write

By the Fundamental Theorem of Calculus Part 1,

\begin{gather*} G'(y)=e^{-y}\, ,\qquad H'(y)= e^{-y^2} \end{gather*}

Hence, by the chain rule,

\begin{gather*} F'(x)=2x G'(x^2)-(-1)H'(-x)=2x e^{-(x^2)}+e^{-(-x)^2} =(2x+1)e^{-x^2} \end{gather*}

(b) Observe that $F'(x) \lt 0$ for $x \lt -1/2$ and $F'(x) \gt 0$ for $x \gt -1/2\text{.}$ Hence $F(x)$ is decreasing for $x \lt -1/2$ and increasing for $x \gt -1/2\text{,}$ and $F(x)$ must take its minimum value when $x=-1/2\text{.}$

Exercise37(*)
Hint

See Example 1.3.6.

$e^{\sin x}-e^{\sin(x^4-x^3)}\big(4x^3-3x^2\big)$

Solution

Define $G(y)=\displaystyle\int_0^ye^{\sin t}\dee{t}\text{.}$ Then:

\begin{align*} F(x) &= \int_0^x e^{\sin t}\,\dee{t} + \int_{x^4-x^3}^0 e^{\sin t}\,\dee{t} = \int_0^x e^{\sin t}\,\dee{t} - \int_0^{x^4-x^3} e^{\sin t}\,\dee{t} \\ &=G(x) - G(x^4-x^3) \end{align*}

By the Fundamental Theorem of Calculus Part 1,

Hence, by the chain rule,

\begin{align*} F'(x) &= G'(x)- G'(x^4-x^3)\ \diff{}{x}\big\{x^4-x^3\big\} \\ &= G'(x)- G'(x^4-x^3)\ (4x^3-3x^2) \\ &=e^{\sin x}-e^{\sin(x^4-x^3)}\big(4x^3-3x^2\big) \end{align*}
Exercise38(*)
Hint

See Example 1.3.6.

$-2x \cos\big(e^{-x^2}\big) -5x^4\cos\big(e^{x^5}\big)$

Solution

Define with $G(y) = \displaystyle\int_0^y \cos\big(e^t\big)\,\dee{t}\text{.}$ Then:

\begin{align*} F(x) &= \int_{x^5}^{-x^2} \cos\big(e^t\big)\,\dee{t} =\int_0^{-x^2} \cos\big(e^t\big)\,\dee{t} + \int_{x^5}^0 \cos\big(e^t\big)\,\dee{t} \\ &=\int_0^{-x^2} \cos\big(e^t\big)\,\dee{t} - \int_0^{x^5} \cos\big(e^t\big)\,\dee{t} \\ &=G(-x^2) - G(x^5) \end{align*}

By the Fundamental Theorem of Calculus,

\begin{gather*} G'(y)=\cos\big(e^y\big) \end{gather*}

Hence, by the chain rule,

\begin{align*} F'(x) &= G'(-x^2)\ \diff{}{x}\big\{-x^2\big\} - G'(x^5)\ \diff{}{x}\big\{x^5\big\} \\ &= G'(-x^2)\ (-2x)- G'(x^5)\ (5x^4) \\ &=-2x \cos\big(e^{-x^2}\big) -5x^4\cos\big(e^{x^5}\big) \end{align*}
Exercise39(*)
Hint

See Example 1.3.6.

$e^x\sqrt{\sin(e^x)} -\sqrt{\sin(x)}$

Solution

Define with $G(y) = \displaystyle\int_0^y \sqrt{\sin t}\,\dee{t}$ Then:

\begin{align*} F(x)&=\int_x^{e^x} \sqrt{\sin t}\,\dee{t} \\ &= \int_0^{e^x} \sqrt{\sin t}\,\dee{t} + \int_x^0 \sqrt{\sin t}\,\dee{t} = \int_0^{e^x} \sqrt{\sin t}\,\dee{t} - \int_0^x \sqrt{\sin t}\,\dee{t} \\ &=G(e^x) - G(x) \end{align*}

By the Fundamental Theorem of Calculus Part 1,

Hence, by the chain rule,

\begin{align*} F'(x) &= G'(e^x)\ \diff{}{x}\big\{e^x\big\} - G'(x)\\ &= e^xG'(e^x) - G'(x)\\ &=e^x\sqrt{\sin(e^x)} -\sqrt{\sin(x)} \end{align*}
Exercise40(*)
Hint

Split up the domain of integration.

$14$

Solution

Splitting up the domain of integration,

\begin{align*} \int_1^5 f(x)\dee{x} &= \int_1^3 f(x)\,\dee{x} + \int_3^5 f(x)\,\dee{x} \\ &= \int_1^3 3\,\dee{x} + \int_3^5 x\,\dee{x} \\ &= 3x\bigg|_{x=1}^{x=3} +\frac{x^2}{2} \bigg|_{x=3}^{x=5} \\ &=14 \end{align*}
Exercise41(*)
Hint

It is possible to guess an antiderivative for $f'(x) f''(x)$ that is expressed in terms of $f'(x)\text{.}$

$\dfrac{5}{2}$

Solution

By the chain rule,

\begin{gather*} \diff{}{x}\big\{\left(f'(x)\right)^2\big\} = 2 f'(x)\,f''(x) \end{gather*}

so $\frac{1}{2} f'(x)^2$ is an antiderivative for $f'(x)\,f''(x)$ and, by the Fundamental Theorem of Calculus Part 2,

\begin{gather*} \int_1^2 f'(x) f''(x)\,\dee{x} =\left[\frac{1}{2}\left(f'(x)\right)^2\right]_{x=1}^{x=2} =\frac{1}{2} f'(2)^2 - \frac{1}{2} f'(1)^2 =\frac{5}{2} \end{gather*}

Remark: evaluating antiderivatives of this type will occupy the next section, Section 1.4.

Exercise42(*)
Hint

When does the car stop? What is the relation between velocity and distance travelled?

$45\,\textrm{m}$

Solution

The car stops when $v(t)=30-10t=0\text{,}$ which occurs at time $t = 3\text{.}$ The distance covered up to that time is

\begin{gather*} \int_0^3 v(t)\,\dee{t} = (30t - 5t^2)\Big|_{0}^{3} = (90-45)-0 = 45\,\textrm{m}. \end{gather*}
Exercise43(*)
Hint

See Example 1.3.5. For the absolute maximum part of the question, study the sign of $f'(x)\text{.}$

$f'(x)=(2-2x)\log\big(1+e^{2x-x^2}\big)$ and $f(x)$ achieves its absolute maximum at $x=1\text{,}$ because $f(x)$ is increasing for $x \lt 1$ and decreasing for $x \gt 1\text{.}$

Solution

Define $g(x) = \displaystyle\int_0^x\log\big(1+e^t\big)\,\dee{t}\text{.}$ By the Fundamental Theorem of Calculus Part 1, $g'(x) = \log\big(1+e^x\big)\text{.}$ But $f(x)=g(2x-x^2)\text{,}$ so by the chain rule,

\begin{gather*} f'(x)=g'(2x-x^2)\cdot \diff{}{x}\{2x-x^2\} =(2-2x)\cdot\log\big(1+e^{2x-x^2}\big) \end{gather*}

Observe that $e^{2x-x^2} \gt 0$ for all $x$ so that $1+e^{2x-x^2} \gt 1$ for all $x$ and $\log\big(1+e^{2x-x^2}\big) \gt 0$ for all $x\text{.}$ Since $2-2x$ is positive for $x \lt 1$ and negative for $x \gt 1\text{,}$ $f'(x)$ is also positive for $x \lt 1$ and negative for $x \gt 1\text{.}$ That is, $f(x)$ is increasing for $x \lt 1$ and decreasing for $x \gt 1\text{.}$ So $f(x)$ achieves its absolute maximum at $x=1\text{.}$

Exercise44(*)
Hint

See Example 1.3.5. For the “minimum value” part of the question, study the sign of $f'(x)\text{.}$

The minimum is $\int_0^{-1} \frac{\dee{t}}{1+t^4}\text{.}$ As $x$ runs from $-\infty$ to $\infty\text{,}$ the function $f(x)= \int_0^{x^2-2x}\frac{\dee{t}}{1+t^4}$ decreases until $x$ reaches 1 and then increases all $x \gt 1\text{.}$ So the minimum is achieved for $x=1\text{.}$ At $x=1\text{,}$ $x^2-2x=-1\text{.}$

Solution

Let $f(x)=\int_0^{x^2-2x}\frac{\dee{t}}{1+t^4}$ and $g(x)= \int_0^{x}\frac{\dee{t}}{1+t^4}\text{.}$ Then $g'(x)=\frac{1}{1+x^4}$ and, since $f(x)=g(x^2-2x)\text{,}$ $f'(x)=(2x-2)g'(x^2-2x)=2\frac{x-1}{1+(x^2-2x)^4}\text{.}$ This is zero for $x=1\text{,}$ negative for $x \lt 1$ and positive for $x \gt 1\text{.}$ Thus as $x$ runs from $-\infty$ to $\infty\text{,}$ $f(x)$ decreases until $x$ reaches 1 and then increases all $x \gt 1\text{.}$ So the minimum of $f(x)$ is achieved for $x=1\text{.}$ At $x=1\text{,}$ $x^2-2x=-1$ and $f(1)=\int_0^{-1}\frac{\dee{t}}{1+t^4}\text{.}$

Exercise45(*)
Hint

See Example 1.3.5. For the “maximum” part of the question, study the sign of $F'(x)\text{.}$

$F$ achieves its maximum value at $x=\pi\text{.}$

Solution

Define $G(x)=\displaystyle\int_0^x\sin(\sqrt{t})\,\dee{t}\text{.}$ By the Fundamental Theorem of Calculus Part 1, $G'(x)=\sin(\sqrt{x})\text{.}$ Since $F(x)=G(x^2)\text{,}$ and since $x \gt 0\text{,}$ we have

\begin{equation*} F'(x)=2xG'(x^2)=2x\sin |x|=2x\sin x. \end{equation*}

Thus $F$ increases as $x$ runs from to $0$ to $\pi$ (since $F'(x) \gt 0$ there) and decreases as $x$ runs from $\pi$ to $4$ (since $F'(x) \lt 0$ there). Thus $F$ achieves its maximum value at $x=\pi\text{.}$

Exercise46(*)
Hint

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11.

$2$

Solution

The given sum is of the form

\begin{gather*} \lim_{n\rightarrow\infty}\sum_{j=1}^n \frac{\pi}{n}\sin\Big(\frac{j\pi}{n}\Big) =\lim_{n\rightarrow\infty}\sum_{j=1}^n f(x_j^*)\De x \end{gather*}

with $\De x=\frac{\pi}{n}\text{,}$ $x_j^*=\frac{j\pi}{n}$ and $f(x)=\sin(x)\text{.}$ Since $x_0^*=0$ and $x_n^*=\pi\text{,}$ the right hand side is the definition (using the right Riemann sum) of

\begin{gather*} \int_0^\pi f(x)\,\dee{x}=\int_0^\pi \sin(x)\,\dee{x} =\left[-\cos(x)\right]_0^\pi=2 \end{gather*}

where we evaluate the definite integral using the Fundamental Theorem of Calculus Part 2.

Exercise47(*)
Hint

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11.

$\log 2$

Solution

The given sum is of the form

\begin{equation*} \lim_{n\rightarrow\infty}\frac{1}{n} \sum_{j=1}^n \frac{1}{1+\frac{j}{n}} =\lim_{n\rightarrow\infty}\sum_{j=1}^n f(x_j)\De x \end{equation*}

with $\De x=\frac{1}{n}\text{,}$ $x_j=\frac{j}{n}$ and $f(x)=\frac{1}{1+x}\text{.}$ The right hand side is the definition (using the right Riemann sum) of

\begin{gather*} \int_0^1 f(x)\,dx=\int_0^1 \frac{1}{1+x}\,\dee{x} =\log|1+x|\Big|_0^1 =\log 2 \end{gather*}
Exercise48
Hint

Carefully check the Fundamental Theorem of Calculus: as written, it only applies directly to $F(x)$ when $x\ge0\text{.}$

Is $F(x)$ even or odd?

In the sketch below, open dots denote inflection points, and closed dots denote extrema.

Solution

• $\mathbf{F(x),\,x \ge 0}\text{:}$ We learned quite a lot last semester about curve sketching. We can use those techniques here. We have to be quite careful about the sign of $x\text{,}$ though. We can only directly apply the Fundamental Theorem of Calculus Part 1 (as it's written in your text) when $x\ge 0\text{.}$ So first, let's graph the right-hand portion. Notice $f(x)$ has even symmetry--so, if we know one half of $F(x)\text{,}$ we should be able to figure out the other half with relative ease.

• $F(0)=\displaystyle\int_0^0 f(t)\dee{t}=0$ (so, $F(x)$ passes through the origin)
• Using the Fundamental Theorem of Calculus Part 1, $F'(x) \gt 0$ when $0 \lt x \lt 1$ and when $3 \lt x \lt 5\text{;}$ $F'(x) \lt 0$ when $1 \lt x \lt 3\text{.}$ So, $F(x)$ is decreasing from 1 to 3, and increasing from 0 to 1 and also from 3 to 5. That gives us a skeleton to work with.

We get the relative sizes of the maxes and mins by eyeballing the area under $y=f(t)\text{.}$ The first lobe (from $x=0$ to $x=1$ has a small positive area, so $F(1)$ is a small positive number. The next lobe (from $x=1$ to $x=3$) has a larger absolute area than the first, so $F(3)$ is negative. Indeed, the second lobe seems to have more than twice the area of the first, so $|F(3)|$ should be larger than $F(1)\text{.}$ The third lobe is larger still, and even after subtracting the area of the second lobe it looks much larger than the first or second lobe, so $|F(3)| \lt F(5)\text{.}$

• We can use $F''(x)$ to get the concavity of $F(x)\text{.}$ Note $F''(x)=f'(x)\text{.}$ We observe $f(x)$ is decreasing on (roughly) $(0,2.5)$ and $(4,5)\text{,}$ so $F(x)$ is concave down on those intervals. Further, $f(x)$ is increasing on (roughly) $(2.5,4)\text{,}$ so $F(x)$ is concave up there, and has inflection points at about $x=2.5$ and $x=4\text{.}$

In the sketch above, closed dots are extrema, and open dots are inflection points.

• $\mathbf{F(x),\,x \lt 0}\text{:}$ Now we can consider the left half of the graph. If you stare at it long enough, you might convince yourself that $F(x)$ is an odd function. We can also show this with the following calculation:

\begin{align*} F(-x)&=\int_0^{-x} f(t)\dee{t} &\text{As in Example }\knowl{./knowl/eg_lefthalfevenfunction.html}{\text{1.2.10}}\text{, since $f(t)$ is even,}\\ &=\int_x^0 f(t)\dee{t}=-\int_0^x f(t)\dee{t}\\ &=-F(x) \end{align*}

Knowing that $F(x)$ is odd allows us to finish our sketch.

Exercise49(*)
Hint

In general, the equation of the tangent line to the graph of $y=f(x)$ at $x=a$ is $y=f(a) + f'(a)\,(x-a)\text{.}$

(a) $3x^2 \displaystyle\int_{0}^{x^3+1} e^{t^3} \dee{t} + 3x^5 e^{(x^3+1)^3}$ \qquad (b) $y = -3(x+1)$

Solution

(a) Using the product rule, followed by the chain rule, followed by the Fundamental Theorem of Calculus Part 1,

\begin{align*} f'(x) & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\diff{}{x}\int_{0}^{x^3+1} e^{t^3} \dee{t} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\ \big[3x^2\big] \Bigg[\frac{d\hfill}{dy}\int_{0}^{y} e^{t^3} \dee{t}\Bigg]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\ \big[3x^2\big] \Big[e^{y^3}\Big]_{y=x^3+1} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + x^3\ \big[3x^2\big] e^{(x^3+1)^3} \\ & = 3x^2 \int_{0}^{x^3+1} e^{t^3} \dee{t} + 3x^5 e^{(x^3+1)^3} \end{align*}

(b) In general, the equation of the tangent line to the graph of $y=f(x)$ at $x=a$ is

\begin{equation*} y=f(a) + f'(a)\,(x-a) \end{equation*}

Substituting in the given $f(x)$ and $a=-1\text{:}$

\begin{align*} f(a)=f(-1)&=(-1)^3\int_0^0 e^{t^3}\dee{t}=0\\ f'(a)=f'(-1)&=3(-1)^2\int_{0}^{0}e^{t^3}\dee{t} + 3(-1)^5e^{0}\\ &=0-3=-3\\ (x-a)=x-(-1)&=x+1\\ \end{align*}

So, the equation of the tangent line is

\begin{align*} y = -3(x+1)\ . \end{align*}
Exercise50
Hint

Recall $\tan^2x+1=\sec^2 x\text{.}$

Both students.

Solution

Recall that “$+C$” means that we can add any constant to the function. Since $\tan^2 x = \sec^2 x - 1\text{,}$ Students A and B have equivalent answers: they only differ by a constant.

So, if one is right, both are right; if one is wrong, both are wrong. We check Student A's work:

\begin{equation*} \diff{}{x}\{\tan^2 x + x + C\}=\diff{}{x}\{\tan^2 x\} +1 + 0 = f(x)-1+1=f(x) \end{equation*}

So, Student A's answer is indeed an anditerivative of $f(x)\text{.}$ Therefore, both students ended up with the correct answer.

Remark: it is a frequent occurrence that equivalent answers might look quite different. As you are comparing your work to others', this is a good thing to keep in mind!

Exercise51
Hint

Since the integration is with respect to $t\text{,}$ the $x^3$ term can be moved outside the integral.

(a) $27(1-\cos 3 )$ \qquad (b) $x^3\sin (x) + 3x^2[1-\cos (x)]$

Solution

1. When $x=3\text{,}$ \begin{align*} F(3)&=\displaystyle\int_0^3 3^3 \sin(t)\dee{t}=27\int_0^3 \sin t \dee{t}\\ \end{align*}

Using the Fundamental Theorem of Calculus Part 2,

\begin{align*} &=27\left[-\cos t\right]_{t=0}^{t=3} =27\left[-\cos 3 - (-\cos 0)\right]\\ &=27(1-\cos 3 ) \end{align*}
2. Since the integration is with respect to $t\text{,}$ the $x^3$ term can be moved outside the integral. That is: for the purposes of the integral, $x^3$ is a constant (although for the purposes of the derivative, it certainly is not). \begin{align*} F(x)&=\displaystyle\int_0^x x^3 \sin(t)\dee{t} = x^3 \int_0^x \sin(t)\dee{t}\\ \end{align*}

Using the product rule and the Fundamental Theorem of Calculus Part 1,

\begin{align*} F'(x)&=x^3\cdot \sin(x) + 3x^2 \int_0^x \sin(t)\dee{t}\\ &=x^3\sin(x)+3x^2\left[-\cos(t)\right]_{t=0}^{t=x}\\ &=x^3\sin(x)+3x^2[-\cos(x)-(-\cos(0))]\\ &=x^3\sin (x) + 3x^2[1-\cos (x)] \end{align*}

Remark: Since $x$ and $t$ play different roles in our problem, it's crucial that they have different names. This is one reason why we should avoid the common mistake of writing $\int_a^x f(x)\dee{x}$ when we mean $\int_a^x f(t)\dee{t}\text{.}$

Exercise52
Hint

Remember that antiderivatives may have a constant term.

If $f(x)=0$ for all $x\text{,}$ then $F(x)$ is even and possibly also odd.

If $f(x) \neq 0$ for some $x\text{,}$ then $F(x)$ is not even. It might be odd, and it might be neither even nor odd.

(Perhaps surprisingly, every antiderivative of an odd function is even.)

Solution

If $F(x)$ is even, then $f(x)$ is odd (by the result of Question 1.2.3.20 in Section 1.2). So, $F(x)$ can only be even if $f(x)$ is both even and odd. By the result in Question 1.2.3.19, Section 1.2, this means $F(x)$ is only even if $f(x)=0$ for all $x\text{.}$ Note if $f(x)=0\text{,}$ then $F(x)$ is a constant function. So, it is certainly even, and it might be odd as well if $F(x)=f(x)=0\text{.}$

Therefore, if $f(x) \neq 0$ for some $x\text{,}$ then $F(x)$ is not even. It could be odd, or it could be neither even nor odd. We can come up with examples of both types: if $f(x)=1\text{,}$ then $F(x)=x$ is an odd antiderivative, and $F(x)=x+1$ is an antiderivative that is neither even nor odd.

Interestingly, the antiderivative of an odd function is always even. The proof is a little beyond what we might ask you, but is given below for completeness. The proof goes like this: First, we'll show that if $g(x)$ is odd, then there is some antiderivative of $g(x)$ that is even. Then, we'll show that every antiderivative of $g(x)$ is even.

So, suppose $g(x)$ is odd and define $G(x)=\displaystyle\int_0^x g(t)dt\text{.}$ By the Fundamental Theorem of Calculus Part 1, $G'(x)=g(x)\text{,}$ so $G(x)$ is an antiderivative of $g(x)\text{.}$ Since $g(x)$ is odd, for any $x\ge 0\text{,}$ the net signed area under the curve along $[0,x]$ is the negative of the net signed area under the curve along $[-x,0]\text{.}$ So,

\begin{align*} \int_0^x g(t)\dee{}t &= - \int_{-x}^0 g(t)\dee{t}&\quad(\mbox{See Example }\knowl{./knowl/eg_areaunderoddfunction.html}{\text{1.2.11}})\\ &=\int_0^{-x} g(t)\dee{t}\\ \end{align*}

By the definition of $G(x)\text{,}$

\begin{align*} G(x)&=G(-x) \end{align*}

That is, $G(x)$ is even. We've shown that there exists some antiderivative of $g(x)$ that is even; it remains to show that all of them are even.

Recall that every antiderivative of $g(x)$ differs from $G(x)$ by some constant. So, any antiderivative of $g(x)$ can be written as $G(x)+C\text{,}$ and $G(-x)+C = G(x)+C\text{.}$ So, every antiderivative of an odd function is even.

Exercises1.4.2Exercises

Exercise1
Hint

One is true, the other false.

(a) true (b) false

Solution

(a) This is true: it is an application of Theorem 1.4.2 with $f(x)=\sin x$ and $u(x)=e^x\text{.}$

(b) This is false: the upper limit of integration is incorrect. Using Theorem 1.4.6, the correct form is

\begin{equation*} \displaystyle\int_0^1 \sin(e^x)\cdot e^x\dee{x} = \displaystyle\int_1^{e} \sin(u)\dee u = -\cos(e) + \cos(1) = \cos(1)-\cos(e). \end{equation*}

Alternately, we can use the Fundamental Theorem of Calculus Part 2, and our answer from (a):

\begin{equation*} \int_0^1 \sin(e^x)\cdot e^x\dee{x}=\left[-\cos(e^x)+C\right]_{0}^1 = \cos(1)-\cos(e)\ . \end{equation*}
Exercise2
Hint

You can check whether the final answer is correct by differentiating.

The reasoning is not sound: when we do a substitution, we need to take care of the differential ($\dee{x}$). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here's the way to do it:

Problem: Evaluate $\displaystyle\int (2x+1)^2 \dee{x}\text{.}$ Work: We use the substitution $u=2x+1\text{.}$ Then $\dee{u}=2\dee{x}\text{,}$ so $\dee{x} = \frac{1}{2}\dee{u}\text{:}$

\begin{align*} \int (2x+1)^2 \dee{x}&=\int u^2\cdot \frac{1}{2}\dee{u}\\ &=\frac{1}{6}u^3+C\\ &=\frac{1}{6}\left(2x+1\right)^3+C \end{align*}
Solution

The reasoning is not sound: when we do a substitution, we need to take care of the differential ($\dee{x}$). Remember the method of substitution comes from the chain rule: there should be a function and its derivative. Here's the way to do it:

Problem: Evaluate $\displaystyle\int (2x+1)^2 \dee{x}\text{.}$ Work: We use the substitution $u=2x+1\text{.}$ Then $\dee{u}=2\dee{x}\text{,}$ so $\dee{x} = \frac{1}{2}\dee{u}\text{:}$

\begin{align*} \int (2x+1)^2 \dee{x}&=\int u^2\cdot \frac{1}{2}\dee{u}\\ &=\frac{1}{6}u^3+C\\ &=\frac{1}{6}\left(2x+1\right)^3+C \end{align*}
Exercise3
Hint

Check the limits.

The problem is with the limits of integration, as in Question 1. Here's how it ought to go:

Problem: Evaluate $\displaystyle\int_{1}^{\pi} \dfrac{\cos(\log t)}{t}\dee{t}\text{.}$ Work: We use the substitution $u=\log t\text{,}$ so $\dee{u}=\frac{1}{t}\dee{t}\text{.}$ When $t=1\text{,}$ we have $u=\log 1 =0$ and when $t=\pi\text{,}$ we have $u=\log(\pi)\text{.}$ Then:

\begin{align*} \int_{1}^{\pi} \dfrac{\cos(\log t)}{t}\dee{t}&=\int_{\log 1}^{\log(\pi)}\cos(u) \dee{u}\\ &=\int_{0}^{\log(\pi)}\cos(u) \dee{u}\\ &=\sin(\log(\pi))-\sin(0)=\sin(\log(\pi)) . \end{align*}
Solution

The problem is with the limits of integration, as in Question 1. Here's how it ought to go:

Problem: Evaluate $\displaystyle\int_{1}^{\pi} \dfrac{\log(\log t)}{t}\dee{t}\text{.}$ Work: We use the substitution $u=\log t\text{,}$ so $\dee{u}=\frac{1}{t}\dee{t}\text{.}$ When $t=1\text{,}$ we have $u=\log 1 =0$ and when $t=\pi\text{,}$ we have $u=\log(\pi)\text{.}$ Then:

\begin{align*} \int_{1}^{\pi} \dfrac{\cos(\log t)}{t}\dee{t}&=\int_{\log 1}^{\log(\pi)}\cos(u) \dee{u}\\ &=\int_{0}^{\log(\pi)}\cos(u) \dee{u}\\ &=\sin(\log(\pi))-\sin(0)=\sin(\log(\pi)) . \end{align*}
Exercise4
Hint

Check every step. Do they all make sense?

This one is OK.

Solution

Perhaps shorter ways exist, but the reasoning here is valid.

Problem: Evaluate $\displaystyle\int_{0}^{\pi/4} x\tan (x^2) \dee{x}\text{.}$ Work: We begin with the substitution $u=x^2\text{,}$ $\dee{u} = 2x\dee{x}\text{:}$ If $u=x^2\text{,}$ then $\diff{u}{x} = 2x\text{,}$ so indeed $\dee{u}=2x\dee{x}\text{.}$

\begin{align*} \int_{0}^{\pi/4} x\tan (x^2) \dee{x}&= \int_{0}^{\pi/4} \frac{1}{2}\tan(x^2)\cdot 2x\dee{x}&\color{red}{\text{algebra}}\\ &=\int_{0}^{\pi^2/16} \frac{1}{2}\tan u\dee{u}\\ \end{align*}

Every piece is changed from $x$ to $u\text{:}$ integrand, differential, limits.

\begin{align*} &=\frac{1}{2}\int_{0}^{\pi^2/16} \dfrac{\sin u}{\cos u}\dee{u} &\color{red}{\tan u = \frac{\sin u}{\cos u}}\\ \end{align*}

Now we use the substitution $v=\cos u\text{,}$ $\dee{v}=-\sin u \dee{u}\text{:}$

\begin{align*} &=\frac{1}{2}\int_{\cos 0}^{\cos(\pi^2/16)} -\dfrac{1}{v}\dee{v}\\ \end{align*}

Every piece is changed from $u$ to $v\text{:}$ integrand, differential, limits.

\begin{align*} &=-\frac{1}{2}\int_{1}^{\cos(\pi^2/16)} \dfrac{1}{v}\dee{v}&\color{red}{ \cos(0)=1}\\ &=-\frac{1}{2}\bigg[\log|v|\bigg]_{1}^{\pi^2/16}&\color{red}{\text{FTC Part 2}}\\ &=-\frac{1}{2}\left(\log\left(\frac{\pi^2}{16}\right)-\log(1)\right)\\ &=-\frac{1}{2}\cdot2\log\left(\frac{\pi}{4}\right)&\color{red}{ \log(a^b)=b\log a}\\ &=-\log\left(\frac{\pi}{4}\right)&\color{red}{\text{algebra}}\\ &=\log\left(\frac{4}{\pi}\right)&\color{red}{\log(a^b)=b\log a} \end{align*}
Exercise5(*)

$\displaystyle\int_{0}^{1} \frac{f(u)}{\sqrt{1-u^2}}\,\dee{u}$

Solution

We substitute:

\begin{align*} u&=\sin x,\\ \dee{u}&=\cos x\,\dee{x},\\ \cos x &= \sqrt{1-\sin^2 x}=\sqrt{1-u^2},\\ \dee{x} &=\dfrac{\dee{u}}{\cos x} = \dfrac{\dee{u}}{\sqrt{1-u^2}}\\ u(0)&=\sin 0 = 0\\ u\left(\frac{\pi}{2}\right)&=\sin\left(\frac{\pi}{2}\right)=1\\ \end{align*}

So,

\begin{align*} \int_{x=0}^{x=\pi/2} f(\sin x)\,\dee{x} &= \int_{u=0}^{u=1} f(u)\,\frac{\dee{u}}{\sqrt{1-u^2}} \end{align*}
Exercise6
Hint

What is $\diff{}{x}\{f(g(x))\}\text{?}$

some constant $C$

Solution

Using the chain rule, we see that

\begin{equation*} \diff{}{x}\{f(g(x))\}=f'(g(x))g'(x) \end{equation*}

So, $\textcolor{red}{f(g(x))}$ is an antiderivative of $\textcolor{red}{f'(g(x))g'(x)}\text{.}$ All antiderivatives of $f'(g(x))g'(x)$ differ by only a constant, so:

\begin{align*} \textcolor{red}{\int f'(g(x))g'(x)\dee{x}} - f(g(x))&=\textcolor{red}{f(g(x))+C}-f(g(x))\\ &=C \end{align*}

That is, our expression simplifies to some constant $C\text{.}$

Remark: since

\begin{equation*} \int f'(g(x))g'(x)\dee{t} - f(g(x))=C \end{equation*}

we conclude

\begin{equation*} \int f'(g(x))g'(x)\dee{t} = f(g(x))+C \end{equation*}

which is precisely how we perform substitution on integrals.

Exercise7(*)
Hint

What is the derivative of the argument of the cosine?

$\dfrac{1}{2}\big( \sin(e) - \sin(1) \big)$

Solution

We write $\textcolor{red}{u(x) = e^{x^2}}$ and find $\textcolor{blue}{\dee{u} = u'(x)\,\dee{x}=2x e^{x^2}\dee{x}}\text{.}$ Note that $u(1)=e^{1^2}=e$ when $x=1\text{,}$ and $u(0)=e^{0^2}=1$ when $x=0\text{.}$ Therefore:

\begin{align*} \int_{0}^{1} \textcolor{blue}{x e^{x^2}} \cos (\textcolor{red}{e^{x^2}}) \,\textcolor{blue}{\dee{x}} &= \textcolor{blue}{\frac{1}{2}}\int_{x=0}^{x=1} \cos (\textcolor{red}{u(x)}) \textcolor{blue}{u'(x)\,\dee{x} }\\ &=\textcolor{blue}{ \frac{1}{2}}\int_{u=1}^{u=e} \cos(\textcolor{red}u)\,\color{blue}{\dee{u}}\\ &= \frac{1}{2}\bigg[\sin(u) \bigg]_1^e = \frac{1}{2}\big( \sin(e) - \sin(1) \big). \end{align*}
Exercise8(*)
Hint

What is the title of the current section?

$\dfrac{1}{3}$

Solution

Substituting $\textcolor{red}{y=x^3}\text{,}$ $\textcolor{blue}{\dee{y}=3x^2\dee{x}}$\ :

\begin{gather*} \int_1^2 \textcolor{blue}{x^2} f(\textcolor{red}{x^3})\,\textcolor{blue}{\dee{x}} =\textcolor{blue}{\frac{1}{3}}\int_1^8 f(\textcolor{red}{y})\,\textcolor{blue}{\dee{y}} =\frac{1}{3} \end{gather*}
Exercise9(*)
Hint

What is the derivative of $x^3+1\text{?}$

$-\dfrac{1}{300{(x^3+1)}^{100}} + C$

Solution

Setting $\textcolor{red}{u=x^3+1}\text{,}$ we have $\textcolor{blue}{\dee{u} = 3x^2\,\dee{x}}$ and so

\begin{align*} \int \frac{\textcolor{blue}{x^2\,\dee{x}}}{{(\textcolor{red}{x^3+1})}^{101}} &= \int \frac{\textcolor{blue}{\dee{u}/3}}{\textcolor{red}{u}^{101}}\\ &=\frac{1}{3}\int u^{-101}\dee{u}\\ &=\frac{1}{3}\cdot\dfrac{u^{-100}}{-100}\\ &= -\frac{1}{3\times 100 u^{100}} + C\\ &=-\frac{1}{300{(x^3+1)}^{100}} + C \end{align*}
Exercise10(*)
Hint

What is the derivative of $\log x\text{?}$

$\log 4$

Solution

Setting $\textcolor{red}{u=\log x}\text{,}$ we have $\textcolor{blue}{\dee{u} = \frac{1}{x}\,\dee{x}}$ and so

\begin{equation*} \int_{e}^{e^4} \frac{\textcolor{blue}{\dee{x}}}{\textcolor{blue}x\cdot\textcolor{red}{\log x}} = \int_{x=e}^{x=e^4} \frac1{\textcolor{red}{\log x}} \cdot \textcolor{blue}{\frac{1}{x}\,\dee{x} } = \int_{u=1}^{u=4} \frac{1}{\textcolor{red}u}\,\textcolor{blue}{ \dee{u}}, \end{equation*}

since $u=\log(e)=1$ when $x=e$ and $u=\log(e^4)=4$ when $x=e^4\text{.}$ Then, by the Fundamental Theorem of Calculus Part 2,

\begin{equation*} \int_{1}^{4} \frac{1}{u}\, \dee{u} = \Big[\log |u| \Big]_{1}^{4} = \log 4 - \log 1 = \log 4. \end{equation*}
Exercise11(*)
Hint

What is the derivative of $1+\sin x\text{?}$

$\log 2$

Solution

Setting $\textcolor{red}{u=1+\sin x}\text{,}$ we have $\textcolor{blue}{\dee{u} = \cos x\dee{x}}$ and so

\begin{equation*} \int_{0}^{\pi/2} \frac{\textcolor{blue}{\cos x}} {\textcolor{red}{1+\sin x}} \textcolor{blue}{\dee{x} } = \int_{x=0}^{x=\pi/2} \frac{1}{\textcolor{red}{1+\sin x}}\, \textcolor{blue}{\cos x \dee{x} } = \int_{u=1}^{u=2} \frac{\textcolor{blue}{\dee{u}}}{\textcolor{red}u} \end{equation*}

since $u=1+\sin 0=1$ when $x=0$ and $u=1+\sin(\pi/2)=2$ when $x=\pi/2\text{.}$ Then, by the Fundamental Theorem of Calculus Part 2,

\begin{equation*} \int_{u=1}^{u=2} \frac{\dee{u}}{u} = \Big[\log|u| \Big]_{1}^{2} = \log 2 \end{equation*}
Exercise12(*)
Hint

$\cos x$ is the derivative of what?

$\dfrac{4}{3}$

Solution

Setting $\textcolor{red}{u=\sin x}\text{,}$ we have $\textcolor{blue}{\dee{u} = \cos x \dee{x}}$ and so

\begin{equation*} \int_{0}^{\pi/2} \textcolor{blue}{\cos x} \cdot (1+\textcolor{red}{\sin}^2 \textcolor{red}x)\textcolor{blue}{\dee{x} } = \int_{x=0}^{x=\pi/2} (1+\textcolor{red}{\sin}^2 \textcolor{red}{x})\cdot \textcolor{blue}{\cos x \dee{x} } = \int_{u=0}^{u=1} (1+\textcolor{red}{u}^2) \,\textcolor{blue}{\dee{u}}, \end{equation*}

since $u=\sin 0=0$ when $x=0$ and $u=\sin(\pi/2)=1$ when $x=\pi/2\text{.}$ Then, by the Fundamental Theorem of Calculus Part 2,

\begin{equation*} \int_{0}^{1} (1+u^2) \,\dee{u} = \left[u+\frac{u^3}{3} \right]_{0}^{1} =\left(1+\frac{1}{3}\right) -0 = \frac{4}{3}. \end{equation*}
Exercise13(*)
Hint

What is the derivative of the exponent?

$e^6-1$

Solution

Substituting $\textcolor{red}{t=x^2-x}\text{,}$ $\textcolor{blue}{\dee{t} = (2x-1)\,\dee{x}}$ and noting that $t=0$ when $x=1$ and $t=6$ when $x=3\text{,}$

\begin{align*} \int_1^3 \textcolor{blue}{(2x-1)}e^{\textcolor{red}{x^2-x}}\textcolor{blue}{ \dee{x}} &= \int_0^6 e^{\textcolor{red}t}\ \textcolor{blue}{\dee{t} } =\big[e^t\big]_0^6 =e^6-1 \end{align*}
Exercise14(*)
Hint

What is the derivative of the argument of the square root?

$\dfrac{1}{3}(4-x^2)^{3/2}+C$

Solution

We use the substitution $\textcolor{red}{u=4-x^2}\text{,}$ for which $\textcolor{blue}{\dee{u}=-2x\,\dee{x}}$\,:

\begin{align*} \int \frac{x^2-4}{\sqrt{4-x^2}}\,x\,\dee{x} &=\int \frac{1}{2}\cdot\frac{\textcolor{red}{4-x^2}}{\sqrt{\textcolor{red}{4-x^2}}}\textcolor{blue}{ ({-}2x)\,\dee{x}}\\ &=\frac{1}{2} \int \frac{\textcolor{red}u}{\sqrt{\textcolor{red}u}}\,\textcolor{blue}{\dee{u} }\\ &=\frac{1}{2}\int \sqrt{u}\,\dee{u}\\ &=\frac{1}{2}\frac{u^{3/2}}{3/2}+C\\ &=\frac{1}{3}(4-x^2)^{3/2}+C \end{align*}
Exercise15
Hint

What is $\diff{}{x}\left\{\sqrt{\log x}\right\}\text{?}$

$e^{\sqrt{\log x}}+C$

Solution

• Solution 1: If we let $\textcolor{red}{u=\sqrt{\log x}}\text{,}$ then $\textcolor{blue}{\dee{u}=\dfrac{1}{2x\sqrt{\log x}}\dee{x}}\text{,}$ and: \begin{align*} \int \dfrac{e^{\textcolor{red}{\sqrt{\log x}}}}{\textcolor{blue}{2x\sqrt{\log x}}}\ \textcolor{blue}{\dee{x}}&=\int e^{\textcolor{red}u}\ \textcolor{blue}{\dee{u}}=e^u+C=e^{\sqrt{\log x}}+C \end{align*}
• Solution 2: In Solution 1, we made a pretty slick choice. We might have tried to work with something a little less convenient. For example, it's not unnatural to think that $\textcolor{red}{u=\log x}\text{,}$ $\textcolor{blue}{\dee{u}=\dfrac{1}{x}\dee{x}}$ would be a good choice. In that case: \begin{align*} \int \dfrac{e^{\sqrt{\textcolor{red}{\log x}}}}{2\textcolor{blue}{x}\sqrt{\textcolor{red}{\log x}}}\ \textcolor{blue}{\dee{x}}&= \int \frac{e^{\sqrt{\textcolor{red}u}}}{2\sqrt{\textcolor{red}u}}\textcolor{blue}{\dee{u}}\\ \end{align*}

Now, we should be able to see that $\textcolor{orange!40!black}{w=\sqrt{u}}\text{,}$ $\textcolor{purple}{\dee{w} = \dfrac{1}{2\sqrt{u}}\dee{u}}$ is a good choice:

\begin{align*} \int\frac{e^{\textcolor{orange!40!black}{\sqrt{u}}}}{\textcolor{purple}{2\sqrt{u}}}\ \textcolor{purple}{\dee{u}} &=\int e^{\textcolor{orange!40!black}{w}}\ \textcolor{purple}{\dee{w}}\\ &=e^{\sqrt{u}}+C\\ &=e^{\sqrt{\log x}}+C \end{align*}
Exercise16(*)
Hint

There is a short, slightly sneaky method — guess an antiderivative — and a really short, still-more-sneaky method.

$0$

Solution

• The slightly sneaky method: We note that $\displaystyle\diff{}{x} \left\{e^{x^2} \right\}= 2x\, e^{x^2}\text{,}$ so that $\dfrac{1}{2} e^{x^2}$ is a antiderivative for the integrand $x e^{x^2}\text{.}$ So \begin{gather*} \int_{-2}^2 xe^{x^2}\,dx = \bigg[\frac{1}{2}e^{x^2}\bigg]_{-2}^2 =\frac{1}{2}e^4-\frac{1}{2}e^4=0 \end{gather*}
• The really sneaky method: The integrand $f(x) = x e^{x^2}$ is an odd function (meaning that $f(-x)=-f(x)$). So by Theorem 1.2.12 every integral of the form $\int_{-a}^a x e^{x^2}\,\dee{x}$ is zero.
Exercise17(*)
Hint

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11.

$\dfrac{1}{2}[\cos 1-\cos 2]\approx0.478$

Solution

The given sum is of the form

\begin{gather*} \lim_{n\rightarrow\infty}\sum_{j=1}^n \frac{j}{n^2}\sin\Big(1+\frac{j^2}{n^2}\Big) =\lim_{n\rightarrow\infty}\sum_{j=1}^n f(x_j^*)\De x \end{gather*}

with $\De x=\frac{1}{n}\text{,}$ $x_j^*=\frac{j}{n}$ and $f(x)=x\sin(1+x^2)\text{.}$ Since $x_0^*=0$ and $x_n^*=1\text{,}$ the right hand side is the definition (using the right Riemann sum) of

\begin{align*} \int_0^1 f(x)\,\dee{x} &=\int_0^1 x\sin(1+x^2)\,\dee{x}\\ &=\frac{1}{2}\int_1^2 \sin(y)\,\dee{y}\qquad\text{with $y=1+x^2$, $\dee{y}=2x\,\dee{x}$}\\ &=\frac{1}{2}\Big[-\cos(y)\Big]_{y=1}^{y=2}\\ &=\frac{1}{2}[\cos 1-\cos 2] \end{align*}

Using a calculator, we see this is close to $0.478\text{.}$

Exercise18
Hint

If $w=u^2+1\text{,}$ then $u^2=w-1\text{.}$

$\dfrac{1}{2}-\dfrac{1}{2}\log 2$

Solution

Often, the denominator of a function is a good guess for the substitution. So, let's try setting $\textcolor{red}{w=u^2+1}\text{.}$ Then $\textcolor{blue}{\dee{w}=2u\dee{u}}\text{:}$

\begin{align*} \int_{0}^1 \dfrac{u^3}{u^2+1}\dee{u}&= \frac{1}{2}\int_{0}^1 \dfrac{u^2}{\textcolor{red}{u^2+1}}\ \textcolor{blue}{2u\dee{u}}\\ \end{align*}

The numerator now is $u^2\text{,}$ and looking at our substitution, we see $\textcolor{red}{u^2=w-1}\text{:}$

\begin{align*} &=\frac{1}{2}\int_{1}^2 \dfrac{\textcolor{red}{w-1}}{\textcolor{red}{w}}\ \textcolor{blue}{\dee{w}}\\ &=\frac{1}{2}\int_{1}^2 \left(1 - \frac{1}{w}\right)\dee{w}\\ &=\frac{1}{2}\left[w - \log|w|\right]_{w=1}^{w=2}\\ &=\frac{1}{2}\left(2-\log 2 - 1\right)=\frac{1}{2}-\frac{1}{2}\log 2 \end{align*}
Exercise19
Hint

Using a trigonometric identity, this is similar (though not identical) to $\int \tan \theta \cdot \sec^2 \theta\dee{\theta}\text{.}$

$\frac{1}{2}\tan^2\theta -\log|\sec \theta|+C$

Solution

The only thing we really have to work with is a tangent, so it's worth considering what would happen if we substituted $\textcolor{red}{u=\tan \theta}\text{.}$ Then $\textcolor{blue}{\dee{u}=\sec^2\theta\dee{\theta}}\text{.}$ This doesn't show up in the integrand as it's written, but we can try and bring it out by using the identity $\tan^2 = \sec^2 \theta - 1\text{:}$

\begin{align*} \int \tan^3 \theta\ \dee{\theta}&= \int \textcolor{red}{\tan \theta}\cdot \tan^2\ \theta \dee{\theta}\\ &= \int \textcolor{red}{\tan \theta}\cdot \left(\sec^2\ \theta-1\right) \dee{\theta}\\ &= \int \textcolor{red}{\tan \theta}\cdot \textcolor{blue}{\sec^2\ \theta \dee{\theta}} -\int \tan \theta \dee{\theta}\\ \end{align*}

In Example 1.4.17, we learned $\int \tan \theta \dee{\theta} = \log |\sec \theta|+C$

\begin{align*} &=\int \textcolor{red}{u}\ \textcolor{blue}{\dee{u}} -\log|\sec \theta|+C\\ &=\frac{1}{2}u^2 -\log|\sec \theta|+C\\ &=\frac{1}{2}\tan^2\theta -\log|\sec \theta|+C \end{align*}
Exercise20
Hint

If you multiply the top and the bottom by $e^x\text{,}$ what does this look like the antiderivative of?

$\arctan(e^x)+C$

Solution

At first glance, it's not clear what substitution to use. If we try the denominator, $u=e^x+e^{-x}\text{,}$ then $\dee{u}=(e^x-e^{-x})\dee{x}\text{,}$ but it's not clear how to make this work with our integral. So, we can try something else.

If we want to tidy things up, we might think to take $\textcolor{red}{u=e^x}$ as a substitution. Then $\textcolor{blue}{\dee{u}=e^x\dee{x}},$ so we need an $e^x$ in the numerator. That can be arranged.

\begin{align*} \int\dfrac{1}{e^x+e^{-x}}\cdot\left(\frac{e^x}{e^x}\right)\dee{x}&= \int\frac{\textcolor{blue}{e^x}}{\left(\textcolor{red}{e^{x}}\right)^2+1}\ \textcolor{blue}{\dee{x}}\\ &=\int \dfrac{1}{\textcolor{red}{u}^2+1}\ \textcolor{blue}{\dee{u}}\\ &=\arctan(u)+C\\ &=\arctan(e^x)+C \end{align*}
Exercise21
Hint

You know methods other than substitution to evaluate definite integrals.

$\dfrac{\pi}{4}-\dfrac{2}{3}$

Solution

We often like to take the “inside” function as our substitution, in this case $\textcolor{red}{u=1-x^2}\text{,}$ so $\textcolor{blue}{\dee{u}=-2x\dee{x}}\text{.}$ This takes care of part of the integral:

\begin{align*} \int_0^1 (1-2x)\sqrt{\textcolor{red}{1-x^2}}\dee{x}&= \int_0^1 \sqrt{1-x^2}\dee{x}+\int_0^1 \textcolor{blue}{(-2x)}\sqrt{\textcolor{red}{1-x^2}}\ \textcolor{blue}{\dee{x}}\\ \end{align*}

The left integral is tough to solve with substitution, but luckily we don't have to--it's the area of a quarter of a circle of radius 1.

\begin{align*} &=\frac{\pi}{4}+\int_1^0 \sqrt{\textcolor{red}{u}}\ \textcolor{blue}{\dee{u}}\\ &=\frac{\pi}{4}+\left[\frac{2}{3}u^{3/2}\right]_{u=1}^{u=0}\\ &=\frac{\pi}{4} + 0 - \frac{2}{3} = \frac{\pi}{4}-\frac{2}{3} \end{align*}
Exercise22
Hint

$\tan x = \dfrac{\sin x}{\cos x}$

$-\frac{1}{2}\left(\log (\cos x)\right)^2+C$

Solution

• Solution 1: We often find it useful to take “inside” functions as our substitutions, so let's try $\textcolor{red}{u=\cos x}\text{,}$ $\textcolor{blue}{\dee{u} = -\sin x\dee{x}}\text{.}$ In order to dig up a sine, we use the identity $\tan x = \dfrac{\sin x}{\cos x}$\ : \begin{align*} \int\tan x \cdot \log\left(\textcolor{red}{\cos x}\right) \dee{x}&= -\int\frac{\textcolor{blue}{-\sin x}}{\textcolor{red}{\cos x}} \cdot \log\left(\textcolor{red}{\cos x}\right) \textcolor{blue}{\dee{x}}\\ &=-\int\frac{1}{\textcolor{red}u}\log(\textcolor{red}u)\textcolor{blue}{\dee{u}}\\ \end{align*}

Now, it is convenient to let $\textcolor{orange!40!black}{w=\log u}\text{,}$ $\textcolor{purple}{\dee{w}=\frac{1}{u}\dee{u}}$\ :

\begin{align*} -\int\textcolor{purple}{\frac{1}{u}}\textcolor{orange!40!black}{\log(u)}\textcolor{purple}{\dee{u}} &=-\int \textcolor{orange!40!black}{w}\ \textcolor{purple}{\dee{w}}\\ &=-\frac{1}{2}w^2+C\\ &=-\frac{1}{2}\left(\log u\right)^2+C\\ &=-\frac{1}{2}\left(\log (\cos x)\right)^2+C \end{align*}
• Solution 2: We might guess that it's useful to have $\textcolor{red}{u=\log(\cos x)}\text{,}$ $\textcolor{blue}{\dee{u}=\dfrac{-\sin x}{\cos x}\dee{x} = -\tan x\dee{x}}\text{:}$ \begin{align*} \int\tan x \cdot \textcolor{red}{\log\left(\cos x\right)} \dee{x}&= -\int\textcolor{blue}{-\tan x} \cdot \textcolor{red}{\log\left(\cos x\right) }\textcolor{blue}{\dee{x}}\\ &=-\int \textcolor{red}{u}\ \textcolor{blue}{\dee{u}}\\ &=-\frac{1}{2}u^2+C\\ &=-\frac{1}{2}\left(\log(\cos x)\right)^2+C \end{align*}
Exercise23(*)
Hint

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11.

$\half\sin(1)$

Solution

The given sum is of the form

\begin{gather*} \lim_{n\rightarrow\infty}\sum_{j=1}^n \frac{j}{n^2}\cos\Big(\frac{j^2}{n^2}\Big) =\lim_{n\rightarrow\infty}\sum_{j=1}^n f(x_j^*)\De x \end{gather*}

with $\De x=\frac{1}{n}\text{,}$ $x_j^*=\frac{j}{n}$ and $f(x)=x\cos(x^2)\text{.}$ Since $x_0^*=0$ and $x_n^*=1\text{,}$ the right hand side is the definition (using the right Riemann sum) of

\begin{align*} \int_0^1 f(x)\,\dee{x} &=\int_0^1 \textcolor{blue}{x}\cos(\textcolor{red}{x^2})\,\textcolor{blue}{\dee{x}}\\ &=\textcolor{blue}{\frac{1}{2}}\int_0^1 \cos(\textcolor{red}{y})\,\textcolor{blue}{\dee{y} }\qquad\text{with } textcolor{red}{y=x^2}, \textcolor{blue}{\dee{y}=2x\,\dee{x}}\\ &=\frac{1}{2}\Big[\sin(y)\Big]_0^1\\ &=\frac{1}{2}\sin 1 \end{align*}
Exercise24(*)
Hint

Review the definition of the definite integral and in particular Definitions 1.1.9 and 1.1.11.

$\dfrac{1}{3}[2\sqrt{2}-1] \approx0.609$

Solution

The given sum is of the form

\begin{gather*} \lim_{n\rightarrow\infty}\sum_{j=1}^n \frac{j}{n^2}\sqrt{1+\frac{j^2}{n^2}} =\lim_{n\rightarrow\infty}\sum_{j=1}^n f(x_j^*)\De x \end{gather*}

with $\De x=\frac{1}{n}\text{,}$ $x_j^*=\frac{j}{n}$ and $f(x)=x\sqrt{1+x^2}\text{.}$ Since $x_0^*=0$ and $x_n^*=1\text{,}$ the right hand side is the definition (using the right Riemann sum) of

\begin{align*} \int_0^1 f(x)\,\dee{x} &=\int_0^1 \textcolor{blue}{x}\sqrt{\textcolor{red}{1+x^2}}\ \textcolor{blue}{\dee{x}}\\ &=\textcolor{blue}{\frac{1}{2}}\int_1^2 \sqrt{\textcolor{red}{y}}\ \textcolor{blue}{\dee{y} }\qquad\text{with } \textcolor{red}{y=1+x^2}, \textcolor{blue}{\dee{y}=2x\,\dee{x}}\\ &=\frac{1}{2}{\left[\frac{2}{3}y^{3/2}\right]}_{y=1}^{y=2}\\ & =\frac{1}{3}[2\sqrt{2}-1] \end{align*}

Using a calculator, we see this is approximately $0.609\text{.}$

Exercise25
Hint

Find the right Riemann sum for both definite integrals.

Using the definition of a definite integral with right Riemann sums:

\begin{align*} \color{red}{\int_a^b 2f(2x)\dee{x}}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x \cdot 2f(2(a+i\Delta x))&\Delta x = \frac{b-a}{n}\\ &=\lim_{n \to \infty}\sum_{i=1}^n \left(\frac{b-a}{n}\right)\cdot2 f\left(2\left(a+i\left(\frac{b-a}{n}\right)\right)\right)\\ &=\lim_{n \to \infty}\sum_{i=1}^n \left(\frac{2b-2a}{n}\right)\cdot f\left(2a+i\left(\frac{2b-2a}{n}\right)\right)\\ \color{blue}{\int_{2a}^{2b} f(x)\dee{x}}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x \cdot f(2a+i\Delta x)&\Delta x = \frac{2b-2a}{n}\\ &=\lim_{n \to \infty}\sum_{i=1}^n \left(\frac{2b-2a}{n}\right) \cdot f\left(2a+i \left(\frac{2b-2a}{n}\right)\right)\\ \end{align*}

Since the Riemann sums are exactly the same,

\begin{align*} \color{red}{\int_a^b 2f(2x)\dee{x}}&= \color{blue}{\int_{2a}^{2b} f(x)\dee{x}} \end{align*}
Solution

Using the definition of a definite integral with right Riemann sums:

\begin{align*} \color{red}{\int_a^b 2f(2x)\dee{x}}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x \cdot 2f(2(a+i\Delta x))&\Delta x = \frac{b-a}{n}\\ &=\lim_{n \to \infty}\sum_{i=1}^n \left(\frac{b-a}{n}\right)\cdot2 f\left(2\left(a+i\left(\frac{b-a}{n}\right)\right)\right)\\ &=\lim_{n \to \infty}\sum_{i=1}^n \left(\frac{2b-2a}{n}\right)\cdot f\left(2a+i\left(\frac{2b-2a}{n}\right)\right)\\ \color{blue}{\int_{2a}^{2b} f(x)\dee{x}}&=\lim_{n \to \infty}\sum_{i=1}^n \Delta x \cdot f(2a+i\Delta x)&\Delta x = \frac{2b-2a}{n}\\ &=\lim_{n \to \infty}\sum_{i=1}^n \left(\frac{2b-2a}{n}\right) \cdot f\left(2a+i \left(\frac{2b-2a}{n}\right)\right)\\ \end{align*}

Since the Riemann sums are exactly the same,

\begin{align*} \color{red}{\int_a^b 2f(2x)\dee{x}}&= \color{blue}{\int_{2a}^{2b} f(x)\dee{x}} \end{align*}

Looking at the Riemann sum in this way is instructive, because it is very clear why the two integrals should be equal (without using substitution). The rectangles in the first Riemann sum are half as wide, but twice as tall, as the rectangles in the second Riemann sum. So, the two Riemann sums have rectangles of the same area.

(Not every substitution corresponds to such a simple picture.)

Exercises1.5.2Exercises

Exercise1
Hint

When we say “area between,” we want positive area, not signed area.

Area between curves $\approx \frac{\pi}{4}\left(2+\sqrt{2}\right)$

Solution

The intervals of our rectangles are $[0,\frac{\pi}{4}]\text{,}$ $[\frac{\pi}{4},\frac{\pi}{2}]\text{,}$ $[\frac{\pi}{2},\frac{3\pi}{4}]\text{,}$ and $[\frac{3\pi}{4},\pi]\text{.}$ Since we're taking a left Riemann sum, we find the height of the rectangles at the left endpoints of the intervals.

• $x = 0\text{:}$ The distance from $\cos 0$ to $\sin 0$ is 1, so our first rectangle has height 1.
• $x = \frac{\pi}{4}\text{:}$ The distance from $\cos \frac{\pi}{4}$ to $\sin \frac{\pi}{4}$ is 0, so our second rectangle has height 0.
• $x = \frac{\pi}{2}\text{:}$ The distance from $\cos \frac{\pi}{2}$ to $\sin \frac{\pi}{2}$ is 1, so our third rectangle has height 1.
• $x = \frac{3\pi}{4}\text{:}$ The distance from $\cos \frac{3\pi}{4}$ to $\sin \frac{3\pi}{4}$ is $\sin(3\pi/4)-\cos(3\pi/4) =\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)=\sqrt{2} \text{,}$ so our fourth rectangle has height $\sqrt{2}\text{.}$

So, our approximation for the area between the two curves is

\begin{equation*} \frac{\pi}{4}\left(1+0+1+\sqrt{2}\right)=\frac{\pi}{4}\left(2+\sqrt{2}\right) \end{equation*}
Exercise2
Hint

We're taking rectangles that reach from one function to the other.

(a) Vertical rectangles:

Solution

1. We are finding the area in the interval from $x=0$ to $x=\frac{\pi}{2}\text{.}$ Since we're taking $n=5$ rectangles, our rectangles cover the following intervals:

2. We are finding the area in the interval from $y=0$ to $y=\frac{\pi}{2}\text{.}$ (In general, when we switch from horizontal rectangles to vertical, the limits of integration will change--it's only coincidence that they are the same in this example.) Since we're taking $n=5$ rectangles, these rectangles cover the following intervals of the $y$-axis:

The question doesn't specify which endpoints we're using. Let's use upper endpoints, to match part (a).

Exercise3(*)
Hint

Draw a sketch first.

$\displaystyle\int_0^{\sqrt{2}}\big[2x-x^3\big]\dee{x}$

Solution

The curves intersect when $y=x$ and $y=x^3-x\text{.}$ To find these points, we set:

For $x\ge 0\text{,}$ the curves intersect at $(0,0)$ and $(\sqrt{2},\sqrt{2})\text{.}$

A handy observation is that, since both curves are continuous and they do not meet each other between $x=0$ and $x=\sqrt{2}\text{,}$ we don't have to worry about dividing our area into two regions: one of the functions is always on the top, and the other is always on the bottom.

Using vertical strips:

The top and bottom boundaries of the specified region are $y=T(x)=x$ and $y=B(x)=x^3-x\text{,}$ respectively. So,

\begin{gather*} {\rm Area} = \int_0^{\sqrt{2}}\big[T(x)-B(x)\big]\dee{x} = \int_0^{\sqrt{2}}\big[x-(x^3-x)\big]\dee{x} = \int_0^{\sqrt{2}} 2x-x^3 \dee{x} \end{gather*}
Exercise4(*)
Hint

Draw a sketch first.

$\displaystyle \int_{-3/2}^{4}\left[\frac{4}{5}(6-y^2)+2y\right]\dee{y}$

Solution

We need to find where the curves intersect.

\begin{align*} \frac{x^2}{4}=y^2&=6-\dfrac{5x}{4}\\ \frac{1}{4}x^2+\frac{5}{4}x-6&=0\\ x^2+5x-24&=0\\ (x+8)(x-3)&=0\\ x=-8,\quad x&=3 \end{align*}

The curves intersect at $(-8,4)$ and $(3,-\frac{3}{2})\text{.}$ Using horizontal strips:

we have

\begin{gather*} \text{Area} = \int_{-3/2}^{4}\Big[\frac{4}{5}(6-y^2)+2y\Big]\dee{y} \end{gather*}
Exercise5(*)
Hint

You can probably find the intersections by inspection.

$\displaystyle\int_0^{4a}\left[\sqrt{4ax}-\frac{x^2}{4a}\right]\dee{x}$

Solution

If the curves intersect at $(x,y)\text{,}$ then

\begin{align*} \left(x^2\right)^2&=\left(4a\right)^2y^2 = (4a)^24ax\\ x^4&=(4a)^3 x\\ x^4&-(4a)^3x=0\\ x(&x^3-(4a)^3)=0\\ x&= 0 \quad\mbox{or}\quad x^3=(4a)^3 \end{align*}

The curves intersect at $(0,0)$ and $(4a,4a)\text{.}$ (It is also possible to find these points by inspection.) Using vertical strips:

We want the $y$-values of the functions. We write the top function as $y =\sqrt{4ax}$ (we care about the positive square root, not the negative one) and we write the bottom function as $y=\frac{x^2}{4a}\text{.}$ Then we have

\begin{gather*} \text{Area} = \int_0^{4a}\left[\sqrt{4ax}-\frac{x^2}{4a}\right]\dee{x} \end{gather*}
Exercise6(*)
Hint

To find the intersection, plug $x=4y^2$ into the equation $x+12y+5=0\text{.}$

$\displaystyle\int_1^{25}\left[-\frac{1}{12}(x+5)+\frac{1}{2}\sqrt{x}\right]\dee{x}$

Solution

The curves intersect when $x=4y^2$ and $0=4y^2+12y+5 =(2y+5)(2y+1)\text{.}$ So, the curves intersect at $(1,-\half)$ and $(25,-\frac{5}{2})\text{.}$ Using vertical strips:

we have

\begin{gather*} \text{Area} = \int_1^{25}\left[-\frac{1}{12}(x+5)+\frac{1}{2}\sqrt{x}\right]\dee{x} \end{gather*}
Exercise7(*)
Hint

If the bottom function is the $x$-axis, this is a familiar question.

$\dfrac{1}{8}$

Solution

The area between the curve $y= \frac{1}{(2x-4)^2}$ and the $x$-axis, with $x$ running from $a=0$ to $b=1\text{,}$ is exactly the definite integral of $\frac{1}{(2x+4)^2}$ with limits $0$ and $1\text{.}$

\begin{align*} \mbox{Area}&=\int_0^1 \frac{\dee{x}}{(2x-4)^2}&u=2x-4,\quad\dee{u}=2\dee{x}\\ &=\frac{1}{2}\int_{-4}^{-2}\frac{1}{u^2}\dee{u} = \frac{1}{2}\left[\frac{-1}{u}\right]_{u=-4}^{u=-2}\\ &=\frac{1}{2}\Big[\frac{1}{2}-\frac{1}{4}\Big] =\frac{1}{8} \end{align*}
Exercise8(*)
Hint

Part of the job is to determine whether $y=x$ lies above or below $y=3x-x^2\text{.}$

$\dfrac{4}{3}$

Solution

If the curves $y=f(x)=x$ and $y=g(x)=3x-x^2$ intersect at $(x,y)\text{,}$ then

Furthermore, $g(x)-f(x) = 2x-x^2 = x(2-x)$ is positive for all $0\le x\le 2\text{.}$ That is, the curve $y=3x-x^2$ lies above the line $y=x$ for all $0\le x\le 2\text{.}$

We therefore evaluate the integral:

\begin{gather*} \int_0^2 \big[ (3x-x^2) - x \big] \,\dee{x} = \int_0^2 [2x-x^2]\,\dee{x} = \bigg[x^2 - \frac{x^3}{3}\bigg]^{2}_{0} = \bigg[ 4-\frac{8}{3} \bigg] -0 = \frac{4}{3} \end{gather*}
Exercise9(*)
Hint

Guess the intersection points by trying small integers.

$\dfrac{5}{3}-\dfrac{1}{\log 2}$

Solution

By inspection, the two curves cross at $(0,1)$ and $(1,2)\text{.}$

To antidifferentiate $2^x\text{,}$ we write $2^x={(e^{\log 2})}^x=e^{x\log 2}\text{.}$

\begin{align*} \text{Area} &= \int_0^1\big[(\sqrt{x}+1)-e^{x\log 2}\big]\,\dee{x} =\left[\frac{2}{3}x^{3/2}+x-\frac{1}{\log 2} 2^x\right]_0^1 \\ &=\frac{2}{3}+1-\frac{1}{\log 2}[2-1] =\frac{5}{3}-\frac{1}{\log 2} \end{align*}
Exercise10(*)
Hint

Draw a sketch first. You can also exploit a symmetry of the region to simplify your solution.

$\dfrac{8}{\pi}-1$

Solution

Here is a sketch of the specified region.

Both functions are even, so the region is symmetric about the $y$--axis. So, we will compute the area of the part with $x\ge 0$ and multiply by $2\text{.}$ The curves $y=\sqrt{2} \cos(\pi x/4)$ and $y=x$ intersect when $x=\sqrt{2} \cos(\pi x/4)$ or $\cos(\pi x/4)=\frac{x}{\sqrt{2}}\text{,}$ which is the case  3 The solution $x=1$ was found by guessing. To guess a solution to $\cos(\pi x/4)=\frac{x}{\sqrt{2}}$ just ask yourself what simple angle has a cosine that involves $\sqrt{2}\text{.}$ This guessing strategy is essentially useless in the real world, but works great on problem sets and exams. when $x=1\text{.}$ So, using vertical strips as in the figure above, the area (including the multiplication by 2) is

\begin{equation*} 2\int_0^1 \big[\sqrt{2} \cos(\pi x/4) - x\big]\,\dee{x} = 2\bigg[\sqrt{2}\,\frac{4}{\pi} \sin(\pi x/4)-\frac{x^2}{2}\bigg]_0^1 = 2\bigg[\frac{4}{\pi}-\frac{1}{2}\bigg] = \frac{8}{\pi}-1 \end{equation*}
Exercise11(*)
Hint

Figure out where the two curves cross. To determine which curve is above the other, try evaluating $f(x)$ and $g(x)$ for some simple value of $x\text{.}$ Alternatively, consider $x$ very close to zero.

$\dfrac{20}{9}$

Solution

For our computation, we will need an antiderivative of $x^2\sqrt{x^3+1}\text{,}$ which can be found using the substitution $u=x^3+1\text{,}$ $\dee{u} = 3x^2\,\dee{x}\text{:}$

\begin{gather*} \int x^2\sqrt{x^3+1} \, \dee{x} = \int \sqrt u \cdot \frac13\,\dee{u} = \frac13\int u^{1/2}\,\dee{u} = \frac13\cdot \frac{u^{3/2}}{3/2}+C = \frac29(x^3+1)^{3/2} + C. \end{gather*}

The two functions $f(x)$ and $g(x)$ are clearly equal at $x=0\text{.}$ If $x\ne0\text{,}$ then the functions are equal when

\begin{align*} 3x^2 &= x^2\sqrt{x^3+1} \\ 3 &= \sqrt{x^3+1} \\ 9 &= x^3+1 \\ 8 &= x^3 \\ 2 &= x. \end{align*}

The function $g(x)=3x^2$ is the larger of the two on the interval $[0,2]\text{,}$ as can be seen by plugging in $x=1\text{,}$ say, or by observing that when $x$ is very small $f(x)=x^2\sqrt{x^3+1}\approx x^2$ and $g(x)=3x^2\text{.}$

The area in question is therefore:

\begin{align*} \int_0^2 \big( 3x^2 - x^2\sqrt{x^3+1} \big) \, \dee{x} &= \bigg( {x^3} - \frac29(x^3+1)^{3/2} \bigg) \bigg|_0^2 \\ &= \bigg( 2^3 - \frac2 9(2^3+1)^{3/2} \bigg) - \bigg( 0^3 - \frac29(0^3+1)^{3/2} \bigg) \\ &= \bigg( 8 - 6 \bigg) - \bigg( 0 - \frac 2 9 \bigg) =\frac{20}9. \end{align*}
Exercise12(*)
Hint

Think about whether it will easier to use vertical strips or horizontal strips.

$\dfrac{1}{6}$

Solution

First, let's figure out what our curve $x=y^2+y=y(y+1)$ looks like.

• The curve intercepts the $y$-axis when $y=0$ and $y=-1\text{.}$
• The $x$-values of the curve are negative when $-1 \lt y \lt 0\text{,}$ and positive elsewhere.

This leads to the figure below. We're evaluating the area from $y=-1$ to $y=0\text{.}$ Since $y^2+y$ is negative there, the length of our (horizontal) slices are $0-(y^2+y)\text{.}$

\begin{gather*} \text{Area}=\int_{-1}^0\big(0-(y^2+y)\big)\,\dee{y} = -\bigg[\frac{y^3}{3}+\frac{y^2}{2}\bigg]_{-1}^0 =-\frac13+\frac12 =\frac{1}{6} \end{gather*}
Exercise13
Hint

Writing an integral for this is nasty. How can you avoid it?

$2\pi$

Solution

Let's begin by sketching our region. Note that $y=\sqrt{1-x^2}$ and $y=\sqrt{9-x^2}$ are the top halves of circles centred at the origin with radii 1 and 3, respectively.

Our region is the difference of two quarter-circles, so we find its area using geometry:

\begin{equation*} \mbox{Area}=\frac{1}{4}\left(\pi\cdot 3^2\right)-\frac{1}{4}\left(\pi\cdot 1^2\right)=2\pi \end{equation*}
Exercise14(*)
Hint

You are asked for the area, not the signed area. Be very careful about signs.

$2\Big[\pi-\frac{1}{4}\pi^2\Big]$

Solution

We will compute the area by using thin vertical strips, as in the sketch below:

By looking at the sketch above, we guess the line $y = 4 + 2\pi - 2x$ intersects the curve $y = 4 + \pi \sin x$ when $x=\frac{\pi}{2},$ $x=\pi\text{,}$ and $x=\frac{3\pi}{2}\text{.}$ Let's make sure these are correct by plugging them into the two equations, and making sure the $y$-values match:

 $x$ $4+2\pi-2x$ $4+\pi\sin(x)$ match? $\frac{\pi}{2}$ $4+\pi$ $4+\pi$ \checkmark $\pi$ $4$ $4$ \checkmark $\frac{3\pi}{2}$ $4-\pi$ $4-\pi$ \checkmark

Also from the sketch, we see that:

• When $\frac{\pi}{2} \le x \le \pi\text{,}$ the top of the strip is at $y = 4 + \pi \sin x$ and the bottom of the strip is at $y = 4 + 2\pi - 2x\text{.}$ So the strip has height $\big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]$ and width $\dee{x}\text{,}$ and hence area $\big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]\dee{x}\text{.}$
• When $\pi \le x \le \frac{3\pi}{2}\text{,}$ the top of the strip is at $y = 4 + 2\pi - 2x$ and the bottom of the strip is at $y = 4 + \pi \sin x\text{.}$ So the strip has height $\big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]$ and width $\dee{x}\text{,}$ and hence area $\big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]\dee{x}\text{.}$

Now we can calculate:

\begin{align*} \hbox{Area} &= \int_{\pi/2}^\pi \big[(4 + \pi \sin x)-(4 + 2\pi - 2x)\big]\dee{x} +\int^{3\pi/2}_\pi \big[(4 + 2\pi - 2x)-(4 + \pi \sin x)\big]\dee{x}\\ &= \int_{\pi/2}^\pi \big[\pi \sin x- 2\pi + 2x\big]\dee{x} +\int^{3\pi/2}_\pi \big[2\pi - 2x- \pi \sin x\big]\dee{x}\\ &=\Big[-\pi \cos x- 2\pi x + x^2\Big]_{\pi/2}^\pi +\Big[2\pi x - x^2+ \pi \cos x\Big]^{3\pi/2}_\pi\\ &=\left[\pi-\pi^2+\frac{3}{4}\pi^2\right] +\left[\pi^2-\frac{5}{4}\pi^2+\pi\right]\\ &=2\Big[\pi-\frac{1}{4}\pi^2\Big] \end{align*}
Exercise15(*)
Hint

You are asked for the area, not the signed area. Draw a sketch of the region and be very careful about signs.

$\dfrac{31}{6}$

Solution

First, here is a sketch of the region. We are not asked for it, but it is crucial for understanding the question.

The two curves $y=x+2$ and $y=x^2$ cross at $(2,4)\text{.}$ The area of the part between them with $0\le x\le 2$ is:

\begin{gather*} \int_0^2 \big[x+2-x^2\big]\,\dee{x}=\Big[\frac{1}{2} x^2+2x-\frac{1}{3}x^3\Big]_0^2 =2+4-\frac{8}{3}=\frac{10}{3} \end{gather*}

The area of the part between the two curves with $2\le x\le 3$ is:

\begin{gather*} \int_2^3 \big[x^2-(x+2)\big]\,\dee{x}=\Big[\frac{1}{3}x^3-\frac{1}{2} x^2-2x\Big]_2^3 =9-\frac{9}{2}-6-\frac{8}{3}+2+4=\frac{11}{6} \end{gather*}

The total area is $\dfrac{10}{3}+\dfrac{11}{6}=\dfrac{31}{6}\text{.}$

Exercise16(*)
Hint

You have to determine whether

• the curve $y = f(x) = x \sqrt{25-x^2}$ lies above the line $y=g(x)=3x$ for all $0\le x\le 4$ or
• the curve $y = f(x)$ lies below the line $y=g(x)$ for all $0\le x\le 4$ or
• $y=f(x)$ and $y=g(x)$ cross somewhere between $x=0$ and $x=4\text{.}$

One way to do so is to study the sign of $f(x)-g(x) = x\big(\sqrt{25-x^2}-3\big)\text{.}$

$\dfrac{26}{3}$

Solution

We need to figure out which curve is on top, when. To do this, set $h(x) = 3x - x\sqrt{25-x^2}\text{.}$ If $h(x) \gt 0\text{,}$ then $y=3x$ is the top curve; if $h(x) \lt 0\text{,}$ then $y=x\sqrt{25-x^2}$ is the top curve.

\begin{align*} h(x) &= 3x - x\sqrt{25-x^2} = x\left[3-\sqrt{25-x^2}\right]\\ \end{align*}

We only care about values of $x$ in $[0,4]\text{,}$ so $x$ is nonnegative. Then $h(x)$ is positive when:

\begin{align*} 3& \gt \sqrt{25-x^2}\\ 9& \gt 25-x^2\\ x^2 & \gt 16\\ x& \gt 4 \end{align*}

That is, $h(x)$ is never positive over the interval $[0,4]\text{.}$ So, $y = x \sqrt{25-x^2}$ lies above $y=3x$ for all $0\le x\le 4\text{.}$

The area we need to calculate is therefore:

\begin{align*} A &= \int_0^4 \left[x \sqrt{25-x^2} - 3x\right]\,\dee{x} \\ &= \int_0^4 x \sqrt{25-x^2}\,\dee{x} - \int_0^4 3x\,\dee{x} \\ &= A_1 - A_2. \end{align*}

To evaluate $A_1\text{,}$ we use the substitution $u(x) = 25-x^2\text{,}$ for which $\dee{u} = u'(x)\,\dee{x}= -2x\,\dee{x}\text{;}$ and $u(4)=25-4^2=9$ when $x=4\text{,}$ while $u(0)=25-0^2=25$ when $x=0\text{.}$ Therefore

\begin{align*} A_1 &= \int_{x=0}^{x=4} x \sqrt{25-x^2}\,\dee{x} = -\frac{1}{2} \int_{u=25}^{u=9} \sqrt{u}\,\dee{u} = \left[-\frac{1}{3} u^{3/2} \right]_{25}^{9} = \frac{125 - 27}{3} = \frac{98}{3} \end{align*}

For $A_2$ we use the antiderivative directly:

\begin{equation*} A_2 = \int_0^4 3x\,\dee{x} =\left[ \frac{3x^2}{2} \right]_0^4 = 24 \end{equation*}

Therefore the total area is:

\begin{gather*} A = \frac{98}{3} - 24 = \frac{26}{3} \end{gather*}
Exercise17
Hint

Flex those geometry muscles.

$\dfrac{7\pi}{8}-\dfrac{1}{2}$

Solution

Let's begin by sketching our region. Note that $y=\sqrt{9-x^2}$ is the top half of a circle centred at the origin with radius 3, while $y=\sqrt{1-(x-1)^2}$ is the top half of a circle of radius 1 centred at $(1,0)\text{.}$

Note $y=x$ intersects $y=\sqrt{1-(x-1)^2}$ at $(1,1)\text{,}$ the highest part of the smaller half-circle.

We can easily take the area of triangles and sectors of circles. With that in mind, we cut up our region the following way:

• The desired area is $A_3-(A_1+A_2)\text{.}$
• $A_1$ is the area of right a triangle with base 1 and height 1, so $A_1 = \frac{1}{2}\text{.}$
• $A_2$ is the area of a quarter circle of radius 1, so $A_2=\frac{\pi}{4}\text{.}$
• $A_3$ is the area of an eighth of a circle of radius 3, so $A_2 = \frac{9\pi}{8}$

So, the area of our region is   $\dfrac{9\pi}{8} - \dfrac{1}{2}-\dfrac{\pi}{4}=\dfrac{7\pi}{8}-\dfrac{1}{2}\text{.}$

Exercise18
Hint

These two functions have three points of intersection. This question is slightly messy, but uses the same concepts we've been practicing so far.

$12\sqrt{2}-\dfrac{13}{4}$

Solution

The first function is a cubic, with intercepts at $x=0,\pm2\text{.}$ The second is a straight line with a positive slope.

We need to figure out what these functions look like in relation to one another, so let's find their points of intersection.

\begin{align*} x(x^2-4)&=x-2\\ x(x+2)(x-2)&=x-2\\ \boxed{\color{blue}{x-2=0}} \quad\mbox{or}\quad x(x+2)&=1\\ x^2+2x-1&=0\\ x &= \dfrac{-2\pm\sqrt{4-4(1)(-1)}}{2}\\ x&=\boxed{\color{red}{-1\pm \sqrt{2}}} \end{align*}

So, our three points of intersection are when $\color{blue}{x=2}$ and when $\color{red}{x=-1\pm\sqrt{2}}\text{.}$ We note

\begin{equation*} \textcolor{red}{-1-\sqrt{2}} \lt \textcolor{red}{ -1+\sqrt{2} } \lt -1+\sqrt{4} \lt \textcolor{blue}{2}\ . \end{equation*}

So, we need to see which function is on top over the two intervals $\left[-1-\sqrt{2},-1+\sqrt{2}\right]$ and $\left[-1+\sqrt{2},2\right]\text{.}$ It suffices to check points in these intervals.

 $x$ $x(x^2-4)$ $x-2$ top function: 0 0 $-2$ $x(x^2-4)$ 1 -3 $-1$ $x-2$

Since 0 is in the interval $\left[-1-\sqrt{2},-1+\sqrt{2}\right]\text{,}$ $x(x^2-4)$ is the top function in that interval. Since 1 is in the interval $\left[-1+\sqrt{2},2\right]\text{,}$ $x-2$ is the top function in that interval. Now we can set up the integral to evaluate the area:

After some taxing but rudimentary algebra:

\begin{align*} &=\left(8\sqrt{2}\right)+\left(4\sqrt{2}-\frac{13}{4}\right)=12\sqrt{2}-\frac{13}{4} \end{align*}

Exercises1.6.2Exercises

Exercise1
Hint

The horizontal cross-sections were discussed in Example 1.6.1.

The horizontal cross-sections are circles, but the vertical cross-sections are not.

Solution

If we take a horizontal slice of a cone, we get a circle. If we take a vertical cross-section, the base is flat (it's a chord on the circular base of the cone), so we know right away it isn't a circle. Indeed, if we slice down through the very centre, we get a triangle. (Other vertical slices have a curvy top, corresponding to a class of curves known as hyperbolas.)

Exercise2
Hint

What are the dimensions of the cross-sections?

The columns have the same volume.

Solution

The columns have the same volume. We can see this by chopping up the columns into horizontal cross-sections. Each cross-section has the same area as the cookie cutter, $A\text{,}$ and height $\dee{y}\text{.}$ Then in both cases, the volume of the column is

\begin{equation*} \int_{0}^h A \dee{y} = hA \mbox{ cubic units} \end{equation*}
Exercise3
Hint

There are two different kinds of washers.

• Washers when $\mathbf{1 \lt y \le 6}\text{:}$ If $y \gt 1\text{,}$ then our washer has inner radius $2+\frac{2}{3}y\text{,}$ outer radius $6-\frac{2}{3}y\text{,}$ and height $\dee{y}\text{.}$

• Washers when $\mathbf{0\le y \lt 1}\text{:}$ When $0 \le y \lt 1\text{,}$ we have a “double washer,” two concentric rings. The inner washer has inner radius $r_1=y$ and outer radius $R_1=2-y\text{.}$ The outer washer has inner radius $r_2=2+\frac{2}{3}y$ and outer radius $R_2=6-\frac{2}{3}y\text{.}$ The thickness of the washers is $\dee{y}\text{.}$

Solution

Notice $f(x)$ is a piecewise linear function, so we can find explicit equations for each of its pieces from the graph. The radii will be determined by the $x$-values, so below we give the $x$-values as functions of $y\text{.}$

If we imagine rotating the region from the picture about the $y$-axis, there will be two kinds of washers formed: when $y \lt 1\text{,}$ we have a “double washer,” two concentric rings. When $y \gt 1\text{,}$ we have a single ring.

• Washers when $\mathbf{1 \lt y \le 6}\text{:}$ If $y \gt 1\text{,}$ then our washer has inner radius $2+\frac{2}{3}y\text{,}$ outer radius $6-\frac{2}{3}y\text{,}$ and height $\dee{y}\text{.}$

• Washers when $\mathbf{0\le y \lt 1}\text{:}$ When $0 \le y \lt 1\text{,}$ we have a “double washer,” two concentric rings corresponding to the two “humps” in the function. The inner washer has inner radius $r_1=y$ and outer radius $R_1=2-y\text{.}$ The outer washer has inner radius $r_2=2+\frac{2}{3}y$ and outer radius $R_2=6-\frac{2}{3}y\text{.}$ The thickness of the washers is $\dee{y}\text{.}$

Exercise4(*)
Hint

Draw sketches. The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which is in the optional section 1.6. The method of washers also works, but requires you to have more patience and also to have a good idea what the specified region looks like. Look at your sketch very careful when identifying the ends of your horizontal strips.

(a) $\pi\displaystyle\int_{0}^{3} xe^{2x^2}\dee{x}$

(b) $\displaystyle\int _0^1 \pi\big[\big(3+\sqrt{y}\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} +\displaystyle\int _ 1^4 \pi\big[\big(5-y\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y}$

Solution

(a) When the strip shown in the figure

is rotated about the $x$--axis, it forms a thin disk of radius $\sqrt{x}e^{x^2}$ and thickness $\dee{x}$ and hence of cross sectional area $\pi xe^{2x^2}$ and volume $\pi xe^{2x^2}\,\dee{x}$ So the volume of the solid is

\begin{gather*} \pi\int_{0}^{3} xe^{2x^2}\dee{x} \end{gather*}

(b) The curves intersect at $(-1,1)$ and $(2,4)\text{.}$

We'll use horizontal washers as in Example 1.6.5.

• We use thin horizontal strips of width $\dee{y}$ as in the figure above.
• When we rotate about the line $x=3\text{,}$ each strip sweeps out a thin washer

• whose inner radius is $r_{in}=3-\sqrt{y}\text{,}$ and
• whose outer radius is $r_{out}=3-(y-2)=5-y$ when $y\ge 1$ (see the red strip in the figure on the right above), and whose outer radius is $r_{out}= 3-(-\sqrt{y})=3+\sqrt{y}$ when $y\le 1$ (see the blue strip in the figure on the right above) and
• whose thickness is $\dee{y}$ and hence
• whose volume is $\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big[\big(5-y\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y}$ when $y\ge 1$ and whose volume is $\pi(r_{out}^2 - r_{in}^2)\dee{y} =\pi\big[\big(3+\sqrt{y}\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y}$ when $y\le 1$ and
• As our bottommost strip is at $y=0$ and our topmost strip is at $y=4\text{,}$ the total volume is \begin{gather*} \int _0^1 \pi\big[\big(3+\sqrt{y}\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} +\int _ 1^4 \pi\big[\big(5-y\big)^2-\big(3-\sqrt{y}\big)^2\big]\dee{y} \end{gather*}
Exercise5(*)
Hint

Draw sketchs.

(a) $\displaystyle\int_{-1}^{1}\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]\,\dee{x}$ \qquad (b) $\displaystyle\int _{-1}^0 \pi\big[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\big]\,\dee{y}$

Solution

(a) The curves intersect at $(1,0)$ and $(-1,0)\text{.}$ When the strip shown in the figure

is rotated about the line $y=-1\text{,}$ it forms a thin washer with:

• inner radius $(1-x^2)-(-1)=2-x^2\text{,}$
• outer radius $(4-4x^2)-(-1)=5-4x^2$ and
• thickness $\dee{x}$\ ; so, it has
• cross sectional area $\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]$ and
• volume $\pi\big[{(5-4x^2)}^2-({2-x^2)}^2\big]\,\dee{x}\text{.}$

So the volume of the solid is

\begin{gather*} \int_{-1}^{1}\pi\big[{(5-4x^2)}^2-{(2-x^2)}^2\big]\,\dee{x} \end{gather*}

(b) The curve $y=x^2-1$ intersects $y=0$ at $(1,0)$ and $(-1,0)\text{.}$

We'll use horizontal washers.

• We use thin horizontal strips of height $\dee{y}$ as in the figure above.
• When we rotate about the line $x=5\text{,}$ each strip sweeps out a thin washer

• whose inner radius is $r_{in}=5-\sqrt{y+1}\text{,}$ and
• whose outer radius is $r_{out}= 5-(-\sqrt{y+1})=5+\sqrt{y+1}$ and
• whose thickness is $\dee{y}$ and hence
• whose volume is $\pi(r_{out}^2 - r_{in}^2)\,\dee{y} = \pi\big[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\big]\,\dee{y}$
• As our topmost strip is at $y=0$ and our bottommost strip is at $y=-1$ (when $x=0$), the total volume is \begin{gather*} \int _{-1}^0 \pi\left[\big(5+\sqrt{y+1}\big)^2-\big(5-\sqrt{y+1}\big)^2\right]\,\dee{y} \end{gather*}
Exercise6(*)
Hint

Draw a sketch.

$\pi\displaystyle\int_{-2}^{2}\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\dee{x}$

Solution

The curves intersect at $(-2,4)$ and $(2,4)\text{.}$ When the strip shown in the figure

is rotated about the line $y=-1\text{,}$ it forms a thin washer (punctured disc) of

• inner radius $x^2+1\text{,}$
• outer radius $9-x^2$ and
• thickness $\dee{x}$ and hence of
• cross sectional area $\pi\big[{(9-x^2)}^2-{(x^2+1)}^2\big]$ and
• volume $\pi\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\,\dee{x}\text{.}$

So the volume of the solid is

\begin{gather*} \pi\int_{-2}^{2}\big[{(9-x^2)}^2-{(x^2+1)}^2\big]\dee{x} \end{gather*}
Exercise7
Hint

If you take horizontal slices (parallel to one face), they will all be equilateral triangles.

Be careful not to confuse the height of a triangle with the height of the tetrahedron.

$\dfrac{\sqrt{2}}{12}\ell^3$

Solution

We'll make horizontal slices, parallel to one of the faces of the tetrahedron. Then our slices will be equilateral triangles, of varying sizes.

For the sake of ease, as in Example 1.6.1, we picture the tetrahedron perched on a tip, one base horizontal on top.

Notice our slice forms the horizontal top of a smaller tetrahedron. The horizontal top of the full tetrahedron has side length $\ell\text{,}$ which is $\sqrt{\frac{3}{2}}$ times the height of the full tetrahedron. Our slice is the horizontal top of a tetrahedron of height $y$ and so has side length $\sqrt{\frac{3}{2}}y\text{.}$ An equilateral triangle with side length $L$ has base $L$ and height $\frac{\sqrt{3}}{2}L\text{,}$ and hence area $\frac{\sqrt{3}}{4}L^2\text{.}$ So, the area of our slice with side length $\sqrt{\frac{3}{2}}y$ is

\begin{equation*} A = \frac{\sqrt{3}}{4}\left(\sqrt{\frac{3}{2}}y\right)^2 = \frac{3\sqrt{3}}{8}y^2 \end{equation*}

So, the volume of a tetrahedron with side length $\ell$ is:

\begin{align*} \mbox{Volume}&=\int_0^{\sqrt{\frac{2}{3}}\ell}\frac{3\sqrt{3}}{8}y^2\dee{y}\\ &=\frac{\sqrt{3}}{8}\cdot \left(\sqrt{\frac{2}{3}}\ell\right)^3=\frac{\sqrt{2}}{12}\ell^3 \end{align*}

You were given the height of a tetrahedron, but for completeness we calculate it here.

Draw a line starting at one tip, and dropping straight down to the middle of the opposite face. It forms a right triangle with one edge of the tetrahedron, and a line from the middle of the face to the corner.

We know the length of the hypotenuse of this right triangle (it's $\ell$), so if we know the length of its base (labeled $Ac$ in the diagram), we can figure out its third side, the height of our tetrahedron. Note by using the Pythagorean theorem, we see that the height of an equilateral triangle with edge length $\ell$ is $\sqrt{\frac{3}{2}}\ell\text{.}$

Here is a sketch of the base of the pyramid:

The triangles $ABC$ and $Abc$ are similar (since $b$ and $B$ are right angles, and also $A$ has the same angle in both). Therefore,

\begin{align*} \frac{{Ac}}{{Ab}}&=\frac{{AC}}{{AB}}\\ \frac{Ac}{\ell/2}&=\frac{\ell}{\sqrt{3}\ell/2}\\ Ac&=\frac{1}{\sqrt{3}}\ell \end{align*}

With this in our pocket, we can find the height of the tetrahedron: $\sqrt{\ell^2 - \left(\frac{1}{\sqrt{3}}\ell\right)^2} =\sqrt{\frac{2}{3}}\ell\text{.}$

Exercise8(*)
Hint

Sketch the region.

$\displaystyle\frac{\pi}{4}\Big(e^{2a^2}-1\Big)$

Solution

Let $f(x)=1+\sqrt{x}e^{x^2}\text{.}$ On the vertical slice a distance $x$ from the $y$-axis, sketched in the figure below, $y$ runs from $1$ to $f(x)\text{.}$ Upon rotation about the line $y=1\text{,}$ this thin slice sweeps out a thin disk of thickness $\dee{x}$ and radius $f(x)-1$ and hence of volume $\pi[f(x)-1]^2\,\dee{x}\text{.}$ The full volume generated (for any fixed $a \gt 0$) is

\begin{gather*} \int_0^a\pi[f(x)-1]^2\,\dee{x} =\pi\int_0^axe^{2x^2}\,\dee{x}. \end{gather*}

Using the substitution $u=2x^2\text{,}$ so that $\dee{u}=4x\,\dee{x}\text{:}$

\begin{gather*} \text{Volume} = \pi\int_0^{2a^2}e^u\,\frac{\dee{u}}{4} =\frac{\pi}{4}e^u\Big|_0^{2a^2} =\frac{\pi}{4}\Big(e^{2a^2}-1\Big) \end{gather*}

Remark: we spent a good deal of time last semester developing highly accurate but time-consuming methods for sketching common functions. For the purposes of questions like this, we don't need a detailed picture of a function--broad outlines suffice. Notice that $\sqrt{x} \gt 0$ whenever $x \gt 0\text{,}$ and $e^{x^2} \gt 0$ for all $x\text{.}$ Therefore, $\sqrt{x}e^{x^2}$ is nonnegative over its entire domain, and so the graph $y=1+\sqrt{x}e^{x^2}$ is always the top function, above the bottom function $y=1\text{.}$ That is the only information we needed to perform our calculation.

Exercise9(*)
Hint

Sketch the region first.

$\pi\left[\dfrac{38}{3}-\dfrac{514}{3^4}\right] = \pi\dfrac{512}{81}$

Solution

The curves $y=1/x$ and $3x+3y=10\text{,}$ i.e. $y =\frac{10}{3}-x$ intersect when

\begin{align*} \frac{1}{x} = \frac{10}{3}-x &\iff 3 = 10x-3x^2 \iff 3x^2-10x+3=0 \\ &\iff(3x-1)(x-3)=0 \\ &\iff x=3\,,\,\frac{1}{3} \end{align*}

When the region is rotated about the $x$--axis, the vertical strip in the figure above sweeps out a washer with thickness $\dee{x}\text{,}$ outer radius $T(x)=\frac{10}{3}-x$ and inner radius $B(x)=\frac{1}{x}\text{.}$ This washer has volume

\begin{equation*} \pi\big(T(x)^2- B(x)^2\big)\,\dee{x} = \pi\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\dee{x} \end{equation*}

Hence the volume of the solid is

\begin{align*} \pi\int_{1/3}^3\Big(\frac{100}{9}-\frac{20}{3}x+x^2-\frac{1}{x^2}\Big)\,\dee{x} &=\pi\Big[\frac{100x}{9}-\frac{10}{3}x^2+\frac{1}{3}x^3 +\frac{1}{x}\Big]_{1/3}^3 \\ &=\pi\Big[\frac{38}{3}-\frac{514}{3^4}\Big] = \pi\frac{512}{81} \end{align*}
Exercise10(*)
Hint

You can save yourself quite a bit of work by interpreting the integral as the area of a known geometric figure.

(a) $8\pi\int_{-1}^1\sqrt{1-x^2}\,\dee{x}$ \qquad (b) $4\pi^2$

Solution

(a) The top and the bottom of the circle have equations $y=T(x)=2+\sqrt{1-x^2}$ and $y=B(x) = 2-\sqrt{1-x^2}\text{,}$ respectively.

When $R$ is rotated about the $x$--axis, the vertical strip of $R$ in the figure above sweeps out a washer with thickness $\dee{x}\text{,}$ outer radius $T(x)$ and inner radius $B(x)\text{.}$ This washer has volume

\begin{equation*} \pi\big(T(x)^2- B(x)^2\big)\,\dee{x} = \pi\big(T(x)+ B(x)\big)\big(T(x)- B(x)\big)\,\dee{x} = \pi\times 4\times 2\sqrt{1-x^2}\,\dee{x} \end{equation*}

Hence the volume of the solid is

\begin{equation*} 8\pi\int_{-1}^1\sqrt{1-x^2}\,\dee{x} \end{equation*}

(b) Since $y=\sqrt{1-x^2}$ is equivalent to $x^2+y^2=1\text{,}$ $y\ge 0\text{,}$ the integral is $8\pi$ times the area of the upper half of the circle $x^2+y^2=1$ and hence is $8\pi\times \frac{1}{2}\pi 1^2 = 4\pi^2\text{.}$

Exercise11(*)
Hint

See Example 1.6.3.

(a) The region $R$ is the region between the blue and red curves, with $3\le x\le 5\text{,}$ in the figures below.

(b) $\frac{4}{3}\pi\approx 4.19$

Solution

(a) The two curves intersect when $x$ obeys $8x=x^2+15$ or $x^2-8x+15=(x-5)(x-3)=0\text{.}$ The points of intersection, in the first quadrant, are $(3,\sqrt{24})$ and $(5, \sqrt{40})\text{.}$ The region $R$ is the region between the blue and red curves, with $3\le x\le 5\text{,}$ in the figures below.

(b) The part of the solid with $x$ coordinate between $x$ and $x+\dee{x}$ is a “washer” shaped region with inner radius $\sqrt{x^2+15}\text{,}$ outer radius $\sqrt{8x}$ and thickness $\dee{x}\text{.}$ The surface area of the washer is $\pi(\sqrt{8x})^2 -\pi(\sqrt{x^2+15})^2=\pi(8x-x^2-15)$ and its volume is $\pi(8x-x^2-15)\,\dee{x}\text{.}$ The total volume is

\begin{align*} \int_3^5 \pi(8x-x^2-15)\,\dee{x} &=\pi\Big[4x^2-\frac{1}{3}x^3-15 x\Big]_3^5 =\pi\Big[100-\frac{125}{3}-75-36+9+45\Big] \\ &=\frac{4}{3}\pi\approx 4.19 \end{align*}
Exercise12(*)
Hint

See Example 1.6.5.

(a) The region $R$ is sketched below.

(b) $\pi\Big[4\log 2 - \frac{3}{2}\Big] \approx 3.998$

Solution

(a) The region $R$ is sketched in the figure on the left below. (The bound $y=0$ renders the bound $x=1$ unnecessary, since the graph $y=\log x$ hits the $x$-axis when $x=1\text{.}$)

(b) We'll use horizontal washers as in Example 1.6.5.

• We cut $R$ into thin horizontal strips of height $\dee{y}$ as in the figure on the right above.
• When we rotate $R$ about the $y$--axis, i.e. about the line $x=0\text{,}$ each strip sweeps out a thin washer

• whose inner radius is $r_{in}= e^y$ and outer radius is $r_{out}=2\text{,}$ and
• whose thickness is $\dee{y}$ and hence
• whose volume $\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big(4-e^{2y}\big)\dee{y}\text{.}$
• As our bottommost strip is at $y=0$ and our topmost strip is at $y=\log 2$ (since at the top $x=2$ and $x=e^y$), the total \begin{align*} \text{Volume} &= \int _0^{\log 2} \pi\big(4-e^{2y}\big)\dee{y} =\pi\big[4y -e^{2y}/2\big]_0^{\log 2} =\pi\Big[4\log 2 - 2 +\frac{1}{2}\Big] \\ &=\pi\Big[4\log 2 - \frac{3}{2}\Big] \end{align*} Using a calculator, we see this is approximately $3.998\text{.}$
Exercise13(*)
Hint

Sketch the region. To find where the curves intersect, look at where $\cos(\frac x2)$ and $x^2 - \pi^2$ both have roots.

$\pi^2 + 8\pi^3 + \frac{8\pi^6}{5}$

Solution

Here is a sketch of the curves $y = \cos(\frac x2)$ and $y = x^2 - \pi^2\text{.}$

By inspection, the curves meet at $x = \pm {\pi}$ where both $\cos(\frac x2)$ and $x^2 - \pi^2$ take the value zero. We'll use vertical washers as specified in the question.

• We cut the specified region into thin vertical strips of width $\dee{x}$ as in the figure above.
• When we rotate about the line $y=-\pi^2\text{,}$ each strip sweeps out a thin washer

• whose inner radius is $r_{in}= (x^2 - {\pi^2} ) - ( {-} {\pi^2} )=x^2$ and outer radius is $r_{out}=\cos(\frac x2) - ( {-} {\pi^2}) =\cos(\frac x2) +\pi^2 \text{,}$ and
• whose thickness is $\dee{x}$ and hence
• whose volume $\pi(r_{out}^2 - r_{in}^2)\dee{x} = \pi\big( {(\cos(\frac x2) +\pi^2)}^2 - {(x^2)}^2\big)\dee{x}\text{.}$
• As our leftmost strip is at $x=-\pi$ and our rightmost strip is at $x=\pi\text{,}$

the total volume is

\begin{align*} &\pi \int_{-\pi}^{\pi} \left( \cos^2 (\tfrac x2) +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\dee{x} \\ & = \pi \int_{-\pi}^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\dee{x} \\ \end{align*}

Because the integrand is even,

\begin{align*} & = 2\pi \int_0^{\pi} \left(\frac{1+\cos(x)}{2} +2{\pi^2}\cos (\tfrac x2) +{\pi^4} -x^4\right)\,\dee{x} \\ & = {2\pi \left[ \frac{1}{2}x + \frac{1}{2}\sin(x) + 4{\pi^2}\sin (\tfrac x2) +{\pi^4} x -\frac{1}{5}x^5 \right]} _0^\pi\\ & = 2\pi \left[\frac{\pi}{2} + 0 + 4{\pi^2} +{\pi^5} - \frac{\pi^5}{5} \right] \\ & = {\pi^2} + 8\pi^3 + \frac{8\pi^6}{5} \end{align*}

We used the fact that the integrand is an even function and the interval of integration $[-\pi, \pi]$ is symmetric, but one can also compute directly.

Exercise14(*)
Hint

See Example 1.6.6.

$\dfrac{8}{3}$

Solution

As in Example 1.6.6, we slice $V$ into thin horizontal “square pancakes”.

• We are told that the pancake at height $x$ is a square of side $\frac{2}{1+x}$ and so
• has cross-sectional area $\big(\frac{2}{1+x}\big)^2$ and thickness $\dee{x}$ and hence
• has volume $\big(\frac{2}{1+x}\big)^2\dee{x}\text{.}$

Hence the volume of $V$ is

\begin{equation*} \int_0^2{\Big[\frac{2}{1+x}\Big]}^2\,\dee{x} =\int_1^3\frac{4}{u^2}\,\dee{u} =4\frac{u^{-1}}{-1}\bigg|_1^3 =-4\Big[\frac{1}{3}-1\Big] =\frac{8}{3} \end{equation*}

We made the change of variables $u=1+x\text{,}$ $\dee{u}=\dee{x}\text{.}$

Exercise15(*)
Hint

See Example 1.6.6. Imagine cross-sections with shadow parallel to the $y$-axis, sticking straight out of the $xy$-plane.

$\dfrac{256\times 8}{15}=136.5\dot3$

Solution

Here is a sketch of the base region.

Consider the thin vertical cross--section resting on the heavy red line in the figure above. It has thickness $\dee{x}\text{.}$ Its face is a square whose side runs from $y=x^2$ to $y=8-x^2\text{,}$ a distance of $8-2x^2\text{.}$ So the face has area ${(8-2x^2)}^2$ and the slice has volume ${(8-2x^2)}^2\,\dee{x}\text{.}$ The two curves cross when $x^2=8-x^2\text{,}$ i.e. when $x^2=4$ or $x=\pm 2\text{.}$ So $x$ runs from $-2$ to $2$ and the total volume is

\begin{align*} \int_{-2}^{2}{(8-2x^2)}^2\,\dee{x}&=2\int_0^2 4{(4-x^2)}^2\,\dee{x} =8\int_0^2\big[16-8x^2+x^4\big]\,\dee{x}\cr &=8\Big[16\times 2-\frac{8}{3}2^3+\frac{1}{5}2^5\Big] =\frac{256\times 8}{15}=136.5\dot3 \end{align*}

In the first simplification step, we used the fact that our integrand was even, but we also could have finished our computation without this step.

Exercise16(*)
Hint

See Example 1.6.1.

$\dfrac{28}{3}\pi h$

Solution

Slice the frustrum into horizontal discs. When the disc is a distance $t$ from the top of the frustrum it has radius $2+2t/h\text{.}$ Note that as $t$ runs from $0$ (the top of the frustrum) to $t=h$ (the bottom of the frustrum) the radius $2+2t/h$ increases linearly from $2$ to $4\text{.}$

Thus the disk has volume $\pi \big(2+2t/h\big)^2 \dee{t}\text{.}$ The total volume of the frustrum is

\begin{gather*} \pi\int_0^h \big(2+2t/h\big)^2 \dee{t} =4\pi\int_0^h \big(1+t/h\big)^2 \dee{t} =4\pi\left[\frac{(1+t/h)^3}{3/h}\right]_0^h =\frac{4}{3}\pi h\times 7 =\frac{28}{3}\pi h \end{gather*}

Remark: we could also solve this problem using the formula for the volume of a cone. Using similar triangles, the frustrum in question is shaped like a right circular cone of height $2h$ and base radius 4 (and hence of volume $\dfrac{1}{3}\pi(4^2)(2h)$), but missing its top, which is a right circular cone of height $h$ and base radius $2$ (and hence volume $\dfrac{1}{3}\pi(2^2)h$). So, the volume of the frustrum is $\dfrac{1}{3}\pi(4^2)(2h) - \dfrac{1}{3}\pi(2^2)h = \dfrac{28}{3}\pi h\text{.}$

Exercise17
Hint

(a) Don't be put off by phrases like “rotating an ellipse about its minor axis.” This is the same kind of volume you've been calculating all section.

(b) Hopefully, you sketched the ellipse in part (a). What was its smallest radius? Its largest? These correspond to the polar and equitorial radii, respectively.

(d) Remember that the absolute error is the absolute difference of your two results--that is, you subtract them and take the absolute value. The relative error is the absolute error divided by the actual value (which we're taking, for our purposes, to be your answer from (c)). When you take the relative error, lots of terms will cancel, so it's easiest to not use a calculator till the end.

• (a) $\dfrac{4\pi}{3b^2a}$ cubic units,
• (b) $a = \dfrac{1}{6356.752}$ and $b=\dfrac{1}{6378.137}\text{,}$
• (c) Approximately $1.08321\times 10^{12} \mathrm{km}^3\text{,}$ \quad or \quad $1.08321\times 10^{21} \mathrm{m}^3\text{,}$
• (d) Absolute error is about $3.64\times 10^{9} \mathrm{km}^3\text{,}$ and relative error is about $0.00336\text{,}$ or $0.336\%\text{.}$
Solution

(a)

We'll want to start by graphing the upper half of the ellipse $(ax)^2+(by)^2=1\text{.}$ Its intercepts will be enough to get us an idea: $(0,\frac{1}{b})$ and $(\pm\frac{1}{a},0)\text{:}$

We note a few things at the outset: first, since $a \geq b\text{,}$ then $\frac{1}{a} \leq \frac{1}{b}\text{,}$ so indeed the $x$-axis is the minor axis. That is, we're rotating about the proper axis to create an oblate spheroid.

Second, if we solve our equation for $y\text{,}$ we get $y=\frac{1}{b}\sqrt{1-(ax)^2}\text{.}$ (Since we only want the upper half of the ellipse, we only need to consider the positive square root.)

Now, we have a standard volume-of-revolution problem. We make vertical slices, of width $\dee{x}$ and height $y=\frac{1}{b}\sqrt{1-(ax)^2}\text{.}$ When we rotate these slices about the $x$-axis, they form thin disks of volume $\pi\left[\frac{1}{b}\sqrt{1-(ax)^2}\right]^2\dee{x}$\ . Since $x$ runs from $-\frac{1}{a}$ to $\frac{1}{a}\text{,}$ the volume of our oblate spheroid is:

\begin{align*} \mbox{Volume}&=\int_{-\frac{1}{a}}^{\frac{1}{a}} \pi \left[\frac{1}{b}\sqrt{1-(ax)^2}\right]^2\dee{x}\\ &=\frac{\pi}{b^2}\int_{-\frac{1}{a}}^{\frac{1}{a}} 1-(ax)^2\dee{x}\\ &=\frac{2\pi}{b^2}\int_{0}^{\frac{1}{a}} 1-(ax)^2\dee{x}&\mbox{(even function)}\\ &=\frac{2\pi}{b^2}\left[x - \frac{a^2x^3}{3}\right]_{0}^{\frac{1}{a}}\\ &=\frac{2\pi}{b^2}\left[\frac{1}{a} - \frac{1}{3a} \right] = \frac{4\pi}{3b^2a} \end{align*}

(b) As we saw in the sketch from part (a), the shortest radius of the ellipse is $\frac{1}{a}\text{,}$ while the largest is $\frac{1}{b}\text{.}$ So, $\frac{1}{a} = 6356.752\text{,}$ and $\frac{1}{b} = 6378.137\text{.}$ That is, $a = \dfrac{1}{6356.752}$ and $b=\dfrac{1}{6378.137}\text{.}$

Note $a \geq b\text{,}$ as specified in part (a).

(c) Combining our answers from (a) and (b), the volume of an oblate spheroid with approximately the same dimensions as the earth is:

\begin{align*} \frac{4\pi}{3b^2a} &= \frac{4\pi}{3}\left(\frac{1}{b}\right)^2\left(\frac{1}{a}\right)\\ &=\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right)\\ &\approx 1.08321\times 10^{12} \quad\mathrm{km}^3\\ &\approx 1.08321\times 10^{21} \quad\mathrm{m}^3 \end{align*}

(d) A sphere of radius 6378.137 has volume

\begin{align*} &\dfrac{4}{3}\pi\left(6378.137\right)^3\\ \end{align*}

So, our absolute error is:

\begin{align*} &\left|\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right) - \dfrac{4}{3}\pi\left(6378.137\right)^3 \right|\\ =&\frac{4\pi}{3}\left(6378.137\right)^2\big|6356.752 - 6378.137\big|\\ =&\frac{4\pi}{3}\left(6378.137\right)^2(21.385)\\ \approx& 3.64 \times 10^{9} \mathrm{km}^3\\ \end{align*}

And our relative error is:

\begin{align*} \frac{\mbox{abs error}}{\mbox{actual value}}&=\frac{\frac{4\pi}{3}\left(6378.137\right)^2\big|6356.752 - 6378.137\big|}{\frac{4\pi}{3}\left(6378.137\right)^2\left(6356.752\right)}\\ &=\frac{\big|6356.752 - 6378.137\big|}{6356.752}\\ &=\frac{6378.137}{6356.752}-1\\ &\approx 0.00336 \end{align*}

That is, about $0.336\%\text{,}$ or about one-third of one percent.

Exercise18(*)
Hint

To find the points of intersection, set $4-(x-1)^2=x+1\text{.}$

(a) $\dfrac{9}{2}$ (b) $\pi\displaystyle\int_{-1}^2 \big[{\big(4-x\big)}^2-{\big(1+(x-1)^2\big)}^2\big]\,\dee{x}$

Solution

(a) The curve $y = 4 - (x - 1)^2$ is an “upside down parabola” and line $y = x + 1$ has slope 1. They intersect at points $(x,y)$ which satisfy both $y=x+1$ and $y=4-(x-1)^2\text{.}$ That is, when $x$ obeys

\begin{align*} x+1&=4-(x-1)^2\\ x+1 &= 4 -x^2+2x-1\\ x^2-x-2&=0 \\ (x-2)(x+1)&=0\\ x&=-1 \quad\mbox{or}\quad x=2 \end{align*}

Thus the intersection points are $(-1,0)$ and $(2,3)\text{.}$ Here is a sketch of $R\text{:}$

The red strip in the sketch above runs from $y=x+1$ to $y=4-(x-1)^2$ and so has area $[4-(x-1)^2 -(x+1)]\,\dee{x} = [2+x-x^2]\,\dee{x}\text{.}$ All together $R$ has

\begin{align*} \text{Area} &= \int_{-1}^2 \big[2+x-x^2\big]\dee{x} \\ &=\bigg[2x+\frac{x^2}{2}-\frac{x^3}{3}\bigg]_{-1}^2 \\ &=6+\frac{3}{2}-\frac{9}{3}=\frac{9}{2} \end{align*}

(b) We'll use vertical washers as in Example 1.6.3. Note that the highest point achieved by $y=4-(x-1)^2$ is $y=4\text{,}$ so rotating around the line $y=5$ causes no unexpected problems.

• We cut $R$ into thin vertical strips of width $\dee{x}$ like the red strip in the figure above.
• When we rotate $R$ about the horizontal line $y=5\text{,}$ each strip sweeps out a thin washer

• whose inner radius is $r_{in}=5-[4-(x-1)^2]=1+(x-1)^2\text{,}$ and
• whose outer radius is $r_{out}= 5-[x+1]=4-x$ and
• whose thickness is $\dee{x}$ and hence
• whose volume is $\pi\big[r_{out}^2-r_{in}^2\big]\,\dee{x} =\pi\big[{\big(4-x\big)}^2-{\big(1+(x-1)^2\big)}^2\big]\,\dee{x}$
• As our leftmost strip is at $x=-1$ and our rightmost strip is at $x=2\text{,}$ the total \begin{align*} {\rm Volume} &= \pi\int_{-1}^2 \big[{\big(4-x\big)}^2-{\big(1+(x-1)^2\big)}^2\big]\,\dee{x} \end{align*}
Exercise19(*)
Hint

You can somewhat simplify your calculations in part (a) (but not part (b)) by using the fact that $\cR$ is symmetric about the line $y=x\text{.}$

When you're solving an equation for $x\text{,}$ be careful about your signs: $x-1$ is negative.

(a) $\dfrac{\pi}{2}-1$ (b) $\dfrac{\pi^2}{2}-\pi\approx 1.793$

Solution

(a) The curves $(x-1)^2+y^2 = 1$ and $x^2+(y-1)^2 = 1$ are circles of radius $1$ centered on $(1,0)$ and $(0,1)$ respectively. Both circles pass through $(0,0)$ and $(1,1)\text{.}$ They are sketched below.

The region $\cR$ is symmetric about the line $y=x\text{,}$ so the area of $\cR$ is twice the area of the part of $\cR$ to the left of the line $y=x\text{.}$ The red strip in the sketch above runs from the edge of the lower circle to $x=y\text{.}$ So, given a value of $y$ in $[0,1]\text{,}$ we need to find the corresponding value of $x$ along the circle. We solve $(x-1)^2+y^2=1$ for $x\text{,}$ keeping in mind that $0 \leq x\leq 1\text{:}$

\begin{align*} (x-1)^2+y^2&=1\\ (x-1)^2&=1-y^2\\ |x-1|&=\sqrt{1-y^2}\\ 1-x&=\sqrt{1-y^2}\\ x&=1-\sqrt{1-y^2}\\ \end{align*}

Now, we calculate:

\begin{align*} \text{Area} &= 2\int_0^1 \big[y-\big(1-\sqrt{1-y^2}\big)\big]\dee{y} \\ &=2\left\{ \int_0^1 y-1\dee{y} + \int_0^1 \sqrt{1-y^2}\dee{y} \right\}\\ &=2\Big\{\Big[\frac{y^2}{2}-y\Big]_0^1 +\int_0^1\sqrt{1-y^2}\dee{y}\Big\} \\ &=\frac{\pi}{2}-1 \end{align*}

Here the integral $\int_0^1\sqrt{1-y^2}\dee{y}$ was evaluated simply as the area of one quarter of a cicular disk of radius $1\text{.}$ It can also be evaluated by substituting $y=\sin\theta\text{,}$ a technique we'll learn more about in Section 1.9.

(b) We'll use horizontal washers as in Example 1.6.5.

• We cut $\cR$ into thin horizontal strips of width $\dee{y}$ like the blue strip in the figure above.
• When we rotate $\cR$ about the $y$--axis, each strip sweeps out a thin washer

• whose inner radius is $r_{in}=1-\sqrt{1-y^2}\text{,}$ and
• whose outer radius is $r_{out}= \sqrt{1-(y-1)^2}$ and
• whose thickness is $\dee{y}$ and hence
• whose volume is \begin{align*} &\pi\big[{\big(\sqrt{1-(y-1)^2}\big)}^2-{\big(1-\sqrt{1-y^2}\,\big)}^2\big]\,\dee{y}\\ =&\pi \big[ 1 - (y-1)^2 -1 + 2\sqrt{1-y^2} - (1-y^2) \big]\\ =&2\pi\big[\sqrt{1-y^2}+y-1\big]\,\dee{y} \end{align*}
• As our bottommost strip is at $y=0$ and our topmost strip is at $y=1\text{,}$ the total \begin{align*} {\rm Volume} &= 2\pi\int_{0}^1\big[\sqrt{1-y^2}+y-1\big]\dee{y} = 2\pi\Big[\frac{\pi}{4}+\frac{1}{2}-1\Big]\\ &=\frac{\pi^2}{2}-\pi\approx 1.793 \end{align*} Here, we again used that $\int_{0}^1 \sqrt{1-y^2}\ dy$ is the area of a quarter circle of radius one, and we used a calculator to approximate the final answer.
Exercise20(*)
Hint

The mechanically easiest way to answer part (b) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what $\cR$ looks like. So it is crucial to first sketch $\cR\text{.}$ Then be very careful in identifying the left end of your horizontal strips.

(a) $V_1=\dfrac{4}{3}\pi c^2$ (b) $V_2 =\dfrac{\pi\,c}{3}\big[4\sqrt{2}-2 \big]$ (c) $c=0\text{ or }c=\sqrt{2}-\frac{1}{2}$

Solution

Before we start, it will be useful to have a reasonable sketch of the graph $y=c\sqrt{1+x^2}$ over the interval $[0,1]\text{.}$ Its endpoints are $(0,c)$ and $(1,c\sqrt{2})\text{.}$ The function is entirely above the $x$-axis, which we need to know for part (a). For part (b), we need to know whether it is always increasing or not: when we're drawing horizontal strips, we need to know their endpoints, and if the function has “humps,” the right endpoint will not be simply the line $x=1\text{.}$

If you're comfortable noticing that $1+x^2$ increases as $x$ increases because we only consider nonnegative values of $x\text{,}$ then you can also be confident that $\sqrt{1+x^2}$ is simply increasing. Alternately, we can consider the derivative:

\begin{align*} \diff{}{x}\left\{c\sqrt{1+x^2}\right\}&=c\cdot \dfrac{1}{2\sqrt{1+x^2}}\cdot 2x = \dfrac{cx}{\sqrt{1+x^2}} \end{align*}

Since we only consider positive values of $x\text{,}$ this derivative is never negative, so the function is never decreasing. This gives us the following basic sketch:

The figures in the solution below use a slightly more detailed rendering of our function, but so much accuracy is not necessary.

(a) Let $\cV_1$ be the solid obtained by revolving $\cR$ about the $x$--axis. The portion of $\cV_1$ with $x$--coordinate between $x$ and $x+\dee{x}$ is obtained by rotating the red vertical strip in the figure on the left below about the $x$--axis. That portion is a disk of radius $c\sqrt{1+x^2}$ and thickness $\dee{x}\text{.}$ The volume of this disk is $\pi(c\sqrt{1+x^2})^2\dee{x}=\pi c^2 (1+x^2)\,\dee{x}\text{.}$ So the total volume of $\cV_1$ is

\begin{gather*} V_1=\int_0^1 \pi c^2 (1+x^2)\,\dee{x} =\pi c^2\Big[x+\frac{x^3}{3}\Big]_0^1 =\frac{4}{3}\pi c^2 \end{gather*}

(b) We'll use horizontal washers as in Example 1.6.5.

• We cut $\cR$ into thin horizontal strips of width $\dee{y}$ as in the figure on the right above.
• When we rotate $\cR$ about the $y$--axis, i.e. about the line $x=0\text{,}$ each strip sweeps out a thin washer

• whose outer radius is $r_{out}=1\text{,}$ and
• whose inner radius is $r_{in}= \sqrt{\frac{y^2}{c^2}-1}$ when $y\ge c\sqrt{1+0^2}=c$ (see the red strip in the figure on the right above), and whose inner radius is $r_{in}= 0$ when $y\le c$ (see the blue strip in the figure on the right above) and
• whose thickness is $\dee{y}$ and hence
• whose volume is $\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big(2-\frac{y^2}{c^2}\big)\dee{y}$ when $y\ge c$ and whose volume is $\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\,\dee{y}$ when $y\le c$ and
• As our bottommost strip is at $y=0$ and our topmost strip is at $y=\sqrt{2}\,c$ (since at the top $x=1$ and $y= c\sqrt{1+x^2}$), the total \begin{align*} V_2 &= \int _c^{\sqrt{2}\,c} \pi\Big(2-\frac{y^2}{c^2}\Big)\dee{y} +\int _ 0^c \pi\,\dee{y} \\ &=\pi{\Big[2y -\frac{y^3}{3c^2}\Big]}_c^{\sqrt{2}\,c} +\pi c\\ &=\pi\,c\Big[\frac{4\sqrt{2}}{3}-\frac{5}{3} \Big]+\pi c\\ &=\frac{\pi\,c}{3}\big[4\sqrt{2}-2 \big] \end{align*}

(c) We have $V_1=V_2$ if and only if

\begin{align*} \frac{4}{3}\pi c^2&=\frac{\pi\,c}{3}\big[4\sqrt{2}-2 \big] \\ 4c^2&=c\left(4\sqrt{2}-2\right)\\ 4c^2-c\left(4\sqrt{2}-2\right)&=0\\ 4c\left(c - \left(\sqrt{2}-\frac{1}{2}\right)\right)&=0\\ c=0 \quad\mbox{or}\quad c&=\sqrt{2}-\frac{1}{2} \end{align*}
Exercise21(*)
Hint

Note that the curves cross. The area of this region was found in Problem 1.5.2.14 of Section 1.5. It would be useful to review that problem.

$\displaystyle\int_{\pi/2}^\pi \pi\big[(5 + \pi \sin x)^2-(5 + 2\pi - 2x)^2\big]\dee{x} +\displaystyle\int^{3\pi/2}_\pi \pi\big[(5 + 2\pi - 2x)^2-(5 + \pi \sin x)^2\big]\ \dee{x}$

Solution

We will compute the volume by rotating thin vertical strips as in the sketch

about the line $y=-1$ to generate thin washers. We need to know when the line $y = 4 + 2\pi - 2x$ intersects the curve $y = 4 + \pi \sin x\text{.}$ Looking at the graph, it appears to be at $\frac{\pi}{2}\text{,}$ $\pi\text{,}$ and $\frac{3\pi}{2}\text{.}$ By plugging in these values of $x$ to both functions, we see they are indeed the points of intersection.

• When $\frac{\pi}{2} \le x \le \pi\text{,}$ the top of the strip is at $y = 4 + \pi \sin x$ and the bottom of the strip is at $y = 4 + 2\pi - 2x\text{.}$ When the strip is rotated, we get a thin washer with outer radius $R_1(x)= 1+ 4 + \pi \sin x=5 + \pi \sin x$ and inner radius $r_1(x) = 1+4 + 2\pi - 2x=5 + 2\pi - 2x\text{.}$
• When $\pi \le x \le \frac{3\pi}{2}\text{,}$ the top of the strip is at $y = 4 + 2\pi - 2x$ and the bottom of the strip is at $y = 4 + \pi \sin x\text{.}$ When the strip is rotated, we get a thin washer with outer radius $R_2(x) = 1+4 + 2\pi - 2x=5 + 2\pi - 2x$ and inner radius $r_2(x) = 1+ 4 + \pi \sin x=5 + \pi \sin x\text{.}$

So, the total

\begin{align*} \hbox{Volume} &= \int_{\pi/2}^\pi \pi\big[R_1(x)^2-r_1(x)^2\big]\dee{x} +\int^{3\pi/2}_\pi \pi\big[R_2(x)^2-r_2(x)^2\big]\dee{x}\\ &= \int_{\pi/2}^\pi \pi\big[(5 + \pi \sin x)^2-(5 + 2\pi - 2x)^2\big]\dee{x}\\ & +\int^{3\pi/2}_\pi \pi\big[(5 + 2\pi - 2x)^2-(5 + \pi \sin x)^2\big]\ \dee{x} \end{align*}
Exercise22
Hint

You can use ideas from this section to answer the question. If you take a very thin slice of the column, the density is almost constant, so you can find the mass. Then you can add up all your little slices. It's the same idea as volume, only applied to mass.

Do be careful about units: in the problem statement, some are given in metres, others in kilometres.

If you're having a hard time with the antiderivative, try writing the exponential function with base $e\text{.}$ Remember $2 = e^{\log 2}\text{.}$

(a) $\dfrac{6000c\pi}{\log 2}\left(1-\dfrac{1}{2^{10}}\right)\text{,}$ which is close to $\dfrac{6000c\pi}{\log 2}\text{.}$

(b) 6km: that is, there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.

Solution

(a)

We use the same ideas for volume, and apply them to mass. We want to take slices of the column, approximate their mass, then add them up. To reconcile our units, let $k=1000h\text{,}$ so $k$ is the height in metres. Then the density of air at height $k$ is $c2^{-k/6000} \frac{\mathrm{kg}}{\mathrm{m}^3}\text{.}$

A horizontal slice of the column is a circular disk with height $\dee{k}$ and radius $1$ m. So, its volume is $\pi \dee{k} \mathrm{m^3}\text{.}$ What we're interested in, though, is its mass. At height $k\text{,}$ its mass is

\begin{align*} (\mathrm{volume})\times (\mathrm{density})&=\left(\pi \dee{k} \mathrm{m^3}\right)\times \left(c2^{-k/6000} \frac{\mathrm{kg}}{\mathrm{m^3}}\right)\\ &=c\pi2^{-k/6000} \dee{k}\quad\mathrm{kg}\\ \end{align*}

Since $k$ runs from 0 to $60,000\text{,}$ the total mass is given by

\begin{align*} \int_0^{60000} c\pi2^{-k/6000} \dee{k}&=c\pi\int_0^{60000} 2^{-k/6000} \dee{k}\\ \end{align*}

To facilitate integration, we can write our exponential function in terms of $e\text{,}$ then use the substitution $u=-\frac{k}{6000}\log 2\text{,}$ $\dee{u} = -\frac{1}{6000}\log 2 \dee{k}\text{.}$

\begin{align*} &=c\pi\int_0^{60000}\left(e^{\log 2}\right)^{-k/6000}\dee{k}\\ &=c\pi\int_0^{60000}e^{-\tfrac{k}{6000}\log 2}\dee{k}\\ &=-\frac{6000c\pi}{\log 2}\int_0^{-10\log 2}e^{u}\dee{u}\\ &=\frac{6000c\pi}{\log 2}\int_{-10\log 2}^0e^{u}\dee{u}\\ &=\frac{6000c\pi}{\log 2}\left(1-\frac{1}{2^{10}}\right) \end{align*}

We note this is fairly close to $\dfrac{6000c\pi}{\log 2}\text{.}$

We also remark that this is a demonstration of the usefulness of integrals. We wanted to know how much of something there was, but the amount of that something was different everywhere: more in some places, less in others. Integration allowed us to account for this gradient. You've seen this behaviour exploited to find distances travelled, areas, volumes, and now mass. In your studies, you will doubtless learn to use it to find still more quantities, and we will discuss other applications in Chapter 2.

(b) We want to find the value of $k$ that gives a mass of $\dfrac{3000c\pi}{\log 2}\text{.}$ By following our reasoning above, the mass of air in the column from the ground to height $k$ is

\begin{align*} \frac{6000c\pi}{\log 2}\left(1-\frac{1}{2^{k/6000}}\right)&\\ \end{align*}

So, we set this equal to the mass we want, and solve for $k\text{.}$

\begin{align*} \frac{6000c\pi}{\log 2}\left(1-\frac{1}{2^{k/6000}}\right)&=\frac{3000c\pi}{\log 2}\\ 2\left(1-\frac{1}{2^{k/6000}}\right)&=1\\ 1&=\frac{2}{2^{k/6000}}\\ 2^{k/6000}&=2^1\\ k&=6000\\ h&=6 \end{align*}

This means that there is roughly the same mass of air in the lowest 6 km of the column as there is in the remaining 54 km.

Exercise23
Hint

You'll want to use horizontal cross sections, like Example 1.6.1. Then the area of your slice can be calculated as a section of a circle, minus a triangle.

$\dfrac{r^2h}{3}\left(\dfrac{\pi}{4}-\dfrac{1}{2}\right)$

Solution

As in Example 1.6.1, we take horizontal slices. If the radius of a slice is $s\text{,}$ then its area can be calculated as one-quarter of the circle of radius $s\text{,}$ minus the right triangle with base and height both $s\text{.}$ So,

\begin{equation*} A = \frac{1}{4}\left(\pi s^2\right) - \frac{1}{2} s^2 = s^2\left(\frac{\pi}{4}-\frac{1}{2}\right). \end{equation*}

As we climb up the cone, the radius of our slices changes. As in the text, let's assume that the base of the cone sits at $y=h\text{,}$ and the tip is at $y=0\text{.}$ Then we use similar triangles to conclude that at height $y\text{,}$ the radius of the cone is $s=\left(\frac{r}{h}\right)y\text{.}$

So:

• Our slices run from $y=0$ to $y=h\text{,}$
• at height $y\text{,}$ the area of our slice is $\left(\frac{r}{h}\cdot y\right)^2\left(\frac{\pi}{4}-\frac{1}{2}\right) = y^2\left(\frac{r}{h}\right)^2\left(\frac{\pi}{4}-\frac{1}{2}\right)\text{,}$
• and the height of our slice is $\dee{y}\text{.}$

So, the volume of our piece of cone is:

\begin{align*} \mbox{Volume}&=\int_0^h y^2\left(\frac{r}{h}\right)^2\left(\frac{\pi}{4}-\frac{1}{2}\right)\dee{y}\\ &=\left(\frac{r}{h}\right)^2\left(\frac{\pi}{4}-\frac{1}{2}\right)\int_0^h y^2\dee{y} \\ &= \left(\frac{r}{h}\right)^2\left(\frac{\pi}{4}-\frac{1}{2}\right)\frac{1}{3}h^3\\ &= \frac{r^2h}{3}\left(\frac{\pi}{4}-\frac{1}{2}\right) \end{align*}

Exercises1.7.2Exercises

Exercise1
Hint

Read back over Sections 1.4 and 1.7. When these methods are introduced, they are justified using the corresponding differentiation rules.

chain; product

Solution

Integration by substitution is just using the chain rule, backwards:

\begin{align*} &&\diff{}{x}\{f(g(x))\}&=f'(g(x))g'(x)\\ &\Leftrightarrow& \int\diff{}{x}\{f(g(x))\}\dee{x}+C&=\int f'(g(x))g'(x)\dee{x}\\ &\Leftrightarrow&\underbrace{ f(g(x))}_{f(u)} +C &=\int \underbrace{f'(g(x))}_{f'(u)}\underbrace{g'(x)\dee{x}}_{\dee{u}}\\ \end{align*}

Similarly, integration by parts comes from the product rule:

\begin{align*} &&\diff{}{x}\{f(x)g(x)\}&=f'(x)g(x)+f(x)g'(x)\\ &\Leftrightarrow&\int\diff{}{x}\{f(x)g(x)\}\dee{x}+C&=\int f'(x)g(x)+f(x)g'(x)\dee{x}\\ &\Leftrightarrow&f(x)g(x)+C&=\int f'(x)g(x)\dee{x}+\int f(x)g'(x)\dee{x}\\ &\Leftrightarrow&\int\underbrace{ f(x)}_{u}\underbrace{g'(x)\dee{x}}_{\dee{v}}&=\underbrace{f(x)}_{u}\underbrace{g(x)}_{v}-\int \underbrace{g(x)}_{v}\underbrace{f'(x)\dee{x}}_{\dee{u}}\\ \end{align*}
Exercise2
Hint

Remember our rule: $\int u \dee{v} = uv - \int v \dee{u}\text{.}$ So, we take $u$ and use it to make $\dee{u}\text{,}$ and we take $\dee{v}$ and use it to make $v\text{.}$

The part chosen as $u$ will be differentiated. The part chosen as $\dee{v}$ will be antidifferentiated.

Solution

Remember our rule: $\int u \dee{v} = uv - \int v \dee{u}\text{.}$ So, we take $u$ and use it to make $\dee{u}$--that is, we differentiate it. We take $\dee{v}$ and use it to make $v$--that is, we antidifferentiate it.

Exercise3
Hint

According to the quotient rule,

\begin{equation*} \diff{}{x}\left\{\dfrac{f(x)}{g(x)}\right\} = \frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}. \end{equation*}

Antidifferentiate both sides of the equation, then solve for the expression in the question.

$\displaystyle\int \frac{f'(x)}{g(x)}\dee{x}= \dfrac{f(x)}{g(x)} +\displaystyle\int\frac{f(x)g'(x)}{g^2(x)}\dee{x}+C$

Solution

We'll use the same ideas that lead to the methods of substitution and integration by parts. (You can review these in your text, or see the solution to Question 1 in this section.) According to the quotient rule,

\begin{align*} \diff{}{x}\left\{\dfrac{f(x)}{g(x)}\right\} &= \frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}.\\ \end{align*}

Antidifferentiating both sides gives us:

\begin{align*} \int\diff{}{x}\left\{\dfrac{f(x)}{g(x)}\right\} \dee{x}+C&= \int\frac{g(x)f'(x)-f(x)g'(x)}{g^2(x)}\dee{x}\\ \dfrac{f(x)}{g(x)} +C&= \int\frac{f'(x)}{g(x)}\dee{x}-\int\frac{f(x)g'(x)}{g^2(x)}\dee{x}\\ \int\frac{f'(x)}{g(x)}\dee{x} &= \dfrac{f(x)}{g(x)}+\int\frac{f(x)g'(x)}{g^2(x)}\dee{x}+C \end{align*}

This isn't quite at catchy as integration by parts--which is probably why it hasn't caught on as a rule with its own name.

Exercise4
Hint

Remember all the antiderivatives differ only by a constant, so you can write them all as $v(x)+C$ for some $C\text{.}$

All the antiderivatives differ only by a constant, so we can write them all as $v(x)+C$ for some $C\text{.}$ Then, using the formula for integration by parts,

\begin{align*} \int u(x)\cdot v'(x) \dee{x}&=\underbrace{u(x)}_u\underbrace{\big[ v(x)+C\big]}_v - \int \underbrace{\big[ v(x)+C\big]}_v \underbrace{u'(x)\dee{x}}_{\dee{u}}\\ &=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - \int Cu'(x)\dee{x}\\ &=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - Cu(x)+D\\ &=u(x)v(x) - \int v(x)u'(x)\dee{x} +D \end{align*}

where $D$ is any constant.

Since the terms with $C$ cancel out, it didn't matter what we chose for $C$--all choices end up the same.

Solution

All the antiderivatives differ only by a constant, so we can write them all as $v(x)+C$ for some $C\text{.}$ Then, using the formula for integration by parts,

\begin{align*} \int u(x)\cdot v'(x) \dee{x}&=\underbrace{u(x)}_u\underbrace{\big[ v(x)+C\big]}_v - \int \underbrace{\big[ v(x)+C\big]}_v \underbrace{u'(x)\dee{x}}_{\dee{u}}\\ &=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - \int Cu'(x)\dee{x}\\ &=u(x)v(x)+Cu(x) - \int v(x)u'(x)\dee{x} - Cu(x)+D\\ &=u(x)v(x) - \int v(x)u'(x)\dee{x} +D \end{align*}

where $D$ is any constant.

Since the terms with $C$ cancel out, it didn't matter what we chose for $C$--all choices end up the same.

Exercise5
Hint

What integral do you have to evaluate, after you plug in your choices to the integration by parts formula?

Suppose we choose $\dee{v} = f(x)\dee{x}\text{,}$ $u=1\text{.}$ Then $v = \displaystyle\int f(x)\dee{x}\text{,}$ and $\dee{u}=\dee{x}\text{.}$ So, our integral becomes:

\begin{align*} \int \underbrace{(1)}_{u}\underbrace{f(x)\dee{x}}_{\dee{v}}&= \underbrace{(1)}_{u}\underbrace{\int f(x)\dee{x}}_{v} - \int\underbrace{ \left(\int f(x)\dee{x}\right)}_{v}\underbrace{\dee{x}}_{\dee{u}} \end{align*}

In order to figure out the first product (and the second integrand), you need to know the antiderivative of $f(x)$--but that's exactly what you're trying to figure out!

Solution

Suppose we choose $\dee{v} = f(x)\dee{x}\text{,}$ $u=1\text{.}$ Then $v = \displaystyle\int f(x)\dee{x}\text{,}$ and $\dee{u}=\dee{x}\text{.}$ So, our integral becomes:

\begin{align*} \int \underbrace{(1)}_{u}\underbrace{f(x)\dee{x}}_{\dee{v}}&= \underbrace{(1)}_{u}\underbrace{\int f(x)\dee{x}}_{v} - \int\underbrace{ \left(\int f(x)\dee{x}\right)}_{v}\underbrace{\dee{x}}_{\dee{u}} \end{align*}

In order to figure out the first product (and the second integrand), you need to know the antiderivative of $f(x)$--but that's exactly what you're trying to figure out! So, using integration by parts has not eased your pain.

We note here that in certain cases, such as $\int\log x \dee{x}$ (Example 1.7.8 in your text), it is useful to choose $\dee{v}=1\dee{x}\text{.}$ This is similar to, but crucially different from, the do-nothing method in this problem.

Exercise6(*)
Hint

You'll probably want to use integration by parts. (It's the title of the section, after all). You'll break the integrand into two parts, integrate one, and differentiate the other. Would you rather integrate $\log x\text{,}$ or differentiate it?

$\dfrac{x^2\log x}{2} - \dfrac{x^2}{4} + C$

Solution

For integration by parts, we want to break the integrand into two pieces, multiplied together. There is an obvious choice for how to do this: one piece is $x\text{,}$ and the other is $\log x\text{.}$ Remember that one piece will be integrated, while the other is differentiated. The question is, which choice will be more helpful. After some practice, you'll get the hang of making the choice. For now, we'll present both choices--but when you're writing a solution, you don't have to write this part down. It's enough to present your choice, and then a successful computation is justification enough.

\begin{equation*} \IBP{x}{x}{1}{\dfrac{1}{2}x^2}{\log x}{\dfrac{1}{x}}{??} \end{equation*}

In Example 1.7.8, we found the antiderivative of logarithm, but it wasn't trivial. We might reasonably want to avoid using this complicated antiderivative. Indeed, Option 2 (differentiating logarithm, antidifferentiating $x$) looks promising--when we multiply the blue equations, we get something easily integrable-- so let's not even bother going deeper into Option 1.

That is, we perform integration by parts with $u=\log x$ and $\dee{v}=x\,\dee{x}\text{,}$ so that $\dee{u}=\frac{\dee{x}}{x}$ and $v = \frac{x^2}{2}\text{.}$

\begin{align*} \int\underbrace{ x}_{u}\underbrace{\log x\,\dee{x}}_{\dee{v}} &= \underbrace{\frac{x^2\log x}{2}}_{uv} - \int \underbrace{\frac{x^2}{2}}_v\ \underbrace{\frac{\dee{x}}{x}}_{\dee{u}} = \frac{x^2\log x}{2} -\frac{1}{2} \int x \dee{x} \\ &= \frac{x^2\log x}{2} - \frac{x^2}{4} + C \end{align*}
Exercise7(*)
Hint

This problem is similar to Question 6.

$- \dfrac{\log x}{6 x^6} - \dfrac{1}{36 x^6} + C$

Solution

Our integrand is the product of two functions, and there's no clear substitution. So, we might reasonably want to try integration by parts. Again, we have two obvious pieces: $\log x\text{,}$ and $x^{-7}\text{.}$ We'll consider our options for assigning these to $u$ and $\dee{v}\text{:}$

\begin{equation*} \IBP{x}{\log x}{\dfrac{1}{x}}{??}{x^{-7}}{-7x^{-8}}{\dfrac{1}{-6}x^{-6}} \end{equation*}

Again, we remember that logarithm has some antiderivative we found in Example 1.7.8, but it was something complicated. Luckily, we don't need to bother with it: when we multiply the red equations in Option 1, we get a perfectly workable integral.

We perform integration by parts with $u=\log x$ and $\dee{v}=x^{-7}\,\dee{x}\text{,}$ so that $\dee{u}=\frac{\dee{x}}{x}$ and $v = -\frac{x^{-6}}{6}\text{.}$

\begin{align*} \int \frac{\log x}{x^7}\,\dee{x} &=\underbrace{ -\log x\ \frac{x^{-6}}{6}}_{uv} + \int \underbrace{\frac{x^{-6}}{6}}_{-v}\ \underbrace{\frac{\dee{x}}{x}}_{du} = - \frac{\log x}{6 x^6} +\frac{1}{6} \int x^{-7} \dee{x} \\ &= - \frac{\log x}{6 x^6} - \frac{1}{36 x^6} + C \end{align*}
Exercise8(*)
Hint

Example 1.7.5 shows you how to find the antiderivative. Then the Fundamental Theorem of Calculus Part 2 gives you the definite integral.

$\pi$

Solution

To integrate by parts, we need to decide what to use as $u\text{,}$ and what to use as $\dee{v}\text{.}$ The salient parts of this integrand are $x$ and $\sin x\text{,}$ so we only need to decide which is $u$ and which $\dee{v}\text{.}$ Again, this process will soon become familiar, but to help you along we show you both options below.

\begin{equation*} \IBP{x}{x}{1}{\dfrac{1}{2}x^2}{\sin x}{\cos x}{-\cos x} \end{equation*}

The derivative and antiderivative of sine are almost the same, but $x$ turns into something simpler when we differentiate it. So, we choose Option 1.

We integrate by parts, using $u = x\text{,}$ $\dee{v} =\sin x \,\dee{x}$ so that $v=-\cos x$ and $\dee{u} = \dee{x}\text{:}$

\begin{gather*} \int_0^\pi x\sin x\,\dee{x} = \underbrace{-x\cos x}_{uv} \Big|_0^\pi -\int_0^\pi \underbrace{(-\cos x)}_v\,\underbrace{\dee{x}}_{\dee{u}} = \Big[-x\cos x +\sin x\Big]_0^\pi =-\pi(-1) =\pi \end{gather*}
Exercise9(*)
Hint

Compare to Question 8. Try to do this one all the way through without peeking at another solution!

$\dfrac{\pi}{2} -1$

Solution

When we have two functions multiplied like this, and there's no obvious substitution, our minds turn to integration by parts. We hope that our integral will be improved by differentiating one part and antidifferentiating the other. Let's see what our choices are:

\begin{equation*} \IBP{x}{x}{1}{\dfrac{1}{2}x^2}{\cos x}{-\sin x}{\sin x} \end{equation*}

Option 1 seems preferable. We integrate by parts, using $u = x\text{,}$ $\dee{v} =\cos x \,\dee{x}$ so that $v=\sin x$ and $\dee{u} = \dee{x}\text{:}$

\begin{gather*} \int_0^{\frac{\pi}{2}} x\cos x\,\dee{x} = \underbrace{x}_u\underbrace{\sin x}_v \Big|_0^{\frac{\pi}{2}} -\int_0^{\frac{\pi}{2}} \underbrace{\sin x}_v\,\underbrace{\dee{x}}_{\dee{u}} = \Big[x\sin x +\cos x\Big]_0^{\frac{\pi}{2}} =\frac{\pi}{2} - 1 \end{gather*}
Exercise10
Hint

If at first you don't succeed, try using integration by parts a few times in a row. Eventually, one part will go away.

$e^x\left(x^3-3x^2+6x - 6\right)+C$

Solution

This integrand is the product of two functions, with no obvious substitution. So, let's try integration by parts, with one part $e^x$ and one part $x^3\text{.}$

\begin{equation*} \IBP{x}{e^x}{e^x}{e^x}{x^3}{3x^2}{\frac{1}{4}x^4} \end{equation*}

At first glance, multiplying the red functions and multiplying the blue functions give largely equivalent integrands to what we started with--none of them with obvious antiderivatives. In previous questions, we were able to choose $u=x\text{,}$ and then $\dee{u}=\dee{x}\text{,}$ so the “$x$” in the integrand effectively went away. Here, we see that choosing $u=x^3$ will lead to $\dee{u}=3x^2\dee{x}\text{,}$ which has a lower power. If we repeatedly perform integration by parts, choosing $u$ to be the power of $x$ each time, then after a few iterations it should go away, because the third derivative of $x^3$ is a constant.

So, we start with Option 2: $u=x^3\text{,}$ $\dee{v}=e^x\dee{x}\text{,}$ $\dee{u}=3x^2\dee{x}\text{,}$ and $v = e^x\text{.}$

\begin{align*} \int x^3 e^x \dee{x}&= \underbrace{x^3}_{u}\underbrace{e^x}_{v} - \int \underbrace{e^x}_v \cdot\underbrace{ 3x^2 \dee{x}}_{\dee{u}}\\ &= {x^3}{e^x} -3 \int {e^x} \cdot x^2 \dee{x}\\ \end{align*}

Now, we take $u=x^2$ and $\dee{v}=e^x\dee{x}\text{,}$ so $\dee{u}=2x\dee{x}$ and $v=e^x\text{.}$ We're only using integration by parts on the actual integral--the rest of the function stays the way it is.

\begin{align*} &=x^3e^x -3 \left[\underbrace{x^2e^x}_{uv} - \int \underbrace{e^x}_{v} \cdot \underbrace{2x\dee{x}}_{\dee{u}}\right]\\ &=x^3e^x - 3x^2e^x + 6\int x e^x\dee{x}\\ \end{align*}

Continuing, we take $u=x$ and $\dee{v}=e^x\dee{x}\text{,}$ so $\dee{u}=\dee{x}$ and $v=e^x\text{.}$ This is the step where the polynomial part of the integrand finally disappears.

\begin{align*} &=x^3e^x - 3x^2e^x + 6\left[\underbrace{xe^x}_{uv} - \int \underbrace{ e^x}_v \underbrace{\dee{x}}_{\dee{u}}\right]\\ &=x^3e^x-3x^2e^x+6xe^x - 6e^x+C\\ &=e^x\left(x^3-3x^2+6x - 6\right)+C \end{align*}

Let's check that this makes sense: the derivative of $e^x\left(x^3-3x^2+6x - 6\right)+C$ should be $x^3e^x\text{.}$ We differentiate using the product rule.

\begin{align*} \diff{}{x}\left\{e^x\left(x^3-3x^2+6x - 6\right)+C\right\} &=e^x\left(x^3-3x^2+6x - 6\right)+e^x\left(3x^2-6x+6\right)\\ &=e^x\left(x^3-3x^2+3x^2+6x-6x-6+6\right)=x^3e^x \end{align*}

Remark: In order to be technically correct in our antidifferentiation, we should add the $+C$ as soon as we do the first integration by parts. However, when we are using integration by parts, we usually end up evaluating an integral at the end, and we add the $+C$ at that point. Since the $+C$ comes up eventually, it is common practice to not clutter our calculations with it until the end.

Exercise11
Hint

Similarly to Question 10, look for a way to use integration by parts a few times to simplify the integrand until it is antidifferentiatable.

$\dfrac{x^2}{2}\log^3x - \dfrac{3x^2}{4}\log^2 x + \dfrac{3x^2}{4}\log x - \dfrac{3x^2}{8}+C$

Solution

Since our integrand is two functions multiplied together, and there isn't an obvious substitution, let's try integration by parts. Here are our salient options.

\begin{equation*} \IBP{x}{x}{1}{\dfrac{1}{2}x^2}{\log^3 x}{3\log^2 x \cdot\frac{1}{x}}{??} \end{equation*}

This calls for some strategizing. Using the template of Example 1.7.8, we could probably figure out the antiderivative of $\log^3 x\text{.}$ Option 1 is tempting, because our $x$-term goes away. So, there might be a benefit there, but on the other hand, the antiderivative of $\log^3 x$ is probably pretty complicated.

Now let's consider Option 2. When we multiply the blue functions together, we get something similar to our original integrand, but the power of logarithm is smaller. If we were to iterate this method (using integration by parts a few times, always choosing $u$ to be the part with a logarithm) then eventually we would end up differentiating logarithm. This seems like a safer plan: let's do Option 2.

We use integration by parts with $u=\log^3 x\text{,}$ $\dee{v} = x\dee{x}\text{,}$ $\dee{u}=\frac{3}{x}\log^2 x \dee{x}\text{,}$ and $v = \frac{1}{2}x^2\text{.}$

\begin{align*} \int x \log^3 x \dee{x}&=\underbrace{\frac{1}{2}x^2\log^3 x}_{uv} - \int\underbrace{ \frac{3}{2}x\log^2 x \dee{x}}_{v\dee{u}}\\ &=\frac{1}{2}x^2\log^3 x- \frac{3}{2}\int x\log^2 x \dee{x}\\ \end{align*}

Continuing our quest to differentiate away the logarithm, we use integration by parts with $u=\log^2x\text{,}$ $\dee{v} = x\dee{x}\text{,}$ $\dee{u} = \dfrac{2}{x}\log x\dee{x}\text{,}$ and $v = \dfrac{1}{2}x^2\text{.}$

\begin{align*} &=\frac{1}{2}x^2\log^3x - \frac{3}{2}\left[ \underbrace{\frac{1}{2}x^2\log^2 x}_{uv} - \int \underbrace{x\log x \dee{x}}_{v\dee{u}} \right]\\ &=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{2}\int x\log x \dee{x}\\ \end{align*}

One last integration by parts: $u=\log x\text{,}$ $\dee{v}=x\dee{x}\text{,}$ $\dee{u}=\dfrac{1}{x}\dee{x}\text{,}$ and $v = \dfrac{1}{2}x^2\text{.}$

\begin{align*} &=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{2}\left[ \underbrace{\frac{1}{2}x^2\log x}_{uv} - \int \underbrace{\frac{1}{2}x\dee{x}}_{v\dee{u}} \right]\\ &=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{4}x^2\log x - \frac{3}{4}\int x\dee{x}\\ &=\frac{1}{2}x^2\log^3x - \frac{3}{4}x^2\log^2 x + \frac{3}{4}x^2\log x - \frac{3}{8}x^2+C \end{align*}

Once again, technically there is a $+C$ in the work after the first integration by parts, but we follow convention by conveniently suppressing it until the final integration.

Exercise12
Hint

Use integration by parts twice to get an integral with only a trigonometric function in it.

$(2-x^2)\cos x + 2x\sin x +C$

Solution

The integrand is the product of two functions, without an obvious substitution, so let's see what integration by parts can do for us.

\begin{equation*} \IBP{x}{x^2}{2x}{\frac{1}{3}x^3}{\sin x}{\cos x}{-\cos x} \end{equation*}

Neither option gives us something immediately integrable, but Option 1 replaces our $x^2$ term with a lower power of $x\text{.}$ If we repeatedly apply integration by parts, we can reduce this power to zero. So, we start by choosing $u=x^2$ and $\dee{v}=\sin x\dee{x}\text{,}$ so $\dee{u}=2x\dee{x}$ and $v=-\cos x\text{.}$

\begin{align*} \int x^2\sin x\dee{x} &=\underbrace{ -x^2\cos x}_{uv} +\underbrace{ \int 2x\cos x \dee{x}}_{-v\dee{u}}\\ &=-x^2\cos x + 2 \int x \cos x \dee{x}\\ \end{align*}

Using integration by parts again, we want to be differentiating (not antidifferentiating) $x\text{,}$ so we choose $u=x\text{,}$ $\dee{v}=\cos x \dee{x}\text{,}$ and then $\dee{u}=\dee{x}$ ($x$ went away!), $v=\sin x\text{.}$

\begin{align*} &=-x^2\cos x + 2\left[\underbrace{x\sin x}_{uv}-\int \underbrace{\sin x \dee{x}}_{v\dee{u}} \right]\\ &=-x^2\cos x + 2x\sin x+2\cos x +C\\ &=(2-x^2)\cos x + 2x\sin x +C \end{align*}
Exercise13
Hint

If you let $u=\log t$ in the integration by parts, then $\dee{u}$ works quite nicely with the rest of the integrand.

$\left( t^3 - \frac{5}{2}t^2+6t \right)\log t -\frac{1}{3}t^3 +\frac{5}{4}t^2-6t+C$

Solution

This problem is similar to Questions 6 and 7: integrating a polynomial multiplied by a logarithm. Just as in these questions, if we use integration by parts with $u=\log t\text{,}$ then $\dee{u} = \dfrac{1}{t}\dee{t}\text{,}$ and our new integrand will consist of powers of $t$--which are easy to antidifferentiate.

So, we use $u=\log t\text{,}$ $\dee{v}= 3t^2-5t+6\text{,}$ $\dee{u}=\frac{1}{t}\dee{t}\text{,}$ and $v = t^3 - \frac{5}{2}t^2+6t\text{.}$

\begin{align*} \int (3t^2-5t+6)\log t\dee{t}&=\underbrace{\log t}_u\left(\underbrace{ t^3 - \frac{5}{2}t^2+6t}_v\right) - \int \underbrace{\frac{1}{t}\left( t^3 - \frac{5}{2}t^2+6t\right)\dee{t}}_{v\dee{u}}\\ &=\left( t^3 - \frac{5}{2}t^2+6t \right)\log t- \int\left( t^2-\frac{5}{2}t+6 \right)\dee{t}\\ &=\left( t^3 - \frac{5}{2}t^2+6t \right)\log t -\frac{1}{3}t^3 +\frac{5}{4}t^2-6t+C \end{align*}
Exercise14
Hint

Those square roots are a little disconcerting-- get rid of them with a substitution.

$e^{\sqrt{s}}\left(2s - 4\sqrt{s} +4\right)+C$

Solution

Before we jump to integration by parts, we notice that the square roots lend themselves to substitution. Let's take $w=\sqrt{s}\text{.}$ Then $\dee{w}=\dfrac{1}{2\sqrt{s}}\dee{s}\text{,}$ so $2w\dee{w}=\dee{s}\text{.}$

\begin{align*} \int \sqrt{s}e^{\sqrt{s}}\dee{s}&=\int w\cdot e^w\cdot 2w\dee{w} = 2\int w^2e^w\dee{w}\\ \end{align*}

Now we have nearly the situation of Question 10. We can repeatedly use integration by parts with $u$ as the power of $w$ to get rid of the polynomial part. We'll start with $u=w^2$ $\dee{v}=e^w\dee{w}\text{,}$ $\dee{u}=2w\dee{w}\text{,}$ and $v=e^w\text{.}$

\begin{align*} &=2\left[\underbrace{w^2e^w}_{uv} - \int \underbrace{2we^w\dee{w}}_{v\dee{u}} \right]\\ &=2w^2e^w - 4\int we^w \dee{w}\\ \end{align*}

We use integration by parts again, this time with $u=w\text{,}$ $\dee{v}=e^w\dee{w}\text{,}$ $\dee{u}=\dee{w}\text{,}$ and $v=e^w\text{.}$

\begin{align*} &=2w^2e^w - 4\left[\underbrace{we^w}_{uv} - \int\underbrace{ e^w\dee{w}}_{v\dee{u}}\right]\\ &=2w^2e^w - 4we^w +4e^w+C\\ &=e^w\left(2w^2 - 4w +4\right)+C\\ &=e^{\sqrt{s}}\left(2s - 4\sqrt{s} +4\right)+C \end{align*}
Exercise15
Hint

This can be solved using the same ideas as Example 1.7.8 in your text.

$x\log^2 x -2x\log x +2x+C$

Solution

Let's use integration by parts. What are our parts? We have a few options.

• Solution 1: Following Example 1.7.8, we choose $u=\log^2 x$ and $\dee{v}=\dee{x}\text{,}$ so that $\dee{u}=\frac{2}{x}\log x \dee{x}$ and $v=x\text{.}$ \begin{align*} \int \log^2 x \dee{x}&= \underbrace{x\log^2x}_{uv} - \int \underbrace{ 2\log x \dee{x}}_{v\dee{u}}\\ \end{align*}

Here we can either use the antiderivative of logarithm from memory, or re-derive it. We do the latter, using integration by parts with $u=\log x\text{,}$ $\dee{v}=2\dee{x}\text{,}$ $\dee{u}=\frac{1}{x}\dee{x}\text{,}$ and $v=2x\text{.}$

\begin{align*} &=x\log^2 x - \left[\underbrace{2x\log x}_{uv} - \int \underbrace{2\dee{x}}_{v\dee{u}}\right]\\ &=x\log^2 x - 2x\log x +2x+C \end{align*}
• Solution 2: Our integrand is two functions multiplied together: $\log x$ and $\log x\text{.}$ So, we will use integration by parts with $u=\log x\text{,}$ $\dee{v}=\log x\text{,}$ $\dee{u}=\frac{1}{x}\dee{x}\text{,}$ and (using the antiderivative of logarithm, found in Example 1.7.8 in the text) $v=x\log x -x\text{.}$ \begin{align*} \int \log^2 x \dee{x}&=(\underbrace{\log x}_u) (\underbrace{x\log x - x}_{v}) -\int (\underbrace{x\log x-x}_v)\underbrace{ \frac{1}{x}\dee{x}}_{\dee{u}}\\ &= x\log^2 x - x\log x - \int \left(\log x -1\right)\dee{x}\\ &= x\log^2 x - x\log x - \left[(x\log x - x )-x\right]+C\\ &=x\log^2 x -2x\log x +2x+C \end{align*}
Exercise16
Hint

Not every integral should be evaluated using integration by parts.

$e^{x^2+1}+C$

Solution

This is your friendly reminder that to a person with a hammer, everything looks like a nail. The integral in the problem is a classic example of an integral to solve using substitution. We have an “inside function,” $x^2+1\text{,}$ whose derivative shows up multiplied to the rest of the integrand. We take $u=x^2+1\text{,}$ then $\dee{u}=2x\dee{x}\text{,}$ so

\begin{equation*} \int 2xe^{x^2+1}\dee{x} = \int e^u\dee{u}=e^u+C = e^{x^2+1}+C \end{equation*}
Exercise17(*)
Hint

You know, or can easily look up, the derivative of arccosine. You can use a similar trick as the book did when antidifferentiating other inverse trigonometric functions in Example 1.7.9.

$y \arccos y - \sqrt{1-y^2} + C$

Solution

In Example 1.7.9, we saw that integration by parts was useful when the integrand has a derivative that works nicely when multiplied by $x\text{.}$ We use the same idea here. Let $u = \arccos y$ and $\dee{v} = \dee{y}\text{,}$ so that $v=y$ and $du = -\frac{\dee{y}}{\sqrt{1-y^2}}\text{.}$

\begin{align*} \int\arccos y\,\dee{y} &= \underbrace{y \arccos y}_{uv} +\int \underbrace{\frac{y}{\sqrt{1-y^2}}\dee{y}}_{-v\dee{u}} \\ \end{align*}

Using the substitution $u=1-y^2\text{,}$ $\dee{u}=-2y\dee{y}\text{,}$

\begin{align*} &=y\arccos y -\frac{1}{2} \int u^{-1/2}\dee{u}\\ &=y\arccos y - u^{1/2} +C\\ &= y \arccos y - \sqrt{1-y^2} + C \end{align*}
Exercise18(*)
Hint

After integrating by parts, do some algebraic manipulation to the integral until it's clear how to evaluate it.

$2y^2\arctan(2y) - y + \frac12\arctan(2y) + C$

Solution

We integrate by parts, using $u = \arctan(2y)\text{,}$ $\dee{v} = 4y \,\dee{y}\text{,}$ so that $v=2y^2$ and $du = \frac{2\, \dee{y}}{1+(2y)^2}\text{:}$

\begin{gather*} \int 4y\arctan(2y) \,\dee{y} = \underbrace{2y^2\arctan(2y)}_{uv} - \int\underbrace{ \frac{4y^2}{(2y)^2+1} \,\dee{y}}_{v\dee{u}} \end{gather*}

The integrand $\frac{4y^2}{(2y)^2+1}$ is a rational function. So the remaining integral can be evaluated using the method of partial fractions, starting with long division. But it is easier to just notice that $\frac{4y^2}{4y^2+1} = \frac{4y^2+1}{4y^2+1} - \frac{1}{4y^2+1}\text{.}$ We therefore have:

\begin{gather*} \int \frac{4y^2}{4y^2+1} \,\dee{y} = \int \left(1 - \frac{1}{4y^2+1}\right)\,\dee{y} = y - \frac12\arctan(2y) + C \end{gather*}

\begin{gather*} \int 4y\arctan(2y) \,\dee{y} = 2y^2\arctan(2y) - y + \frac{1}{2}\arctan(2y) + C \end{gather*}
Exercise19
Hint

After integration by parts, use a substitution.

$\dfrac{x^3}{3}\arctan x- \dfrac{1}{6}(1+x^2) + \dfrac{1}{6}\log(1+x^2)+C$

Solution

We've got an integrand that consists of two functions multiplied together, and no obvious substitution. So, we think about integration by parts. Let's consider our options. Note in Example 1.7.9, we found that the antiderivative of arctangent is $x\arctan x -\frac{1}{2}\log(1+x^2)+C\text{.}$

\begin{equation*} \IBP{x}{\arctan x}{\dfrac{1}{1+x^2}}{x\arctan x -\dfrac{1}{2}\log(1+x^2)}{x^2}{2x}{\dfrac{1}{3}x^3} \end{equation*}

• Option 1: Option 1 seems likelier. Let's see how it plays out. We use integration by parts with $u=\arctan x\text{,}$ $\dee{v}=x^2 \dee{x}\text{,}$ $\dee{u}=\frac{\dee{x}}{1+x^2}\text{,}$ and $v=\frac{1}{3}x^3\text{.}$

\begin{align*} \displaystyle\int x^2\arctan x \dee{x}&=\underbrace{\frac{x^3}{3}\arctan x}_{uv} - \int \underbrace{\frac{x^3}{3(1+x^2)}\dee{x}}_{v\dee{u}}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{3}\int \frac{x^3}{1+x^2}\dee{x}\\ \end{align*}

This is starting to look like a candidate for a substitution! Let's try the denominator, $s =1+x^2\text{.}$ Then $\dee{s} = 2x\dee{x}\text{,}$ and $x^2 = s-1\text{.}$

\begin{align*} &= \frac{x^3}{3}\arctan x- \frac{1}{6}\int \frac{x^2}{\textcolor{red}{1+x^2}}\cdot \textcolor{blue}{2x\dee{x}}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6}\int \frac{s-1}{\textcolor{red}{s}}\textcolor{blue}{\dee{s}}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6} \int 1 - \frac{1}{s}\dee{s}\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6}s + \frac{1}{6}\log|s|+C\\ &=\frac{x^3}{3}\arctan x- \frac{1}{6}(1+x^2) + \frac{1}{6}\log(1+x^2)+C \end{align*}
• Option 2: What if we had tried the other option? That is, $u=x^2\text{,}$ $\dee{u} = 2x\dee{x}\text{,}$ $\dee{v} = \arctan x\text{,}$ and $v = x\arctan x - \frac{1}{2}\log(1+x^2)\text{.}$ It's not always the case that both options work, but sometimes they do. (They are almost never of equal difficulty.) This solution takes advantage of two previously hard-won results: the antiderivatives of logarithm and arctangent.
\begin{align*} \int x^2\arctan x \dee{x}&=\underbrace{x^2}_{u}\underbrace{\left(x\arctan x - \frac{1}{2}\log(1+x^2)\right)}_v - \int \underbrace{\left(x\arctan x - \frac{1}{2}\log(1+x^2)\right)}_v\cdot \underbrace{2x\dee{x}}_{\dee{u}}\\ &=x^3\arctan x - \frac{x^2}{2}\log(1+x^2)- 2\int x^2\arctan x \dee{x} + \int x\log(1+x^2)\dee{x}\\ \end{align*}

Adding \quad $2\displaystyle\int x^2\arctan x \dee{x}$ \quad to both sides:

\begin{align*} 3\int x^2\arctan x \dee{x}&=x^3\arctan x - \frac{x^2}{2}\log(1+x^2) + \int x\log(1+x^2)\dee{x}\\ \int x^2\arctan x \dee{x}&=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{3}\int x\log(1+x^2)\dee{x}\\ \end{align*}

Using the substitution $s=1+x^2\text{,}$ $\dee{s}=2x\dee{x}\text{:}$

\begin{align*} &=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{6}\int\log s\dee{s}\\ \end{align*}

Using the antiderivative of logarithm found in Example 1.7.8,

\begin{align*} &=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{6}\left(s\log s - s\right)+C\\ &=\frac{x^3}{3}\arctan x - \frac{x^2}{6}\log(1+x^2) + \frac{1}{6}\left((1+x^2)\log (1+x^2) - (1+x^2)\right)+C\\ &=\frac{x^3}{3}\arctan x + \left[-\frac{x^2}{6}+\frac{1+x^2}{6}\right]\log(1+x^2)-\frac{1}{6}(1+x^2)+C\\ &=\frac{x^3}{3}\arctan x +\frac{1}{6}\log(1+x^2)-\frac{1}{6}(1+x^2)+C \end{align*}
Exercise20
Hint

This example is similar to Example 1.7.10 in the text. The functions $e^{x/2}$ and $\cos(2x)$ both do not substantially alter when we differentiate or antidifferentiate them. If we use integration by parts twice, we'll end up with an expression that includes our original integral. Then we can just solve for the original integral in the equation, without actually integrating.

$\dfrac{2}{17}e^{x/2}\cos(2x)+\dfrac{8}{17}e^{x/2}\sin(2x)+C$

Solution

This example is similar to Example 1.7.10 in the text. The functions $e^{x/2}$ and $\cos(2x)$ both do not substantially alter when we differentiate or antidifferentiate them. If we use integration by parts twice, we'll end up with an expression that includes our original integral. Then we can just solve for the original integral in the equation, without actually antidifferentiating.

Let's use $u=\cos(2x)$ and $\dee{v} = e^{x/2}\dee{x}\text{,}$ so $\dee{u}=-2\sin(2x)\dee{x}$ and $v=2e^{x/2}\text{.}$

\begin{align*} \int e^{x/2}\cos(2x)\dee{x}&= \underbrace{2e^{x/2}\cos(2x)}_{uv}- \int\underbrace{-4e^{x/2}\sin(2x)\dee{x}}_{v\dee{u}}\\ &=2e^{x/2}\cos(2x)+4\int e^{x/2}\sin(2x)\dee{x}\\ \end{align*}

Similarly to our first integration by parts, we use $u=\sin(2x)\text{,}$ $\dee{v}=e^{x/2}\dee{x}\text{,}$ $\dee{u}=2\cos(2x)\dee{x}\text{,}$ and $v=2e^{x/2}\text{.}$

\begin{align*} &=2e^{x/2}\cos(2x)+4\left[ \underbrace{2e^{x/2}\sin(2x)}_{uv} -\int\underbrace{4e^{x/2}\cos(2x)\dee{x}}_{v\dee{u}} \right]\\ \end{align*}

So, we've found the equation

\begin{align*} \color{red}{\int e^{x/2}\cos(2x)\dee{x}}&=2e^{x/2}\cos(2x)+8e^{x/2}\sin(2x)-\color{red}{16\int e^{x/2}\cos(2x)\dee{x}} +C\\ \end{align*}

We add \quad$16\displaystyle\int e^{x/2}\cos(2x)\dee{x}$ \quad to both sides.

\begin{align*} \color{red}{17\int e^{x/2}\cos(2x)\dee{x}}&=2e^{x/2}\cos(2x)+8e^{x/2}\sin(2x)+C\\ \int e^{x/2}\cos(2x)\dee{x}&=\frac{2}{17}e^{x/2}\cos(2x)+\frac{8}{17}e^{x/2}\sin(2x)+C \end{align*}

Remark: remember that $C$ is a stand-in for “we can add any real constant”. Since $C$ can be any number in $(-\infty,\infty)\text{,}$ also $\frac{C}{17}$ can be any number in $(-\infty,\infty)\text{.}$ So, rather than write $\frac{C}{17}$ in the last line, we re-named $\frac{C}{17}$ to $C\text{.}$

Exercise21
Hint

This looks a bit like a substitution problem, because we have an “inside function.”

It might help to review Example 1.7.11.

$\dfrac{x}{2} \big[\sin(\log x) - \cos (\log x)\big]+C$

Solution

• Solution 1: This question looks like a substitution, since we have an “inside function.” So, let's see where that leads: let $u=\log x\text{.}$ Then $\dee{u}=\dfrac{1}{x}\dee{x}\text{.}$ We don't see this right away in our function, but we can bring it into the function by multiplying and dividing by $x\text{,}$ and noting from our substitution that $e^u=x\text{.}$ \begin{align*} \int \sin(\log x)\dee{x}&=\int \frac{x\sin(\log x)}{x}\dee{x}\\ &=\int e^u \sin u \dee{u}\\ \end{align*}

Using the result of Example 1.7.11:

\begin{align*} &=\frac{1}{2}e^u\left(\sin u - \cos u\right)+C\\ &=\frac{1}{2}e^{\log x}\left(\sin(\log x) - \cos (\log x)\right)+C\\ &=\frac{1}{2}x \left(\sin(\log x) - \cos (\log x)\right)+C \end{align*}
• Solution 2: It's not clear how to antidifferentiate the integrand, but we can certainly differentiate it. So, keeping in mind the method of Example 1.7.11 in the text, we take $u=\sin(\log x)$ and $\dee{v}=\dee{x}\text{,}$ so $\dee{u}=\frac{1}{x}\cos(\log x)\dee{x}$ and $v=x\text{.}$

\begin{align*} \int \sin(\log x)\dee{x}&= \underbrace{x\sin(\log x)}_{uv} - \int \underbrace{ \cos(\log x)\dee{x} }_{v\dee{u}}\\ \end{align*}

Continuing on, we again use integration by parts, with $u=\cos(\log x)\text{,}$ $\dee{v}=\dee{x}\text{,}$ $\dee{u}=-\frac{1}{x}\sin(\log x)\dee{x}\text{,}$ and $v=x\text{.}$

\begin{align*} &=x\sin(\log x) - \bigg[\underbrace{x\cos(\log x)}_{uv} + \int\underbrace{\sin(\log x)}_{-v\dee{u}}\dee{x} \bigg]\\ \end{align*}

That is, we have

\begin{align*} \int \sin(\log x)\dee{x}&=x\left[\sin(\log x) - \cos(\log x)\right] - \int \sin(\log x)\dee{x}+C\\ \end{align*}

Adding \quad$\int \sin(\log x)\dee{x}$ \quad to both sides,

\begin{align*} 2\int \sin(\log x)\dee{x}&=x\left[\sin(\log x) - \cos(\log x)\right]+C\\ \int \sin(\log x)\dee{x}&=\frac{x}{2}\left[\sin(\log x) - \cos(\log x)\right]+C \end{align*}

Remark: remember that $C$ is a stand-in for “we can add any real constant”. Since $C$ can be any number in $(-\infty,\infty)\text{,}$ also $\frac{C}{2}$ can be any number in $(-\infty,\infty)\text{.}$ So, rather than write $\frac{C}{2}$ in the last line, we re-named $\frac{C}{2}$ to $C\text{.}$

Exercise22
Hint

Start by simplifying.

$\dfrac{2^x}{\log 2}\left(x - \dfrac{1}{\log 2}\right)+C$

Solution

We begin by simplifying the integrand.

\begin{align*} \int 2^{x+\log_2 x} \dee{x}&= \int 2^{x}\cdot 2^{\log_2 x} \dee{x}= \int 2^{x}\cdot x \dee{x}\\ \end{align*}

This is similar to the integral $\displaystyle\int xe^x \dee{x}\text{,}$ which we saw in Example 1.7.1. Let's write $2=e^{\log 2}$ to take advantage of the easy integrability of $e^x\text{.}$

\begin{align*} &=\int x\cdot e^{x\log 2} \dee{x}\\ \end{align*}

We use integration by parts with $u=x\text{,}$ $\dee{v}=e^{x\log 2}\dee{x}\text{;}$ $\dee{u}=\dee{x}\text{,}$ $v = \frac{1}{\log 2}e^{x\log 2}\text{.}$ (Remember $\log 2$ is a constant. If you'd prefer, you can do a substitution with $s= x\log 2$ first, to have a simpler exponent of $e\text{.}$)

\begin{align*} &=\underbrace{\frac{x}{\log 2}e^{x\log 2}}_{uv} - \int \underbrace{\frac{1}{\log 2}e^{x\log 2}\dee{x}}_{v\dee{u}}\\ &=\frac{x}{\log 2}e^{x\log 2} - \frac{1}{(\log 2)^2}e^{x\log 2}+C\\ &=\frac{x}{\log 2}2^x - \frac{1}{(\log 2)^2}2^x+C \end{align*}
Exercise23
Hint

$\sin(2x) = 2\sin x \cos x$

$2e^{\cos x}[1-\cos x]+C$

Solution

It's not obvious where to start, but in general it's nice to have the arguments of our trig functions the same. So, we use the identity $\sin(2x)=2\sin x \cos x\text{.}$

\begin{align*} \int e^{\cos x}\sin(2x)\dee{x}&=2\int e^{\cos x}\cos x \sin x\\ \end{align*}

Now we can use the substitution $w=\cos x\text{,}$ $\dee{w}=-\sin x \dee{x}\text{.}$

\begin{align*} &=-2\int we^w\dee{w}\\ \end{align*}

From here the integral should look more familiar. We can use integration by parts with $u=w\text{,}$ $\dee{v} = e^w\dee{w}\text{,}$ $\dee{u}=\dee{w}\text{,}$ and $v=e^w\text{.}$

\begin{align*} &=-2\left[\underbrace{we^w}_{uv} - \int \underbrace{e^w\dee{w}}_{v\dee{u}}\right]\\ &=2e^w\left[1-w\right]+C\\ &=2e^{\cos x}[1-\cos x]+C \end{align*}
Exercise24(*)
Hint

You'll want to do an integration by parts for (a)--check the end result to get a guess as to what your parts should be. A trig identity and some amount of algebraic manipulation will be necessary to get the final form.

(a) We integrate by parts with $u=\sin^{n-1}x$ and $\dee{v}=\sin x\,\dee{x}\text{,}$ so that $\dee{u}=(n-1)\sin^{n-2}x\cos x$ and $v=-\cos x\text{.}$

\begin{align*} \int\sin^nx\,\dee{x} &=\underbrace{-\sin^{n-1}x\ \cos x}_{uv}+\underbrace{(n-1)\int \cos^2x\ \sin^{n-2}x\ \dee{x}}_{-\int v\dee{u}}\\ \end{align*}

Using the identity $\sin^2 x + \cos^2 x =1 \text{,}$

\begin{align*} &=-\sin^{n-1}x\ \cos x+(n-1)\int (1-\sin^2 x)\sin^{n-2}x\ \dee{x}\\ &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} -(n-1)\int\sin^{n}x\ \dee{x} \end{align*}

Moving the last term on the right hand side to the left hand side gives

\begin{align*} n\int\sin^nx\,\dee{x} &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} \end{align*}

Dividing across by $n$ gives the desired reduction formula.

\qquad (b) $\dfrac{35}{256}\pi\approx0.4295$

Solution

(a) The “parts” in the integrand are powers of sine. Looking at the right hand side of the reduction formula, we see that it looks a little like the derivative of $\sin^{n-1}x\text{,}$ although not exactly. So, let's integrate by parts with $u=\sin^{n-1}x$ and $\dee{v}=\sin x\,\dee{x}\text{,}$ so that $\dee{u}=(n-1)\sin^{n-2}x\cos x$ and $v=-\cos x\text{.}$

\begin{align*} \int\sin^nx\,\dee{x} &=\underbrace{-\sin^{n-1}x\ \cos x}_{uv}+\underbrace{(n-1)\int \cos^2x\ \sin^{n-2}x\ \dee{x}}_{-\int v\dee{u}}\\ \end{align*}

Using the identity $\sin^2 x + \cos^2 x =1 \text{,}$

\begin{align*} &=-\sin^{n-1}x\ \cos x+(n-1)\int (1-\sin^2 x)\sin^{n-2}x\ \dee{x}\\ &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} -(n-1)\int\sin^{n}x\ \dee{x} \end{align*}

Moving the last term on the right hand side to the left hand side gives

\begin{align*} n\int\sin^nx\,\dee{x} &=-\sin^{n-1}x\ \cos x+(n-1)\int\sin^{n-2}x\ \dee{x} \end{align*}

Dividing across by $n$ gives the desired reduction formula.

(b) By the reduction formula of part (a), if $n \ge 2\text{,}$

\begin{gather*} \int_0^{\pi/2}\sin^n(x)\,\dee{x}= \frac{n-1}{n}\int_0^{\pi/2}\sin^{n-2}(x)\,\dee{x} \end{gather*}

since $\sin 0=\cos\frac{\pi}{2}=0\text{.}$ Applying this reduction formula, with $n=8,6,4,2\text{:}$

\begin{align*} \int_0^{\pi/2}\sin^8(x)\,\dee{x} &=\frac{7}{8}\int_0^{\pi/2}\sin^6(x)\,\dee{x} =\frac{7}{8}\cdot\frac{5}{6}\int_0^{\pi/2}\sin^4(x)\,\dee{x} =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\int_0^{\pi/2}\sin^2(x)\,\dee{x} \\ &=\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\int_0^{\pi/2}\,\dee{x} =\frac{7}{8}\cdot\frac{5}{6}\cdot\frac{3}{4}\cdot\frac{1}{2}\cdot\frac{\pi}{2} =\frac{35}{256}\pi \end{align*}

Using a calculator, we see this is approximately $0.4295\text{.}$

Exercise25(*)
Hint

See Examples 1.7.9 and 1.6.5 for refreshers on integrating arctangent, and using washers.

Remember $\tan^2 x + 1 = \sec^2 x\text{,}$ and $\sec^2x$ is easy to integrate.

(a) Area: $\dfrac{\pi}{4}-\dfrac{\log 2}{2}$

(b) Volume: $\dfrac{\pi^2}{2}-\pi$

Solution

(a) The sketch is the figure on the left below. By integration by parts with $u=\arctan x\text{,}$ $\dee{v}=\dee{x}\text{,}$ $v=x$ and $\dee{u}=\frac{1}{1+x^2}\,\dee{x}\text{,}$ and then the substitution $s=1+x^2\text{,}$

\begin{align*} A&=\int_0^1\arctan x\dee{x}=\underbrace{ x\arctan x}_{uv}\Big|_0^1-\int_0^1\underbrace{\frac{x}{1+x^2}\,\dee{x}}_{v\dee{u}} =\arctan 1-\half\log(1+x^2)\Big|_0^1\\ &=\frac{\pi}{4}-\frac{\log 2}{2} \end{align*}

(b) We'll use horizontal washers as in Example 1.6.5.

• We cut $R$ into thin horizontal strips of width $\dee{y}$ as in the figure on the right above.
• When we rotate $R$ about the $y$--axis, each strip sweeps out a thin washer

• whose inner radius is $r_{in}=\tan y$ and outer radius is $r_{out}=1\text{,}$ and
• whose thickness is $\dee{y}$ and hence
• whose volume $\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi(1-\tan^2 y)\dee{y}\text{.}$
• As our bottommost strip is at $y=0$ and our topmost strip is at $y=\frac{\pi}{4}$ (since at the top $x=1$ and $x=\tan y$), the total \begin{align*} \text{Volume} &= \int _0^{\frac{\pi}{4}} \pi(1-\tan^2 y)\dee{y} = \int _0^{\frac{\pi}{4}} \pi(2-\sec^2 y)\dee{y} =\pi\big[2y -\tan y\big]_0^{\frac{\pi}{4}} \\ &= \frac{\pi^2}{2}-\pi \end{align*}
Exercise26(*)
Hint

Your integral can be broken into two integrals, which yield to two different integration methods.

$\pi \left( \dfrac{17 e^{18}-4373}{36} \right)$

Solution

For a fixed value of $x\text{,}$ if we rotate about the $x$-axis, we form a washer of inner radius $B(x)$ and outer radius $T(x)$ and hence of area $\pi [T(x)^2 - B(x)^2]\text{.}$ We integrate this function from $x=0$ to $x=3$ to find the total volume $V\text{:}$

\begin{align*} V &= \int_0^3 \pi [T(x)^2 - B(x)^2]\,\dee{x} \\ &= \pi \int_0^3 (\sqrt{x}e^{3x})^2 - (\sqrt{x}(1+2x))^2 \,\dee{x} \\ &= \pi \int_0^3 \big( xe^{6x} - (x+4x^2+4x^3) \big) \,\dee{x} \\ &= \pi \int_0^3 xe^{6x} \,\dee{x} - \pi \Big[ \frac{x^2}{2} + \frac{4x^3}{3} + x^4\Big]_{0}^3 \\ &= \pi \int_0^3 xe^{6x} \,\dee{x} - \pi \Big[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4\Big] \end{align*}

For the first integral, we use integration by parts with $u(x) = x\text{,}$ $\dee{v} = e^{6x}\dee{x}\text{,}$ so that $\dee{u}=\dee{x}$ and $v(x)=\frac16e^{6x}\text{:}$

\begin{align*} \int_0^3 xe^{6x} \,\dee{x} &=\underbrace{ \frac{xe^{6x}}{6}}_{uv}\bigg|_{0}^3 - \int_0^3 \underbrace{\frac{1}{6}e^{6x} \,\dee{x} }_{v\dee{u}}\\ &= \frac{3e^{18}}{6} - 0 - \frac{1}{36} e^{6x} \bigg|_{0}^3 = \frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg). \end{align*}

Therefore, the total volume is

\begin{gather*} V = \pi \bigg[\frac{e^{18}}{2} - \bigg( \frac{e^{18}}{36} - \frac1{36} \bigg) \bigg] - \pi \bigg[\frac{3^2}{2} + \frac{4\cdot3^3}{3} + 3^4 \bigg] = \pi \bigg( \frac{17 e^{18}-4373}{36} \bigg). \end{gather*}
Exercise27(*)
Hint

Think, first, about how to get rid of the square root in the argument of $f''\text{,}$ and, second, how to convert $f''$ into $f'\text{.}$ Note that you are told that $f'(2) = 4$ and $f(0) = 1\text{,}$ $f(2) = 3\text{.}$

$12$

Solution

To get rid of the square root in the argument of $f''\text{,}$ we make the change of variables (also called “substitution”) $x=t^2,\dee{x}=2t\,\dee{t}\text{.}$

\begin{align*} \int_0^4 f''\big(\sqrt{x}\big)\,\dee{x} &= 2\int_0^2 tf''(t)\,\dee{t} \end{align*}

Then, to convert $f''$ into $f'\text{,}$ we integrate by parts with $u=t\text{,}$ $\dee{v}=f''(t)\,\dee{t}\text{,}$ $v=f'(t) \text{.}$

\begin{align*} \int_0^4 f''\big(\sqrt{x}\big)\,\dee{x} &= 2\bigg\{\Big[\underbrace{tf'(t)}_{uv}\Big]_0^2-\int_0^2\!\!\!\underbrace{ f'(t)\,\dee{t}}_{v\dee{u}}\bigg\} \\ &=2\Big[tf'(t)-f(t)\Big]_0^2 \\ &=2\big[2f'(2)-f(2)+f(0)\big]=2\big[2\times 4-3+1\big]\\ &=12 \end{align*}
Exercise28
Hint

Interpret the limit as a right Riemann sum.

$\dfrac{2}{e}$

Solution

As we saw in Section 1.1, there are many different ways to interpret a limit as a Riemann sum. In the absence of instructions that restrain our choices, we go with the most convenient interpretations.

With that in mind, we choose:

• that our Riemann sum is a right Riemann sum (because we see $i\text{,}$ not $i-1$ or $i-\frac{1}{2}$)
• $\Delta x = \frac{2}{n}$ (because it is multiplied by the rest of the integrand, and also shows up multiplied by $i$),
• then $x_i = a+i\Delta x = \frac{2}{n}i-1\text{,}$ which leads us to $a=-1$ and
• $f(x) = xe^x\text{.}$
• Finally, since $\Delta x = \frac{b-a}{n}=\frac{2}{n}$ and $a=-1\text{,}$ we have $b=1\text{.}$

So, the limit is equal to the definite integral

\begin{align*} \lim_{n \to \infty}\sum_{i=1}^n \frac{2}{n}\left(\frac{2}{n}i-1\right)e^{\frac{2}{n}i-1}&=\int_{-1}^1x e^x\dee{x}\\ \end{align*}

which we evaluate using integration by parts with $u=x\text{,}$ $\dee{v}=e^x\dee{x}\text{,}$ $\dee{u}=\dee{x}\text{,}$ and $v=e^x\text{.}$

\begin{align*} &=\Big[\underbrace{xe^x}_{uv} \Big]_{-1}^1 - \int_{-1}^1 \underbrace{e^x\dee{x}}_{v\dee{u}}\\ &=\left(e+\frac{1}{e}\right) - \left(e-\frac{1}{e}\right) = \frac{2}{e} \end{align*}

Exercises1.8.4Exercises

Exercise1
Hint

(e)

Solution

If $u=\cos x\text{,}$ then $-\dee{u}=\dee{x}\text{.}$ If $n \neq -1\text{,}$ then

\begin{align*} \int_0^{\pi/4} \sin x \cos^n x \dee{x}&= - \int_1^{1/\sqrt2} u^n \dee{u} = \left[-\frac{1}{n+1}u^{n+1}\right]_1^{1/\sqrt2}=\frac{1}{n+1}\left(1-\frac{1}{\sqrt{2}^{n+1}}\right)\\ \end{align*}

If $n=-1\text{,}$ then

\begin{align*} \int_0^{\pi/4} \sin x \cos^n x \dee{x}&= - \int_1^{1/\sqrt2} u^n \dee{u} = - \int_1^{1/\sqrt2} \frac{1}{u} \dee{u} =\bigg[ -\log|u|\bigg]_1^{1/\sqrt2} \\ & =-\log\left(\frac{1}{\sqrt2}\right) = \frac{1}{2}\log 2 \end{align*}

So, (e) $n$ can be any real number.

Exercise2
Hint

Use the substitution $u=\sec x\text{.}$

$\dfrac{1}{n}\sec^n x +C$

Solution

We use the substitution $u=\sec x\text{,}$ $\dee{u}=\sec x \tan x \dee{x}\text{.}$

\begin{align*} \int \sec^n x \tan x \dee{x}&=\int \sec^{n-1}x \cdot \sec x \tan x \dee{x} = \int u^{n-1}\dee{u}\\ \end{align*}

Since $n$ is positive, $n-1 \neq -1\text{,}$ so we antidifferentiate using the power rule.

\begin{align*} &=\frac{u^n}{n}+C = \frac{1}{n}\sec^n x +C \end{align*}
Exercise3
Hint

Divide both sides of the second identity by $\cos^2 x\text{.}$

We divide both sides by $\cos^2 x\text{,}$ and simplify.

\begin{align*} \sin^2x+\cos^2 x &=1\\ \frac{\sin^2x+\cos^2 x }{\cos^2 x}&=\frac{1}{\cos^2 x}\\ \frac{\sin^2x}{\cos^2 x}+1&=\sec^2 x\\ \tan^2 x+1&=\sec^2 x \end{align*}
Solution

We divide both sides by $\cos^2 x\text{,}$ and simplify.

\begin{align*} \sin^2x+\cos^2 x &=1\\ \frac{\sin^2x+\cos^2 x }{\cos^2 x}&=\frac{1}{\cos^2 x}\\ \frac{\sin^2x}{\cos^2 x}+1&=\sec^2 x\\ \tan^2 x+1&=\sec^2 x \end{align*}
Exercise4(*)
Hint

See Example 1.8.6. Note that the power of cosine is odd, and the power of sine is even (it's zero).

$\sin x-\dfrac{\sin^3 x}{3} +C$

Solution

The power of cosine is odd, and the power of sine is even (zero). Following the strategy in the text, we make the substitution $u=\sin x\text{,}$ so that $\dee{u}=\cos x\,\dee{x}$ and $\cos^2 x = 1-\sin^2 x = 1-u^2\text{:}$

\begin{align*} \int \cos^3x\,\dee{x} &=\int (1-\sin^2x)\cos x\,\dee{x} =\int (1-u^2)\,\dee{u}\\ &=u-\frac{u^3}3+C =\sin x-\frac{\sin^3 x}{3}+C \end{align*}
Exercise5(*)
Hint

See Example 1.8.7. All you need is a helpful trig identity.

$\dfrac{\pi}{2}$

Solution

Using the trig identity $\cos^2 x=\dfrac{1+\cos(2x)}{2}\text{,}$ we have \begin{alignat*}{3} \int \cos^2 x\dee{x} &= \frac{1}{2}\int_0^\pi \big[1+\cos(2x)\big]\dee{x} &= \frac{1}{2} \Big[x+\frac{1}{2}\sin(2x)\Big]_0^\pi &=\frac{\pi}{2} \end{alignat*}

Exercise6(*)
Hint

The power of cosine is odd, so we can reserve one cosine for $\dee{u}\text{,}$ and turn the rest into sines using the identity $\sin^2 x + \cos^2 x =1\text{.}$

$\dfrac{\sin^{37}t}{37}-\dfrac{\sin^{39}t}{39}+C$

Solution

Since the power of cosine is odd, following the strategies in the text, we make the substitution $u=\sin t\text{,}$ so that $\dee{u}=\cos t\,\dee{t}$ and $\cos^2 t = 1-\sin^2 t = 1-u^2\text{.}$

\begin{align*} \int\sin^{36}t\cos^3t\,\dee{t} &=\int\sin^{36}t \, (1-\sin^2t)\cos t\,\dee{t} =\int u^{36}(1-u^2)\,\dee{u}\\ &=\frac{u^{37}}{37}-\frac{u^{39}}{39}+C =\frac{\sin^{37}t}{37}-\frac{\sin^{39}t}{39}+C \end{align*}
Exercise7
Hint

Since the power of sine is odd (and positive), we can reserve one sine for $\dee{u}\text{,}$ and turn the rest into cosines using the identity $\sin^2 + \cos^2 x =1\text{.}$

$\dfrac{1}{3\cos^3 x} - \dfrac{1}{\cos x}+C$

Solution

Since the power of sine is odd (and positive), we can reserve one sine for $\dee{u}\text{,}$ and turn the rest into cosines using the identity $\sin^2 + \cos^2 x =1\text{.}$ This allows us to use the substitution $u=\cos x\text{,}$ $\dee{u}=-\sin x\dee{x}\text{,}$ and $\sin^2 x = 1-\cos^2 x = 1-u^2\text{.}$

\begin{align*} \int \dfrac{\sin^3 x}{\cos ^4 x} \dee{x}&=\int \frac{\sin^2 x}{\cos^4 x}\sin x\dee{x} =\int -\frac{1-u^2}{u^4}\dee{u}\\ &=\int\left( -\frac{1}{u^{4}}+\frac{1}{u^{2}}\right)\dee{u} = \frac{1}{3u^3}-\frac{1}{u}+C\\ &=\frac{1}{3\cos^3 x} - \frac{1}{\cos x}+C \end{align*}
Exercise8
Hint

When we have even powers of sine and cosine both, we use the identities in the last two lines of Equation 1.8.3.

$\displaystyle\frac{\pi}{8} -\frac{9\sqrt3}{64}$

Solution

Both sine and cosine have even powers (four and zero, respectively), so we don't have the option of using a substitution like $u=\sin x$ or $u=\cos x\text{.}$ Instead, we use the identity $\sin^2 \theta = \dfrac{1-\cos(2\theta)}{2}\text{.}$

\begin{align*} \int_0^{\pi/3} \sin^{4}x \dee{x} &= \int_0^{\pi/3} \left(\sin^{2}x\right)^2 \dee{x} = \int_0^{\pi/3} \left(\frac{1-\cos(2x)}{2}\right)^2 \dee{x}\\ &=\frac{1}{4}\int_0^{\pi/3}\left(1-2\cos(2x)+\cos^2(2x)\right) \dee{x}\\ &=\frac{1}{4}\int_0^{\pi/3}\left(1-2\cos(2x)\right) \dee{x}+\frac{1}{4}\int_0^{\pi/3}\cos^2(2x) \dee{x}\\ \end{align*}

We can antidifferentiate the first integral right away. For the second integral, we use the identity \quad $\cos^2 \theta = \dfrac{1+\cos(2\theta)}{2}\text{,}$ with $\theta=2x\text{.}$

\begin{align*} &=\frac{1}{4}\Big[x - \sin(2x)\Big]_{0}^{\pi/3} + \frac{1}{8}\int_0^{\pi/3}(1+\cos(4x)) \dee{x}\\ &=\frac{1}{4}\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]+\frac{1}{8}\Big[x+\frac{1}{4}\sin(4x)\Big]_0^{\pi/3}\\ &=\frac{1}{4}\left[\frac{\pi}{3}-\frac{\sqrt{3}}{2}\right]+\frac{1}{8}\left[\frac{\pi}{3}-\frac{\sqrt{3}}{8}\right]\\ &=\frac{\pi}{8} -\frac{9\sqrt3}{64} \end{align*}
Exercise9
Hint

Since the power of sine is odd, you can use the substitution $u=\cos x\text{.}$

$-\cos x + \dfrac{2}{3}\cos^3 x - \dfrac{1}{5}\cos^5 x +C$

Solution

Since the power of sine is odd, we can reserve one sine for $\dee{u}\text{,}$ and change the remaining four into cosines. This sets us up to use the substitution $u=\cos x\text{,}$ $\dee{u}=-\sin x \dee{x}\text{.}$

\begin{align*} \int \sin^{5}x \dee{x}&=\int \sin^4 x \cdot \sin x \dee{x} = \int (1-\cos^2 x)^2 \sin x \dee{x}\\ &=-\int (1-u^2)^2 \dee{u} = -\int (1-2u^2+u^4)\dee{u}\\ &=-u+\frac{2}{3}u^3-\frac{1}{5}u^5+C\\ &=-\cos x + \frac{2}{3}\cos^3 x - \frac{1}{5}\cos^5 x +C \end{align*}
Exercise10
Hint

Which substitution will work better: $u=\sin x\text{,}$ or $u=\cos x\text{?}$

$\dfrac{1}{2.2}\sin^{2.2}x+C$

Solution

If we use the substitution $u=\sin x\text{,}$ then $\dee{u}=\cos x \dee{x}\text{,}$ which very conveniently shows up in the integrand.

\begin{align*} \int \sin^{1.2}x\cos x \dee{x}&=\int u^{1.2}\dee{u} = \frac{u^{2.2}}{2.2}+C = \frac{1}{2.2}\sin^{2.2}x+C \end{align*}

Note this is exactly the strategy described in the text when the power of cosine is odd. The non-integer power of sine doesn't cause a problem.

Exercise11
Hint

Try a substitution.

$\dfrac{1}{2}\tan^2 x+C\text{,}$ or equivalently, $\dfrac{1}{2}\sec^2 +C$

Solution

• Solution 1: Let's use the substitution $u=\tan x\text{,}$ $\dee{u} = \sec^2 x \dee{x}\text{:}$ \begin{equation*} \int \tan x \sec^2 x \dee{x} = \int u \dee{u} = \frac{1}{2}u^2+C = \frac{1}{2}\tan^2 x +C \end{equation*}
• Solution 2: We can also use the substitution $u=\sec x\text{,}$ $\dee{u} = \sec x \tan x \dee{x}\text{:}$ \begin{equation*} \int\tan x \sec^2 x \dee{x} = \int u \dee{u} = \frac{1}{2}u^2+C = \frac{1}{2}\sec^2 x +C \end{equation*}

We note that because $\tan^2x$ and $\sec^2 x$ only differ by a constant, the two answers are equivalent.

Exercise12(*)
Hint

For practice, try doing this in two ways, with different substitutions.

$\dfrac{1}{7}\sec^7 x -\dfrac{1}{5}\sec^5 x + C$

Solution

• Solution 1: Substituting $u=\cos x\text{,}$ $\dee{u}=-\sin x\,\dee{x}\text{,}$ $\sin^2 x= 1-\cos^2x=1-u^2\text{,}$ gives \begin{align*} \int \tan^3 x \sec^5x \,\dee{x} &=\int\frac{\sin^3 x}{\cos^8 x}\,\dee{x} =\int\frac{(1-\cos^2 x)\sin x}{\cos^8 x}\,\dee{x} =-\int\frac{1-u^2}{u^8}\,\dee{u} \\ &=-\Big[\frac{u^{-7}}{-7}-\frac{u^{-5}}{-5}\Big]+C =\frac{1}{7}\sec^7 x -\frac{1}{5}\sec^5 x + C \end{align*}
• Solution 2: Alternatively, substituting $u=\sec x\text{,}$ $\dee{u}=\sec x\tan x\,\dee{x}\text{,}$ $\tan^2 x= \sec^2x-1=u^2-1\text{,}$ gives \begin{align*} \int \tan^3 x \sec^5x \,\dee{x} &=\int \tan^2 x \sec^4x\ (\tan x\sec x)\,\dee{x} =\int (u^2-1) u^4\,\dee{u} \\ &=\Big[\frac{u^{7}}{7}-\frac{u^{5}}{5}\Big]+C =\frac{1}{7}\sec^7 x -\frac{1}{5}\sec^5 x + C \end{align*}
Exercise13(*)
Hint

A substitution will work. See Example 1.8.14 for a template for integrands with even powers of secant.

$\displaystyle\frac{\tan^{49}x}{49}+\frac{\tan^{47}x}{47}+C$

Solution

Use the substitution $u=\tan x\text{,}$ so that $\dee{u}=\sec^2 x\,\dee{x}\text{:}$

\begin{align*} \int\sec^4x\,\tan^{46}x\,\dee{x} &=\int(\tan^2x+1) \tan^{46}x\, \sec^2 x\,\dee{x} =\int (u^2+1)u^{46}\,\dee{u} \\ &=\frac{u^{49}}{49}+\frac{u^{47}}{47}+C =\frac{\tan^{49}x}{49}+\frac{\tan^{47}x}{47}+C \end{align*}
Exercise14
Hint

Try the substitution $u=\sec x\text{.}$

$\dfrac{1}{3.5}\sec^{3.5}x - \dfrac{1}{1.5}\sec^{1.5}x+C$

Solution

We use the substitution $u=\sec x\text{,}$ $\dee{u} = \sec x \tan x \dee{x}\text{.}$ Then $\tan^2 x = \sec^2 x - 1 = u^2-1\text{.}$

\begin{align*} \int \tan^3 x \sec^{1.5} x \dee{x} &= \int \tan^2 x \cdot \sec^{0.5}x \cdot \sec x \tan x \dee{x}\\ &=\int(u^2-1)u^{0.5} \dee{u} = \int \left(u^{2.5} - u^{0.5}\right) \dee{u}\\ &=\frac{u^{3.5}}{3.5} - \frac{u^{1.5}}{1.5}+C\\ &=\frac{1}{3.5}\sec^{3.5}x - \frac{1}{1.5}\sec^{1.5}x+C \end{align*}

Note this solution used the same method as Example 1.8.13 for the case that the power of tangent is odd and there is at least one secant.

Exercise15
Hint

Compare to Question 14.

$\dfrac{1}{4}\sec^4 x - \dfrac{1}{2}\sec^2 x +C$

Solution

As in Question 14, we have an odd power of tangent and at least one secant. So, we can use the substitution $u=\sec x\text{,}$ $\dee{u}=\sec x \tan x \dee{x}\text{,}$ and $\tan^2 x = \sec^2 x -1=u^2-1\text{.}$

\begin{align*} \int \tan^3x\sec^2x \dee{x}&=\int \tan^2 x \sec x \cdot \sec x \tan x \dee{x}\\ &=\int(u^2-1)u \dee{u} = \int \left(u^3-u\right) \dee{u}\\ &=\frac{1}{4}u^4 - \frac{1}{2}u^2+C\\ &=\frac{1}{4}\sec^4 x - \frac{1}{2}\sec^2 x +C \end{align*}
Exercise16
Hint

What is the derivative of tangent?

$\dfrac{1}{5}\tan^5 x +C$

Solution

In contrast to Questions 14 and 15, we do not have an odd power of tangent, so we should consider a different substitution. Luckily, if we choose $u=\tan x\text{,}$ then $\dee{u}=\sec^2 x \dee{x}\text{,}$ and this fits our integrand nicely.

\begin{align*} \int \tan^4 x \sec^2 x \dee{x}&=\int u^4 \dee{u}=\frac{1}{5}u^5+C = \frac{1}{5}\tan^5 x +C \end{align*}
Exercise17
Hint

Don't be scared off by the non-integer power of secant. You can still use the strategies in the notes for an odd power of tangent.

$\dfrac{1}{1.3}\sec^{1.3}x + \dfrac{1}{0.7}\cos^{0.7}x+C$

Solution

• Solution 1: Since the power of tangent is odd, let's try to use the substitution $u=\sec x\text{,}$ $\dee{u} = \sec x \tan x \dee{x}\text{,}$ and $\tan^2 x = \sec^2 x -1 = u^2-1\text{,}$ as in Questions 14 and 15. In order to make this work, we need to see $\sec x \tan x \dee{x}$ in the integrand, so we do a little algebraic manipulation. \begin{align*} \int \tan^3 x \sec^{-0.7}x \dee{x}&= \int \dfrac{\tan^3 x}{\sec^{0.7 x}} \dee{x} = \int \dfrac{\tan^3 x}{\sec^{1.7 x}}\sec x \dee{x}\\ &=\int \frac{\tan^2x}{\sec^{1.7}x}\cdot \sec x \tan x \dee{x}\\ &=\int \frac{u^2-1}{u^{1.7}} \dee{u} = \int \left(u^{0.3}-u^{-1.7}\right) \dee{u}\\ &=\frac{u^{1.3}}{1.3} + \frac{1}{0.7u^{0.7}}+C\\ &=\frac{1}{1.3}\sec^{1.3}x + \frac{1}{0.7\sec^{0.7}x}+C\\ &=\frac{1}{1.3}\sec^{1.3}x + \frac{1}{0.7}\cos^{0.7}x+C \end{align*}
• Solution 2: Let's convert the secants and tangents to sines and cosines. \begin{align*} \int \tan^3 x \sec^{-0.7}x \dee{x}&= \int \frac{\sin^3 x}{\cos^3 x}\cdot \cos^{0.7}x \dee{x}\\ &=\int\frac{\sin^3 x}{\cos^{2.3}x} \dee{x}=\int \frac{\sin^2 x}{\cos^{2.3}x}\cdot\sin x \dee{x}\\ \end{align*}

Using the substitution $u=\cos x\text{,}$ $\dee{u}=-\sin \dee{x}\text{,}$ and $\sin^2 x = 1-\cos^2 x = 1-u^2\text{:}$

\begin{align*} & = -\int\frac{1-u^2}{u^{2.3}} \dee{u} = \int \left(-u^{-2.3}+u^{-0.3}\right) \dee{u}\\ &=\frac{1}{1.3}u^{-1.3} + \frac{1}{0.7}u^{0.7}+C\\ &=\dfrac{1}{1.3}\sec^{1.3}x + \dfrac{1}{0.7}\cos^{0.7}x+C \end{align*}
Exercise18
Hint

Since there are no secants in the problem, it's difficult to use the substitution $u=\sec x$ that we've enjoyed in the past. Example 1.8.12 in the text provides a template for antidifferentiating an odd power of tangent.

$=\dfrac{1}{4}\sec^4 x - \sec^2 x + \log|\sec x|+C$

Solution

We replace $\tan x$ with $\dfrac{\sin x}{\cos x}\text{.}$

\begin{align*} \int \tan^5 x \dee{x}&=\int\left(\frac{\sin x}{\cos x}\right)^5 \dee{x} = \int\frac{\sin^4 x}{\cos^5 x}\cdot \sin x \dee{x}\\ \end{align*}

Now we use the substitution $u=\cos x\text{,}$ $\dee{u}=-\sin x \dee{x}\text{,}$ and $\sin^2 x = 1-\cos^2 x = 1-u^2\text{.}$

\begin{align*} &=-\int\frac{(1-u^2)^2}{u^5} \dee{u} = \int \left(-u^{-5}+2u^{-3}-u^{-1}\right) \dee{u}\\ &=\frac{1}{4}u^{-4} - u^{-2}-\log|u|+C\\ &=\dfrac{1}{4}\sec^4 x - \sec^2 x - \log|\cos x|+C\\ &=\dfrac{1}{4}\sec^4 x - \sec^2 x + \log|\sec x|+C \end{align*}

where in the last line, we used the logarithm rule $\log(b^a) = a\log b\text{,}$ with $b^a = \cos x = \left(\sec x\right)^{-1}\text{.}$

Exercise19
Hint

Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. You'll want to use the identity $\tan^2x = \sec^2 x -1\text{,}$ then use the substitution $u=\tan x\text{,}$ $\dee{u}=\sec^2 x\dee{x}$ on (parhaps only a part of) the resulting integral. Example 1.8.16 show you how this can be accomplished.

$\dfrac{41}{45\sqrt{3}} - \dfrac{\pi}{6}$

Solution

Integrating even powers of tangent is surprisingly different from integrating odd powers of tangent. For even powers, we use the identity $\tan^2x = \sec^2 x -1\text{,}$ then use the substitution $u=\tan x\text{,}$ $\dee{u}=\sec^2 x\dee{x}$ on (parhaps only a part of) the resulting integral.

\begin{align*} \int_0^{\pi/6} \tan^6 x \dee{x}&=\int_0^{\pi/6} \tan^4 x(\sec^2 x -1) \dee{x}\\ &=\int_0^{\pi/6} \bigg(\underbrace{\tan^4 x \sec^2 x}_{u^4\dee{u}} - \tan^4 x\bigg)\dee{x}\\ &=\int_0^{\pi/6} \bigg(\tan^4 x \sec^2 x - \tan^2 x(\sec^2 x-1)\bigg)\dee{x}\\ &=\int_0^{\pi/6}\bigg( \tan^4 x \sec^2 x - \underbrace{\tan^2 x\sec^2 x}_{u^2\dee{u}}+\tan^2 x\bigg)\dee{x}\\ &=\int_0^{\pi/6}\bigg( \tan^4 x \sec^2 x - \tan^2 x\sec^2 x+(\underbrace{\sec^2x}_{\dee{u}}-1)\bigg)\dee{x}\\ &=\int_0^{\pi/6}\left( \tan^4 x - \tan^2 x+1\right)\sec^2 x\dee{x} - \int_0^{\pi/6} 1\dee{x}\\ \end{align*}

Note $\tan(0)=0\text{,}$ and $\tan(\pi/6)=1/\sqrt{3}\text{.}$

\begin{align*} &=\int_0^{1/\sqrt{3}}(u^4-u^2+1)\dee{u} - \big[ x\big]_0^{\pi/6}\\ &=\left[\frac{1}{5}u^5 - \frac{1}{3}u^3+u\right]_0^{1/\sqrt{3}} - \frac{\pi}{6}\\ &=\frac{1}{5\sqrt{3}^5} - \frac{1}{3\sqrt{3}^3}+\frac{1}{\sqrt{3}}-\frac{\pi}{6}\\ &=\dfrac{41}{45\sqrt{3}} - \dfrac{\pi}{6} \end{align*}
Exercise20
Hint

Since there is an even power of secant in the integrand, we can use the substitution $u=\tan x\text{.}$

$\dfrac{1}{11}+\dfrac{1}{9}$

Solution

Since there is an even power of secant in the integrand, we can reserve two secants for $\dee{u}$ and change the rest to tangents. That sets us up nicely to use the substitution $u=\tan x\text{,}$ $\dee{u}=\sec^2 x\dee{x}\text{.}$ Note $\tan(0)=0$ and $\tan(\pi/4)=1\text{.}$

\begin{align*} \int_0^{\pi/4} \tan^8 x \sec^4 x \dee{x}&=\int_0^{\pi/4} \tan^8 x( \tan^2 x+1) \sec^2 x\dee{x}\\ &=\int_0^{1} u^8 ( u^2 +1) \dee{u}\\ &=\int_0^1 u^{10}+u^8\dee{u}\\ &=\frac{1}{11}+\frac{1}{9} \end{align*}
Exercise21
Hint

How have we handled integration in the past that involved an odd power of tangent?

$2\sqrt{\sec x}+C$

Solution

• Solution 1: Let's use the substitution $u=\sec x\text{,}$ $\dee{u}=\sec x \tan x\dee{x}\text{.}$ In order to make this work, we need to see $\sec x \tan x$ in the integrand, so we start with some algebraic manipulation. \begin{align*} \int \tan x \sqrt{\sec x}\left(\frac{\sqrt{\sec x}}{\sqrt{\sec x}}\right) \dee{x}&=\int \frac{1}{\sqrt{\sec x}}\sec x\tan x\dee{x}\\ &=\int \frac{1}{\sqrt{u}}\dee{u}=2\sqrt{u}+C\\ &=2\sqrt{\sec x}+C \end{align*}
• Solution 2: Let's turn our secants and tangents into sines and cosines. \begin{align*} \int \tan x \sqrt{\sec x}\dee{x}&=\int \frac{\sin x}{\cos x\cdot\sqrt{\cos x}}\dee{x}=\int \frac{\sin x}{\cos^{1.5}x}\dee{x}\\ \end{align*}

We use the substitution $u=\cos x\text{,}$ $\dee{u}=-\sin x\dee{x}\text{.}$

\begin{align*} &=\int -u^{-1.5}\dee{u}=\frac{2}{\sqrt{u}}+C\\ &=2\sqrt{\sec x}+C \end{align*}
Exercise22
Hint

Remember $e$ is some constant. What are our strategies when the power of secant is even and positive? We've seen one such substitution in Example 1.8.15.

$\tan^{e+1}\theta\left( \dfrac{\tan^{6}\theta}{7+e}+\dfrac{3\tan^4\theta}{5+e}+\dfrac{3\tan^2\theta}{3+e}+\dfrac{1}{1+e} \right)+C$

Solution

Since the power of secant is even and positive, we can reserve two secants for $\dee{u}\text{,}$ and change the rest into tangents, setting the stage for the substitution $u = \tan \theta\text{,}$ $\dee{u}=\sec^2 \theta\dee{\theta}\text{.}$

\begin{align*} \int \sec^{8}\theta \tan^{e}\theta \dee{\theta}&=\int \sec^6 \theta \tan^e \theta \sec^2 \theta\dee{\theta}\\ &=\int (\tan^2 \theta +1)^3 \tan^e \theta \sec^2 \theta\dee{\theta}\\ &=\int (u^2+1)^3 \cdot u^e \dee{u}\\ &=\int (u^6+3u^4 +3u^2+1) \cdot u^e \dee{u}\\ &=\int (u^{6+e}+3u^{4+e} +3u^{2+e}+ u^e) \dee{u}\\ &=\frac{1}{7+e}u^{7+e}+\frac{3}{5+e}u^{5+e}+\frac{3}{3+e}u^{3+e}+\frac{1}{1+e}u^{1+e}+C\\ &=\frac{1}{7+e}\tan^{7+e}\theta+\frac{3}{5+e}\tan^{5+e}\theta+\frac{3}{3+e}\tan^{3+e}\theta+\frac{1}{1+e}\tan^{1+e}\theta+C\\ &=\tan^{1+e}\theta\left( \frac{\tan^{6}\theta}{7+e}+\frac{3\tan^4\theta}{5+e}+\frac{3\tan^2\theta}{3+e}+\frac{1}{1+e} \right)+C \end{align*}
Exercise23(*)
Hint

See Example 1.8.16 for a strategy for integrating powers of tangent.

(a) Using the trig identity $\tan^2x=\sec^2 x-1$ and the substitution $y=\tan x\text{,}$ $\dee{y}=\sec^2 x\ \dee{x}\text{,}$ \begin{alignat*}{3} \int\tan^nx\dee{x} &=\int\tan^{n-2}x\ \tan^2x\dee{x} &&=\int\tan^{n-2}x\ \sec^2x\dee{x}-\int\tan^{n-2}x\dee{x}\\ &=\int y^{n-2}\,\dee{y}-\int\tan^{n-2}x\dee{x} &&=\frac{y^{n-1}}{n-1}-\int\tan^{n-2}x\dee{x}\\ &=\frac{\tan^{n-1}x}{n-1} -\int\tan^{n-2}x\dee{x} \end{alignat*}

(b) $\displaystyle\frac{13}{15}-\frac{\pi}{4}\approx0.0813$

Solution

(a) Using the trig identity $\tan^2x=\sec^2 x-1$ and the substitution $y=\tan x\text{,}$ $\dee{y}=\sec^2 x\ \dee{x}\text{,}$ \begin{alignat*}{3} \int\tan^nx\dee{x} &=\int\tan^{n-2}x\ \tan^2x\dee{x} &&=\int\tan^{n-2}x\ \sec^2x\dee{x}-\int\tan^{n-2}x\dee{x}\\ &=\int y^{n-2}\,\dee{y}-\int\tan^{n-2}x\dee{x} &&=\frac{y^{n-1}}{n-1}-\int\tan^{n-2}x\dee{x}\\ &=\frac{\tan^{n-1}x}{n-1} -\int\tan^{n-2}x\dee{x} \end{alignat*}

(b) By the reduction formula of part (a),

\begin{align*} \int_0^{\pi/4}\tan^n(x)\,\dee{x}&= \left[\frac{\tan^{n-1}x}{n-1}\right]_{0}^{\pi/4}-\int_0^{\pi/4}\tan^{n-2}(x)\,\dee{x}\\ &=\frac{1}{n-1}-\int_0^{\pi/4}\tan^{n-2}(x)\,\dee{x} \end{align*}

for all integers $n\ge 2\text{,}$ since $\tan 0=0$ and $\tan\frac{\pi}{4}=1\text{.}$ We apply this reduction formula, with $n=6,4,2\text{.}$

\begin{align*} \int_0^{\pi/4}\tan^6(x)\,\dee{x} &=\frac{1}{5}-\int_0^{\pi/4}\tan^4(x)\,\dee{x} =\frac{1}{5}-\frac{1}{3}+\int_0^{\pi/4}\tan^2(x)\,\dee{x} =\frac{1}{5}-\frac{1}{3}+1-\int_0^{\pi/4}\,\dee{x}\cr &=\frac{1}{5}-\frac{1}{3}+1-\frac{\pi}{4} =\frac{13}{15}-\frac{\pi}{4} \end{align*}

Using a calculator, we see this is approximately $0.0813\text{.}$

Notice how much faster this was than the method of Question 19.

Exercise24
Hint

Write $\tan x = \dfrac{\sin x}{\cos x}\text{.}$

$\dfrac{1}{2\cos^2 x}+2\log|\cos x|-\dfrac{1}{2}\cos^2 x +C$

Solution

Recall $\tan x = \dfrac{\sin x}{\cos x}\text{.}$

\begin{align*} \int \tan^5 x \cos^2 x \dee{x}&=\int \frac{\sin^5 x}{\cos^5 x}\cos^2 x \dee{x} =\int \frac{\sin^5 x}{\cos^3 x}\dee{x}\\ \end{align*}

Substitute $u=\cos x\text{,}$ so $\dee{u}=-\sin x\dee{x}$ and $\sin^2 x = 1-\cos^2 x = 1-u^2\text{.}$

\begin{align*} &=\int \frac{\sin^4 x}{\cos^3 x} \sin x\dee{x} =-\int \frac{(1-u^2)^2}{u^3}\dee{u}\\ &=-\int \frac{1-2u^2+u^4}{u^3}\dee{u}= \int \left(-\frac{1}{u^3}+\frac{2}{u}-u\right)\dee{u}\\ &=\frac{1}{2u^2}+2\log|u|-\frac{1}{2}u^2+C\\ &=\frac{1}{2\cos^2 x}+2\log|\cos x|-\frac{1}{2}\cos^2 x +C \end{align*}
Exercise25
Hint

$\dfrac{1}{\cos \theta} = \sec \theta$

$\tan \theta +C$

Solution

We can use the definition of secant to make this integral look more familiar.

\begin{equation*} \int \frac{1}{\cos^2 \theta}\dee{\theta} = \int \sec^2\theta\dee{\theta} = \tan \theta +C \end{equation*}
Exercise26
Hint

$\cot x = \dfrac{\cos x}{\sin x}$

$\log|\sin x|+C$

Solution

We re-write $\cot x = \dfrac{\cos x}{\sin x}\text{,}$ and use the substitution $u=\sin x\text{,}$ $\dee{u}=\cos x\dee{x}\text{.}$

\begin{align*} \int \cot x\dee{x}&= \int \frac{\cos x}{\sin x}\dee{x} = \int \frac{1}{u}\dee{u}\\ &=\log|u|+C = \log|\sin x|+C \end{align*}
Exercise27
Hint

Try substituting.

$\dfrac{1}{2}\sin^2(e^x)+C$

Solution

• Solution 1: We begin with the obvious substitution, $w=e^x\text{,}$ $\dee{w}=e^x \dee{w}\text{.}$ \begin{align*} \int e^x\sin(e^x)\cos(e^x) \dee{x}&= \int \sin w \cos w \dee{w}\\ \end{align*}

Now we see another substitution, $u=\sin w\text{,}$ $\dee{u}=\cos w\dee{w}\text{.}$

\begin{align*} &=\int u\dee{u}=\frac{1}{2}u^2+C=\frac{1}{2}\sin^2 w +C\\ &=\frac{1}{2}\sin(e^x)+C \end{align*}
• Solution 2: Notice that $\diff{}{x}\{\sin(e^x)\} = e^x \cos(e^x)\text{.}$ This suggests to us the substitution $u=\sin(e^x)\text{,}$ $\dee{u} = e^x \cos(e^x)\dee{x}\text{.}$ \begin{align*} \int e^x\sin(e^x)\cos(e^x) \dee{x}&= \int u\dee{u} =\frac{1}{2}u^2+C = \frac{1}{2}\sin^2(e^x)+C \end{align*}
Exercise28
Hint

$(\sin^2x+2)\cos (\cos x) + 2\cos x\sin (\cos x) +C$

Solution

Since we have an “inside function,” we start with the substitution $s=\cos x\text{,}$ so $-\dee{s}=\sin x \dee{x}$ and $\sin^2 x = 1-\cos^2 x = 1-s^2\text{.}$

\begin{align*} \int \sin(\cos x)\sin^3 x \dee{x}&=\int \sin(\cos x) \cdot \sin^2 x \cdot \sin x \dee{x}\\ &=-\int \sin(s)\cdot (1-s^2) \dee{s}\\ \end{align*}

We use integration by parts with $u=(1-s^2)\text{,}$ $\dee{v}=\sin s \dee{s}\text{;}$ $\dee{u}=-2s\dee{s}\text{,}$ and $v = -\cos s\text{.}$

\begin{align*} &=-\left[-(1-s^2)\cos s - \int 2s\cos s \dee{s}\right]\\ &= (1-s^2)\cos s + \int 2s\cos s \dee{s}\\ \end{align*}

We integrate by parts again, with $u=2s\text{,}$ $\dee{v}=\cos s \dee{s}\text{;}$ $\dee{u}=2\dee{s}\text{,}$ and $v=\sin s\text{.}$

\begin{align*} &=(1-s^2)\cos s + 2s\sin s - \int 2\sin s\dee{s}\\ &=(1-s^2)\cos s + 2s\sin s +2\cos s +C\\ &=\sin^2 x\cdot\cos (\cos x) + 2\cos x\cdot\sin (\cos x) +2\cos (\cos x) +C\\ &=(\sin^2x+2)\cos (\cos x) + 2\cos x\cdot\sin (\cos x) +C \end{align*}
Exercise29
Hint

Try an integration by parts.

$\dfrac{x}{2}\sin^2 x - \dfrac{x}{4} +\dfrac{1}{4}\sin x \cos x+C$

Solution

Since the integrand is the product of polynomial and trigonometric functions, we suspect it might yield to integration by parts. There are a number of ways this can be accomplished.

• Solution 1: Before we choose parts, let's use the identity $\sin(2x) = 2\sin x \cos x\text{.}$ \begin{align*} \int x\sin x \cos x \dee{x}&=\frac{1}{2}\int x \sin(2x)\dee{x}\\ \end{align*}

Now let $u= x\text{,}$ $\dee{v}=\sin(2x)\dee{x}\text{;}$ $\dee{u}=\dee{x}\text{,}$ and $v=-\frac{1}{2}\cos (2x) \text{.}$ Using integration by parts:

\begin{align*} &=\frac{1}{2}\left[-\frac{x}{2}\cos (2x) +\frac{1}{2} \int \cos (2x) \dee{x}\right]\\ &=-\frac{x}{4}\cos (2x) +\frac{1}{8}\sin (2x) +C\\ &=-\frac{x}{4}(1-2\sin^2x) +\frac{1}{4}\sin x\cos x +C\\ &=-\frac{x}{4} + \frac{x}{2}\sin^2x+\frac{1}{4}\sin x \cos x +C \end{align*}
• Solution 2: If we let $u=x\text{,}$ then $\dee{u}=\dee{x}\text{,}$ and this seems desirable for integration by parts. If $u=x\text{,}$ then $\dee{v} = \sin x \cos x \dee{x}\text{.}$ To find $v$ we can use the substitution $u=\sin x\text{,}$ $\dee{u}=\cos x \dee{x}\text{.}$ \begin{align*} v=\int \sin x \cos x \dee{x}&=\int u \dee{u} = \frac{1}{2}u^2+C = \frac{1}{2}\sin^2 x +C\\ \end{align*}

So, we take $v = \frac{1}{2}\sin^2 x\text{.}$ Now we can apply integration by parts to our original integral.

\begin{align*} \int x\sin x \cos x \dee{x}&=\frac{x}{2}\sin^2 x - \int \frac{1}{2}\sin^2 x \dee{x}\\ \end{align*}

Apply the identity $\sin^2x = \dfrac{1-\cos(2x)}{2}\text{.}$

\begin{align*} &=\frac{x}{2}\sin^2 x - \frac{1}{4}\int 1-\cos(2 x) \dee{x}\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} +\frac{1}{8}\sin(2 x)+C\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} +\frac{1}{4}\sin x \cos x+C \end{align*}
• Solution 3: Let $u=x\sin x$ and $\dee{v}=\cos x \dee{x}\text{;}$ then $\dee{u} = (x\cos x + \sin x)\dee{x}$ and $v = \sin x\text{.}$ \begin{align*} \int x\sin x \cos x \dee{x}&=x\sin^2 x - \int \sin x(x\cos x + \sin x)\dee{x}\\ &=x\sin^2 x - \int x \sin x \cos x\dee{x} - \int \sin^2 x\dee{x}\\ \end{align*}

Apply the identity $\sin^2 x = \dfrac{1-\cos(2x)}{2}$ to the second integral.

\begin{align*} &=x\sin^2 x - \int x \sin x \cos x\dee{x} - \int \dfrac{1-\cos(2x)}{2}\dee{x}\\ &=x\sin^2 x - \int x \sin x \cos x\dee{x} - \frac{x}{2} +\frac{1}{4}\sin(2x)+C\\ \end{align*}

So, we have the equation

\begin{align*} \color{red}{\int x\sin x \cos x \dee{x}}&=x\sin^2 x -\textcolor{red}{ \int x \sin x \cos x\dee{x}} - \frac{x}{2} + \frac{1}{4}\sin(2x)+C\\ \color{red}{2\int x\sin x \cos x \dee{x}} &=x\sin^2 x - \frac{x}{2} + \frac{1}{4}\sin(2x)+C\\ \int x\sin x \cos x \dee{x}&=\frac{x}{2}\sin^2 x - \frac{x}{4} + \frac{1}{8}\sin(2x)+\frac{C}{2}\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} + \frac{1}{4}\sin x\cos x+\frac{C}{2} \end{align*} Since $C$ is an arbitrary constant that can take any number in $(-\infty,\infty)\text{,}$ also $\frac{C}{2}$ is an arbitrary constant that can take any number in $(-\infty,\infty)\text{,}$ so we're free to rename $\frac{C}{2}$ to $C\text{.}$

Exercises1.9.2Exercises

Exercise1(*)
Hint

The beginning of this section has a template for choosing a substitution. Your goal is to use a trig identity to turn the argument of the square root into a perfect square, so you can cancel $\sqrt{(\mbox{something})^2}=(\mbox{something})\text{.}$

(a) $x=\dfrac{4}{3}\sec\theta$

(b) $x=\dfrac{1}{2}\sin\theta$

(c) $x=5\tan\theta$

Solution

In the text, there is a template for choosing an appropriate substitution, but for this problem we will explain the logic of the choices.

The trig identities that we can use are:

\begin{align*} &1-\sin^2\theta=\cos^2\theta & &\tan^2 \theta +1 = \sec^2 \theta & &\sec^2 \theta - 1 =\tan^2 \theta\\ \end{align*}

They have the following forms:

\begin{align*} &\mbox{constant } - \mbox{ function} &&\mbox{function } + \mbox{ constant} & &\mbox{function } - \mbox{ constant} \end{align*}

In order to cancel out the square root, we should choose a substitution that will match the argument under the square root with the trig identity of the corresponding form.

(a) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function $9x^2-16\text{,}$ which has the form (function) $-$ (constant). This form matches the trig identity $\sec^2 \theta - 1 = \tan^2 \theta\text{.}$ We can set $x$ to be whatever we need it to be, but we don't have the same control over the constant, 16. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 16, we get

\begin{align*} 16\sec^2\theta - 16 &= 16\tan^2\theta\\ \end{align*}

What we want is a substitution that gives us

\begin{align*} 9x^2-16&=16\sec^2\theta - 16\\ \mbox{So,}\qquad 9x^2&=16\sec^2\theta\\ x &= \frac{4}{3}\sec\theta\\ \end{align*}

Using this substitution,

\begin{align*} \sqrt{9x^2-16}&=\sqrt{16\sec^2\theta-16}\\ &=\sqrt{16\tan^2\theta}\\ &=4\tan\theta \end{align*}

So, we eliminated the square root.

(b) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the square root in the denominator. Under the square root is the function $1-4x^2\text{,}$ which has the form (constant) $-$ (function). This form matches the trig identity $1-\sin^2\theta = \cos^2\theta\text{.}$ What we want is a substitution that gives us

\begin{align*} 1-4x^2&=1-\sin^2\theta\\ \mbox{So,}\qquad 4x^2&=\sin^2\theta\\ x &= \frac{1}{2}\sin\theta\\ \end{align*}

Using this substitution,

\begin{align*} \sqrt{1-4x^2}&=\sqrt{1-\sin^2\theta}\\ &=\sqrt{\cos^2\theta}\\ &=\cos\theta \end{align*}

So, we eliminated the square root.

(c) There's not an obvious non-trig substitution for evaluating this problem, so we want a trigonometric substitution to get rid of the fractional power. (That is, we want to eliminate the square root.) The function under the power is $25+x^2\text{,}$ which has the form (constant) $+$ (function). This form matches the trig identity $\tan^2 \theta + 1 = \sec^2 \theta\text{.}$ We can set $x$ to be whatever we need it to be, but we don't have the same control over the constant, 25. So, to make the substitution work, we use a different form of the trig identity: multiplying both sides by 25, we get

\begin{align*} 25\tan^2 \theta + 25 &= 25\sec^2 \theta\\ \end{align*}

What we want is a substitution that gives us

\begin{align*} 25+x^2&=25\tan^2\theta+25\\ \mbox{So,}\qquad x^2&=25\tan^2\theta\\ x &= 5\tan\theta\\ \end{align*}

Using this substitution,

\begin{align*} (25+x^2)^{-5/2}&=(25+25\tan^2\theta)^{-5/2}\\ &=(25\sec^2\theta)^{-5/2}\\ &=(5\sec\theta)^{-5} \end{align*}

So, we eliminated the square root.

Exercise2
Hint

You want to do the same thing you did in Question 1, but you'll have to complete the square first.

(a) $x-2=\sqrt{3}\sec u$

(b) $x-1=\sqrt{5}\sin u$

(c) $\left(2x+\dfrac{3}{2}\right) =\dfrac{\sqrt{31}}{2}\tan u$

(d) $x - \dfrac{1}{2}=\dfrac{1}{2}\sec u$

Solution

Just as in Question 1, we want to choose a trigonometric substitution that will allow us to eliminate the square roots. Before we can make that choice, though, we need to complete the square. In subsequent problems, we won't show the algebra behind completing the square, but for this problem we'll work it out explicitly. After some practice, you'll be able to do this step in your head for many cases.

After the squares are completed, the choice of trig substitution follows the logic outlined in the solutions to Question 1, or (equivalently) the template in the text.

1. The quadratic function under the square root is $x^2-4x+1\text{.}$ To complete the square, we match the non-constant terms to those of a perfect square.

\begin{align*} (ax+b)^2&=a^2x^2+2abx+b^2\\ \textcolor{red}{x^2}-\textcolor{blue}{4x}+1&=\textcolor{red}{a^2x^2} + \textcolor{blue}{2abx} +b^2 + c \quad\mbox{for some constant } c \end{align*}
• Looking at the leading term tells us $a=1\text{.}$
• Then the second term tells us $-4=2ab=2b\text{,}$ so $b=-2\text{.}$
• Finally, the constant terms give us $1=b^2+c=4+c\text{,}$ so $c=-3\text{.}$
\begin{equation*} \displaystyle\int \dfrac{1}{\sqrt{x^2-4x+1}}\dee{x}=\displaystyle\int \dfrac{1}{\sqrt{(x-2)^2-3}}\dee{x}=\displaystyle\int \dfrac{1}{\sqrt{\left(x-2\vphantom{\sqrt{3}}\right)^2-\sqrt{3}^2}}\dee{x} \end{equation*}

So we use the substitution $(x-2) = \sqrt{3}\sec u\text{,}$ which eliminates the square root:

\begin{equation*} \sqrt{\left(x-2\right)^2-3}=\sqrt{3\sec^2 u - 3} = \sqrt{3\tan^2 u} = \sqrt{3}\tan u \end{equation*}
2. The quadratic function under the square root is $-x^2+2x+4=-[x^2-2x-4]\text{.}$ To complete the square, we match the non-constant terms to those of a perfect square. We factored out the negative to make things a little easier — don't forget to put it back in before choosing a substitution!

\begin{align*} (ax+b)^2&=a^2x^2+2abx+b^2\\ \textcolor{red}{x^2}-\textcolor{blue}{2x}-4&=\textcolor{red}{a^2x^2} + \textcolor{blue}{2abx} +b^2 + c \quad\mbox{for some constant } c \end{align*}
• Looking at the leading term tells us $a=1\text{.}$
• Then the second term tells us $-2=2ab=2b\text{,}$ so $b=-1\text{.}$
• Finally, the constant terms give us $-4=b^2+c=1+c\text{,}$ so $c=-5\text{.}$
• Then $-x^2+2x+4 = -[x^2-2x-4]=-[(x-1)^2-5]=5-(x-1)^2\text{.}$
\begin{equation*} \displaystyle\int \dfrac{(x-1)^6}{(-x^2+2x+4)^{3/2}}\dee{x}=\displaystyle\int \dfrac{(x-1)^6}{(5-(x-1)^2)^{3/2}}\dee{x}=\displaystyle\int \dfrac{(x-1)^6}{\left(\sqrt{5}^2-\left(x-1\vphantom{\sqrt{3}}\right)^2\right)^{3/2}}\dee{x} \end{equation*}

So we use the substitution $(x-1) = \sqrt{5}\sin u\text{,}$ which eliminates the square root (fractional power):

\begin{equation*} (5-\left(x-1\right)^2)^{3/2}=\left(5-5\sin^2u\right)^{3/2} = \left(5\cos^2 u\right)^{3/2} = 5\sqrt{5}\cos^3 u \end{equation*}
3. The quadratic function under the square root is $4x^2+6x+10\text{.}$ To complete the square, we match the non-constant terms to those of a perfect square.

\begin{align*} (ax+b)^2&=a^2x^2+2abx+b^2\\ \textcolor{red}{4x^2}+\textcolor{blue}{6x}+10&=\textcolor{red}{a^2x^2} + \textcolor{blue}{2abx} +b^2 + c \quad\mbox{for some constant $c$} \end{align*}
• Looking at the leading term tells us $a=2\text{.}$
• Then the second term tells us $6=2ab=4b\text{,}$ so $b=\frac{3}{2}\text{.}$
• Finally, the constant terms give us $10=b^2+c=\frac{9}{4}+c\text{,}$ so $c=\frac{31}{4}\text{.}$
\begin{equation*} \displaystyle\int \dfrac{1}{\sqrt{4x^2+6x+10}}\dee{x}=\displaystyle\int \dfrac{1}{\sqrt{\left(2x+\frac{3}{2}\right)^2+\frac{31}{4}}}\dee{x} =\displaystyle\int \dfrac{1}{\sqrt{\left(2x+\frac{3}{2}\right)^2+\left(\frac{\sqrt{31}}{2}\right)^2}}\dee{x} \end{equation*}

So we use the substitution $\left(2x+\frac{3}{2}\right) =\frac{\sqrt{31}}{2}\tan u\text{,}$ which eliminates the square root:

\begin{equation*} \sqrt{\left(2x+\frac{3}{2}\right)^2+\frac{31}{4}}=\sqrt{\frac{31}{4}\tan^2 u +\frac{31}{4}} = \sqrt{\frac{31}{4}\sec^2 u} =\frac{\sqrt{31}}{2}\sec u \end{equation*}
4. The quadratic function under the square root is $x^2-x\text{.}$ To complete the square, we match the non-constant terms to those of a perfect square.

\begin{align*} (ax+b)^2&=a^2x^2+2abx+b^2\\ \textcolor{red}{x^2}-\textcolor{blue}{x}&=\textcolor{red}{a^2x^2} + \textcolor{blue}{2abx} +b^2 + c \quad\mbox{for some constant $c$} \end{align*}
• Looking at the leading term tells us $a=1\text{.}$
• Then the second term tells us $-1=2ab=2b\text{,}$ so $b=-\frac{1}{2}\text{.}$
• Finally, the constant terms give us $0=b^2+c=\frac{1}{4}+c\text{,}$ so $c=-\frac{1}{4}\text{.}$
\begin{equation*} \displaystyle\int \sqrt{x^2-x}\dee{x}=\displaystyle\int \sqrt{\left(x-\frac{1}{2}\right)^2 -\frac{1}{4}}\dee{x}=\displaystyle\int \sqrt{\left(x-\frac{1}{2}\right)^2 -\left(\frac{1}{2}\right)^2}\dee{x} \end{equation*}

So we use the substitution $(x-1/2) = \frac{1}{2}\sec u\text{,}$ which eliminates the square root:

\begin{equation*} \sqrt{\left(x-\frac{1}{2}\right)^2-\frac{1}{4}}=\sqrt{\frac{1}{4}{\sec\vphantom{|}}^2 u - \frac{1}{4}} = \sqrt{\frac{1}{4}\tan^2 u} =\frac{1}{2}\tan u \end{equation*}
Exercise3
Hint

Since $\theta$ is acute, you can draw it as an angle of a right triangle. The given information will let you label two sides of the triangle, and the Pythagorean Theorem will lead you to the third.

(a) $\dfrac{\sqrt{399}}{20}$

(b) $\dfrac{5\sqrt{2}}{7}$

(c) $\dfrac{\sqrt{x-5}}{2}$

Solution

1. If $\sin\theta=\dfrac{1}{20}$ and $\theta$ is between 0 and $\pi/2\text{,}$ then we can draw a right triangle with angle $\theta$ that has opposite side length 1, and hypotenuse length 20. By the Pythagorean Theorem, the adjacent side has length $\sqrt{20^2-1^2}=\sqrt{399}\text{.}$ So, $\cos\theta = \dfrac{\mathrm{adj}}{\mathrm{hyp}}=\dfrac{\sqrt{399}}{20}\text{.}$

We can do a quick “reasonableness” check here: $\frac{1}{20}$ is pretty close to 0, so we might expect $\theta$ to be pretty close to 0, and so $\cos \theta$ should be pretty close to 1. Indeed it is: $\dfrac{\sqrt{399}}{20}\approx \dfrac{\sqrt{400}}{20}=\dfrac{20}{20}=1\text{.}$

Alternately, we can solve this problem using identities.

\begin{align*} \sin^2 \theta + \cos^2 \theta &=1\\ \left(\frac{1}{20}\right)^2+ \cos^2 \theta &=1\\ \cos\theta &= \pm\sqrt{1-\frac{1}{400}}=\pm\frac{\sqrt{399}}{20}\\ \end{align*}

Since $0 \leq \theta \leq \frac{\pi}{2}\text{,}$ $\cos\theta \geq 0\text{,}$ so

\begin{align*} \cos\theta &= \frac{\sqrt{399}}{20} \end{align*}
2. If $\tan\theta=7$ and $\theta$ is between 0 and $\pi/2\text{,}$ then we can draw a right triangle with angle $\theta$ that has opposite side length 7 and adjacent side length 1. By the Pythagorean Theorem, the hypotenuse has length $\sqrt{7^2+1^2} = \sqrt{50}=5\sqrt{2}\text{.}$ So, $\csc\theta = \dfrac{\mathrm{hyp}}{\mathrm{opp}}=\dfrac{5\sqrt{2}}{7}\text{.}$

Again, we can do a quick reasonableness check. Since 7 is much larger than 1, the triangle we're thinking of doesn't look much like the triangle in our standardized picture above: it's really quite tall, with a small base. So, the opposite side and hypotenuse are pretty close in length. Indeed, $\dfrac{5\sqrt{2}}{7}\approx 7.071\text{,}$ so this dimension seems reasonable.

3. If $\sec\theta=\dfrac{\sqrt{x-1}}{2}$ and $\theta$ is between 0 and $\pi/2\text{,}$ then we can draw a right triangle with angle $\theta$ that has hypotenuse length $\sqrt{x-1}$ and adjacent side length 2. By the Pythagorean Theorem, the opposite side has length $\sqrt{\sqrt{x-1}^2 - 2^2} = \sqrt{x-1-4}=\sqrt{x-5}\text{.}$ So, $\tan\theta = \dfrac{\mathrm{opp}}{\mathrm{adj}}=\dfrac{\sqrt{x-5}}{2}\text{.}$

We can also solve this using identities. Note that since $\sec\theta$ exists, $\theta \neq \frac{\pi}{2}\text{.}$

\begin{align*} \tan^2\theta+1&=\sec^2\theta\\ \tan^2\theta+1&=\left(\frac{\sqrt{x-1}}{2}\right)^2=\frac{x-1}{4}\\ \tan\theta &= \pm\sqrt{\frac{x-1}{4}-1} = \pm\frac{\sqrt{x-5}}{2}\\ \end{align*}

Since $0 \leq \theta \lt \frac{\pi}{2}\text{,}$ $\tan\theta \geq 0\text{,}$ so

\begin{align*} \tan\theta &= \frac{\sqrt{x-5}}{2} \end{align*}
Exercise4
Hint

You can draw a right triangle with angle $\theta\text{,}$ and use the given information to label two of the sides. The Pythagorean Theorem gives you the third side.

(a) $\dfrac{\sqrt{4-x^2}}{2}$

(b) $\dfrac{1}{2}$

(c) $\dfrac{1}{\sqrt{1-x}}$

Solution

1. Let $\theta = \arccos \left(\frac{x}{2}\right)\text{.}$ That is, $\cos(\theta) = \frac{x}{2}\text{,}$ and $0 \leq \theta \leq \pi\text{.}$ Then we can draw the corresponding right triangle with angle $\theta$ with adjacent side of signed length $x$ (we note that if $\theta \gt \frac{\pi}{2}\text{,}$ then $x$ is negative) and hypotenuse of length $2\text{.}$ By the Pythagorean Theorem, the opposite side of the triangle has length $\sqrt{4-x^2}\text{.}$

So,

\begin{equation*} \sin\left(\arccos \left(\frac{x}{2}\right)\right)=\sin \theta = \frac{\mathrm{opp}}{\mathrm{hyp}} = \frac{\sqrt{4-x^2}}{2} \end{equation*}
2. Let $\theta = \arctan \left(\frac{1}{\sqrt{3}}\right)\text{.}$ That is, $\tan(\theta) = \frac{1}{\sqrt{3}}\text{,}$ and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\text{.}$

• Solution 1: Then $\theta = \dfrac{\pi}{6}\text{,}$ so $\sin\theta = \dfrac{1}{2}\text{.}$
• Solution 2: Then we can draw the corresponding right triangle with angle $\theta$ with opposite side of length $1$ and adjacent side of length $\sqrt{3}\text{.}$ By the Pythagorean Theorem, the hypotenuse of the triangle has length $\sqrt{\sqrt{3}^2+1^2}=2\text{.}$

So,

\begin{equation*} \sin\left(\arctan \left(\frac{1}{\sqrt{3}}\right)\right)=\sin \theta = \frac{\mathrm{opp}}{\mathrm{hyp}} = \frac{1}{2} \end{equation*}
3. Let $\theta = \arcsin \left(\sqrt{x}\right)\text{.}$ That is, $\sin(\theta) = \sqrt{x}\text{,}$ and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}\text{.}$ Then we can draw the corresponding right triangle with angle $\theta$ with opposite side of length $\sqrt{x}$ and hypotenuse of length $1\text{.}$ By the Pythagorean Theorem, the adjacent side of the triangle has length $\sqrt{1-x}\text{.}$

So,

\begin{equation*} \sec\left(\arcsin \left(\sqrt{x}\right)\right)=\sec \theta = \frac{\mathrm{hyp}}{\mathrm{adj}} = \frac{1}{\sqrt{1-x}} \end{equation*}
Exercise5(*)
Hint

As in Question 1, choose an appropriate substitution. Your answer should be in terms of your original variable, $x\text{,}$ which can be achieved using the methods of Question 3.

$\dfrac14\cdot \dfrac x{\sqrt{x^2+4}} + C$

Solution

Let $x = 2\tan\theta\text{,}$ so that $x^2+4 = 4\tan^2\theta+4=4\sec^2\theta$ and $\dee{x} = 2\sec^2\theta\,\dee{\theta}\text{.}$ Then

\begin{align*} \int \frac1{(x^2+4)^{3/2}} \,\dee{x} &= \int \frac1{(4\sec^2\theta)^{3/2}} \cdot 2\sec^2\theta\,\dee{\theta} \\ &= \int \frac{2\sec^2\theta}{8\sec^3\theta} \,\dee{\theta} \\ &=\frac14 \int \cos\theta\,\dee{\theta} \\ &= \frac14\sin\theta+C = \frac14 \frac x{\sqrt{x^2+4}} + C \end{align*}

To find $\sin\theta$ in terms of $x\text{,}$ we construct the right triangle above. Since $\tan\theta = \dfrac{x}{2} = \dfrac{\mbox{opp}}{\mbox{adj}}\text{,}$ we label the opposite side $x$ and the adjacent side $2\text{.}$ By the Pythagorean Theorem, the hypotenuse has length $\sqrt{x^2+4}\text{.}$ Then $\sin\theta = \dfrac{\mbox{opp}}{\mbox{hyp}} = \dfrac{x}{\sqrt{x^2+4}}\text{.}$

Exercise6(*)
Hint

As in Question 1, choose an appropriate substitution. Your answer will be a number, so as long as you change your limits of integration when you substitute, you don't need to bother changing the antiderivative back into the original variable $x\text{.}$ However, you might want to use the techniques of Question 4 to simplify your final answer.

$\dfrac{1}{2\sqrt{5}}$

Solution

• Solution 1: As in Question 5, substitute $x=2\tan u\text{,}$ $\dee{x}=2 \sec^2u\,\dee{u}\text{.}$ Note that when $x=4$ we have $4=2\tan u\text{,}$ so that $\tan u=2\text{.}$

\begin{align*} \int_0^4 \frac{1}{{(4+x^2)}^{3/2}}\,\dee{x} &=\int_0^{\arctan 2} \frac{1}{{(4+4\tan^2 u)}^{3/2}}\,2\sec^2 u\,\dee{u}\\ &=\int_0^{\arctan 2} \frac{2\sec^2 u}{{(2\sec u)}^{3}}\,\dee{u}\\ &=\frac{1}{4}\int_0^{\arctan 2} \frac{\sec^2u}{\sec^3u}\,\dee{u}\\ &=\frac{1}{4}\int_0^{\arctan 2} \cos u\,\dee{u}\\ &=\bigg[\frac{1}{4}\sin u \bigg]_0^{\arctan2}\\ &=\frac{1}{4} \big( \sin(\arctan 2) - 0 \big) = \frac{1}{2\sqrt{5}} \end{align*}

To find $\sin(\arctan 2)\text{,}$ we use the right triangle above, with angle $u=\arctan 2\text{.}$ Since $\tan u=2 = \dfrac{\mbox{opp}}{\mbox{adj}}\text{,}$ we label the opposite side as 2, and the adjacent side as 1. The Pythagorean Theorem tells us the hypotenuse has length $\sqrt{5}\text{,}$ so $\sin u = \dfrac{\mbox{opp}}{\mbox{hyp}} = \dfrac{2}{\sqrt{5}}\text{.}$

• Solution 2: Using our result from Question 5, \begin{align*} \int_0^4 \frac{1}{{(4+x^2)}^{3/2}}\,\dee{x}&=\frac{1}{4}\left[ \dfrac x{\sqrt{x^2+4}}\right]_0^4\\ &=\frac{1}{4}\cdot \dfrac{4}{\sqrt{4^2+4}}=\frac{1}{2\sqrt{5}} \end{align*}
Exercise7(*)
Hint

Question 1 guides the way to finding the appropriate substitution. Since the integral is definite, your final answer will be a number. Your limits of integration should be common reference angles.

$\dfrac{\pi}{6}$

Solution

Make the change of variables $x=5\sin\theta\text{,}$ $\dee{x}=5\cos\theta\,\dee{\theta}\text{.}$ Since $x=0$ corresponds to $\theta=0$ and $x=\frac{5}{2}$ correponds to $\sin\theta=\half$ or $\theta =\frac{\pi}{6}\text{,}$

\begin{gather*} \int_0^{5/2} \frac{\dee{x}}{\sqrt{25-x^2}} =\int_0^{\pi/6} \frac{5\cos\theta\,\dee{\theta}}{\sqrt{25-25\sin^2\theta}} =\int_0^{\pi/6} \dee{\theta} =\frac{\pi}{6} \end{gather*}
Exercise8(*)
Hint

Question 1 guides the way to finding the appropriate substitution. Since you have in indefinite integral, make sure to get your answer back in terms of the original variable, $x\text{.}$ Question 3 gives a reliable method for this.

$\displaystyle\log\left|\sqrt{1+\frac{x^2}{25}}+\frac{x}{5}\right|+C$

Solution

Substitute $x=5\tan u\text{,}$ so that $\dee{x}=5 \sec^2u\,\dee{u}\text{.}$

\begin{align*} \int\frac{1}{\sqrt{x^2+25}}\,\dee{x} &=\int \frac{1}{\sqrt{25\tan^2 u+25}}\,5\sec^2 u\,\dee{u} \\ &=\int \frac{5\sec^2u}{5\sec u}\,\dee{u} =\int \sec u\,\dee{u} \\ &= \log\big|\sec u+\tan u\big|+C\\ &= \log\Big|\sqrt{1+\frac{x^2}{25}}+\frac{x}{5}\Big|+C \end{align*}

To find $\sec u$ and $\tan u\text{,}$ we have two options. One is to set up a right triangle with angle $u$ and $\tan u = \frac{x}{5}\text{.}$ Then we can label the opposite side $x$ and the adjacent side 5, and use Pythagorus to find that the hypotenuse is $\sqrt{x^2+25}\text{.}$

Another option is to look back at our work a little more closely — in fact, we've already found what we're looking for. Since we used the substitution $x=5\tan u\text{,}$ this gives us $\tan u = \frac{x}{5}\text{.}$ In the denominator of the integrand, we simplified $\sqrt{x^2+25} = 5\sec u\text{,}$ so $\sec u = \frac{1}{5}\sqrt{x^2+25} = \sqrt{1+\frac{x^2}{25}}\text{.}$

Exercise9
Hint

A trig substitution is not the easiest path.

$\dfrac{1}{2}\sqrt{2x^2+4x}+C$

Solution

The quadratic formula underneath the square root makes us think of a trig substitution, but in the interest of developing good habits, let's check for an easier way first. If we let $u=2x^2+4x\text{,}$ then $\dee{u} = (4x+4)\dee{x}\text{,}$ so $\frac{1}{4}\,\dee{u}=(x+1)\,\dee{x}\text{.}$ This substitution looks easier than a trig substitution (which would start with completing the square).

\begin{align*} \int\frac{x+1}{\sqrt{2x^2+4x}} \, \dee{x}&=\frac{1}{4}\int \frac{1}{\sqrt{u}}\,\dee{u} = \frac{1}{2}\sqrt{u}+C = \frac{1}{2}\sqrt{2x^2+4x}+C \end{align*}
Exercise10(*)
Hint

To antidifferentiate, change your trig functions into sines and cosines.

$-\displaystyle\frac{1}{16}\dfrac{\sqrt{x^2+16}}{x}+C$

Solution

Substitute $x=4\tan u\text{,}$ $\dee{x}=4 \sec^2u\,\dee{u}\text{.}$

\begin{align*} \int\frac{1}{x^2\sqrt{x^2+16}}\,\dee{x} &=\int \frac{1}{16\tan^2 u \sqrt{16\tan^2 u+16}}\,4\sec^2 u\,\dee{u} \\ &=\int \frac{\sec^2u}{16\tan^2u\sec u}\,\dee{u} =\frac{1}{16}\int \frac{\sec u}{\tan^2u}\,\dee{u}\\ &= \frac{1}{16}\int\frac{\cos u}{\sin^2 u}\,\dee{u} \end{align*}

To finish off the integral, we'll substitute $v=\sin u\text{,}$ $\dee{v}=\cos u\,\dee{u}\text{.}$

\begin{align*} \int\frac{1}{x^2\sqrt{x^2+16}}\,\dee{x} &=\frac{1}{16} \int\frac{\cos u}{\sin^2 u}\,\dee{u} = \frac{1}{16}\int\frac{\dee{v}}{v^2} =-\frac{1}{16v} +C \\ &=-\frac{1}{16\sin u} +C =-\frac{1}{16}\dfrac{\sqrt{x^2+16}}{x}+C \end{align*}

To find $\sin u\text{,}$ we draw a right triangle with angle $u$ and $\tan u = \frac{x}{4}\text{.}$ We label the opposite side $x$ and the adjacent side $4\text{,}$ and then from Pythagorus we find that the hypotenuse has length $\sqrt{x^2+16}\text{.}$ So, $\sin u = \dfrac{\sqrt{x^2+16}}{x}\text{.}$

Exercise11(*)
Hint

The integrand should simplify quite far after your substitution.

$\displaystyle\frac{\sqrt{x^2-9}}{9x} +C$

Solution

Substituting $x=3\sec u\text{,}$ so that $\dee{x}= 3\sec u\tan u\,\dee{u}$ and $x^2-9=9\sec^2 u-9 = 9\tan^2 u\text{,}$ gives

\begin{align*} \int \frac{\dee{x}}{x^2\sqrt{x^2-9}} &=\int \frac{3\sec u\tan u\dee{u}} {9\sec^2u\sqrt{9\tan^2u}} \\ &=\frac{1}{9}\int\frac{\dee{u}}{\sec u} \\ &=\frac{1}{9}\int \cos u\dee{u} =\frac{1}{9}\sin u +C. \end{align*}

To evaluate $\sin u\text{,}$ we make a right triangle with angle $u\text{.}$ Since $\sec u = \dfrac{x}{3} = \dfrac{\mbox{hyp}}{\mbox{adj}}\text{,}$ we label the hypotenuse $x$ and the adjacent side $3\text{.}$

Using the Pythagorean Theorem, the opposite side has length $\sqrt{x^2-9}\text{.}$ So, $\sin u = \dfrac{\sqrt{x^2-9}}{x}$ and

\begin{align*} \int \frac{\dee{x}}{x^2\sqrt{x^2-9}} &= \frac{\sqrt{x^2-9}}{9x} +C. \end{align*}
Exercise12(*)
Hint

In part (a) you are asked to integrate an even power of $\cos x\text{.}$ For part (b) you can use a trigonometric substitution to reduce the integral of part (b) almost to the integral of part (a).

(a) We'll use the trig identity $\cos2\theta=2\cos^2\theta-1\text{.}$ It implies that

\begin{align*} \cos^2\theta=\frac{\cos2\theta+1}{2} \implies \cos^4\theta &=\frac{1}{4}\big[\cos^22\theta+2\cos2\theta+1\big] =\frac{1}{4}\Big[\frac{\cos4\theta+1}{2}+2\cos2\theta+1\Big]\\ &=\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8}\\ \end{align*}

So,

\begin{align*} \int_0^{\pi/4}\cos^4\theta\dee{\theta} &=\int_0^{\pi/4}\Big(\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8}\Big) \dee{\theta} \\ &=\left[\frac{\sin4\theta}{32}+\frac{\sin2\theta}{4}+\frac{3}{8}\theta\right]_0^{\pi/4}\\ &= \frac{1}{4}+\frac{3}{8}\cdot \frac{\pi}{4}\\ &=\frac{8+3\pi}{32} \end{align*}

as required. \\ (b) $\dfrac{8+3\pi}{16}$

Solution

(a) We'll use the trig identity $\cos2\theta=2\cos^2\theta-1\text{.}$ It implies that

\begin{align*} \cos^2\theta=\frac{\cos2\theta+1}{2} \implies \cos^4\theta &=\frac{1}{4}\big[\cos^22\theta+2\cos2\theta+1\big] =\frac{1}{4}\Big[\frac{\cos4\theta+1}{2}+2\cos2\theta+1\Big]\\ &=\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8}\\ \end{align*}

So,

\begin{align*} \int_0^{\pi/4}\cos^4\theta\dee{\theta} &=\int_0^{\pi/4}\Big(\frac{\cos4\theta}{8}+\frac{\cos2\theta}{2}+\frac{3}{8}\Big) \dee{\theta} \\ &=\left[\frac{\sin4\theta}{32}+\frac{\sin2\theta}{4}+\frac{3}{8}\theta\right]_0^{\pi/4}\\ &= \frac{1}{4}+\frac{3}{8}\cdot \frac{\pi}{4}\\ &=\frac{8+3\pi}{32} \end{align*}

as required.

(b) We'll use the trig substitution $x=\tan\theta\text{,}$ $\dee{x}=\sec^2\theta\dee{\theta}\text{.}$ Note that when $\theta=\pm\frac{\pi}{4}\text{,}$ we have $x=\pm 1\text{.}$ Also note that dividing the trig identity $\sin^2\theta+\cos^2\theta=1$ by $\cos^2\theta$ gives the trig identity $\tan^2\theta+1=\sec^2\theta\text{.}$ So

\begin{align*} \int_{-1}^1\frac{\dee{x}}{{(x^2+1)}^3} &=2\int_0^1\frac{\dee{x}}{{(x^2+1)}^3}&\mbox{(even integrand)}\\ &=2\int_0^{\pi/4}\frac{\sec^2\theta\dee{\theta}}{{(\tan^2\theta+1)}^3}\\ &=2\int_0^{\pi/4}\frac{\sec^2\theta\dee{\theta}}{{(\sec^2\theta)}^3}\\ &=2\int_0^{\pi/4}\cos^4\theta\dee{\theta}\\ &=\frac{8+3\pi}{16} \end{align*}

by part (a).

Exercise13
Hint

What is the symmetry of the integrand?

0

Solution

The integrand is an odd function, and the limits of integration are symmetric, so $\displaystyle\int_{-\pi/12}^{\pi/12} \dfrac{15x^3}{(x^2+1)\sqrt{9-x^2}^5}\dee{x}=0\text{.}$

Exercise14(*)
Hint

See Example 1.9.3.

$\displaystyle2\arcsin\frac{x}{2}+\frac{x}{2}\sqrt{4-x^2}+ C$

Solution

Substitute $x=2\sin u\text{,}$ so that $\dee{x}=2 \cos u\,\dee{u}\text{.}$

\begin{align*} \int \sqrt{4-x^2}\,\dee{x} &=\int \sqrt{4-4\sin^2u}\ 2\cos u\,\dee{u} \\ &=\int \sqrt{4\cos^2u}\ 2\cos u\,\dee{u} \\ &=\int 4\cos^2 u\,\dee{u} =2\int \big(1+\cos(2u)\big)\,\dee{u} \\ &= 2u +\sin(2u) + C \\ &=2u + 2\sin u\cos u + C \\ &=2\arcsin\frac{x}{2} + \frac{ x}{2}\sqrt{4-x^2} + C \end{align*}

We used the substitution $x = 2\sin u\text{,}$ so we know $\sin u = \frac{x}{2}$ and $u=\arcsin(\frac{x}{2})\text{.}$ We have three options for finding $\cos u\text{.}$

First, we can draw a right triangle with angle $u\text{.}$ Since $\sin u = \frac{x}{2}\text{,}$ we label the opposite side $x$ and the hypotenuse 2, then by the Pythagorean Theorem the adjacent side has length $\sqrt{4-x^2}\text{.}$ So, $\cos u = \dfrac{\mbox{adj}}{\mbox{hyp}} = \dfrac{\sqrt{4-x^2}}{2}\text{.}$

Second, we can look back carefully at our work. We simplified $\sqrt{4-x^2} = 2\cos u\text{,}$ so $\cos u = \dfrac{\sqrt{4-x^2}}{2}\text{.}$

Third, we could use the identity $\sin^2 u + \cos^2 u =1\text{.}$ Then $\cos u = \pm\sqrt{1-\sin^2 u} = \pm\sqrt{1-\frac{x^2}{4}}\text{.}$ Since $u = \arcsin (x/2)\text{,}$ $u$ is in the range of arcsine, which means $-\frac{\pi}{2} \leq u \leq \frac{\pi}{2}\text{.}$ Therefore, $\cos u \geq 0\text{,}$ so $\cos u = \sqrt{1-\frac{x^2}{4}} = \frac{\sqrt{4-x^2}}{2}\text{.}$

So,

\begin{equation*} \int \sqrt{4-x^2}\,\dee{x}=2u + 2\sin u\cos u + C= 2\arcsin\frac{x}{2}+x\cdot\frac{\sqrt{4-x^2}}{2}+C \end{equation*}
Exercise15(*)
Hint

To integrate an even power of tangent, use the identity $\tan^2 x = \sec^2 x - 1\text{.}$

$\sqrt{25x^2-4}-2\arcsec\frac{5x}{2} + C$

Solution

Substitute $x=\frac{2}{5}\sec u\text{,}$ so that $\dee{x}=\frac{2}{5} \sec u\,\tan u\,\dee{u}$ and $25 x^2-4 = 4(\sec^2u-1) = 4\tan^2u\text{.}$

\begin{align*} \int \frac{\sqrt{25x^2-4}}{x}\,\dee{x} &=\int \frac{2\tan u}{\frac{2}{5}\sec u}\cdot\frac{2}{5}\sec u\tan u\,\dee{u} \\ &=2\int \tan^2 u\,\dee{u} =2\int \big(\sec^2 u -1\big)\,\dee{u} \\ &= 2\tan u -2u + C \\ &=\sqrt{25x^2-4}-2\arcsec\tfrac{5x}{2} + C \end{align*}

To find $\tan u\text{,}$ we draw a right triangle with angle $u\text{.}$ Since $\sec u =\dfrac{5x}{2}\text{,}$ we label the hypotenuse $5x$ and the adjacent side 2. Then the Pythagorean Theorem gives us the opposite side as length $\sqrt{25x^2-4}\text{.}$ Then $\tan u = \dfrac{\mbox{opp}}{\mbox{adj}} = \dfrac{\sqrt{25x^2-4}}{2}\text{.}$

Alternately, we can notice that in our work, we already showed $2\tan u = \sqrt{25x^2-4}\text{,}$ so $\tan u = \frac{1}{2}\sqrt{25x^2-4} .$

Exercise16
Hint

A trig substitution is not the easiest path.

$\dfrac{40}{3}$

Solution

The integrand has a quadratic polynomial under a square root, which makes us think of trig substitutions. However, it's good practice to look for simpler methods before we jump into more complicated ones, and in this case we find something nicer than a trig substitution: the substitution $u=x^2-1\text{,}$ $\dee{u}=2x\,\dee{x}\text{.}$ Then $x\dee{x} = \frac{1}{2}\dee{u}\text{,}$ and $x^2 ={u+1}\text{.}$ When $x=\sqrt{10}\text{,}$ $u=9\text{,}$ and when $x=\sqrt{17}\text{,}$ $u=16\text{.}$

\begin{align*} \int_{\sqrt{10}}^{\sqrt{17}} \frac{x^3}{\sqrt{x^2-1}}\, \dee{x}&= \int_{\sqrt{10}}^{\sqrt{17}} \frac{x^2}{\sqrt{x^2-1}}\, \cdot x\dee{x}\\ &=\frac{1}{2} \int_{9}^{16} \frac{u+1}{\sqrt{u}}\,\dee{u}\\ &=\frac{1}{2} \int_{9}^{16}\left(u^{1/2}+u^{-1/2}\right)\,\dee{u}\\ &=\frac{1}{2}\left[\frac{2}{3}u^{3/2} + 2u^{1/2} \right]_{9}^{16}\\ &=\frac{1}{2}\left[\frac{2}{3}\cdot 4^3 + 2\cdot 4 -\frac{2}{3}\cdot 3^3 -2\cdot 3 \right]\\ &=\frac{40}{3} \end{align*}
Exercise17(*)
Hint

Complete the square. Your final answer will have an inverse trig function in it.

$\arcsin\dfrac{x+1}{2} + C$

Solution

This integrand looks very different from those above. But it is only slightly disguised. If we complete the square

\begin{gather*} \int \frac{\dee{x}}{\sqrt{3-2x-x^2}} = \int \frac{\dee{x}}{\sqrt{4-(x+1)^2}} \end{gather*}

and make the substitution $y=x+1\text{,}$ $\dee{y}=\dee{x}$

\begin{gather*} \int \frac{\dee{x}}{\sqrt{3-2x-x^2}} = \int \frac{\dee{x}}{\sqrt{4-(x+1)^2}} = \int \frac{\dee{y}}{\sqrt{4-y^2}} \end{gather*}

we get a typical trig substitution integral. So, we substitute $y=2\sin\theta\text{,}$ $\dee{y}=2\cos\theta\,\dee{\theta}$ to get

\begin{align*} \int \frac{\dee{x}}{\sqrt{3-2x-x^2}} &= \int \frac{\dee{y}}{\sqrt{4-y^2}} = \int\frac{2\cos\theta\,\dee{\theta}}{\sqrt{4-4\sin^2\theta}} = \int\frac{2\cos\theta\,\dee{\theta}}{\sqrt{4\cos^2\theta}}\\ &=\int\dee{\theta} =\theta +C =\arcsin\frac{y}{2} + C \\ &=\arcsin\frac{x+1}{2} + C \end{align*}

An experienced integrator would probably substitute $x+1 = 2\sin\theta$ directly, without going through $y\text{.}$

Exercise18
Hint

To antidifferentiate even powers of cosine, use the formula $\cos^2\theta = \frac{1}{2}(1+\cos(2\theta))\text{.}$ Then, remember $\sin(2\theta)=2\sin\theta\cos\theta\text{.}$

$\displaystyle\frac{1}{4}\left(\arccos\left(\frac{1}{2x-3}\right) + \frac{\sqrt{4x^2-12x+8}}{(2x-3)^2}\right)+C\text{,}$ or equivalently,\\ $\displaystyle\frac{1}{4}\left(\arcsec\left({2x-3}\right) + \frac{\sqrt{4x^2-12x+8}}{(2x-3)^2}\right)+C$

Solution

Completing the square, we see $4x^2-12x+8 = (2x-3)^2-1\text{.}$

\begin{align*} \int \dfrac{1}{(2x-3)^3\sqrt{4x^2-12x+8}}\dee{x}&= \int \dfrac{1}{(2x-3)^3\sqrt{(2x-3)^2-1}}\dee{x}\\ \end{align*}

We use the substitution $2x-3 = \sec \theta\text{,}$ so $2\dee{x}=\sec\theta\tan\theta\dee{\theta}\text{.}$

\begin{align*} &=\frac{1}{2}\int\frac{1}{\sec^3\theta\sqrt{\sec^2\theta-1}}\sec\theta\tan\theta\dee{\theta}\\ &=\frac{1}{2}\int\frac{1}{\sec^3\theta\tan\theta}\sec\theta\tan\theta\dee{\theta}\\ &=\frac{1}{2}\int\frac{1}{\sec^2\theta}\dee{\theta}\\ &=\frac{1}{2}\int{\cos^2\theta}\dee{\theta}\\ &=\frac{1}{4}\int{\left(1+\cos(2\theta)\right)}\dee{\theta}\\ &=\frac14\left(\theta + \frac{1}{2}\sin(2\theta)\right)+C\\ &=\frac14\left(\theta + \sin\theta\cos\theta\right)+C\\ &=\frac{1}{4}\left(\arccos\left(\frac{1}{2x-3}\right) + \frac{\sqrt{4x^2-12x+8}}{(2x-3)^2}\right)+C \end{align*}

Since $2x-3=\sec\theta\text{,}$ we know $\cos\theta = \frac{1}{2x-3}$ and $\theta = \arccos\left(\frac{1}{2x-3}\right)\text{.}$ (Equivalently, $\theta = \arcsec(2x-3)\text{.}$) To find $\sin\theta\text{,}$ we draw a right triangle with adjacent side of length 1, and hypotenuse of length $2x-3\text{.}$ By the Pythagorean Theorem, the opposite side has length $\sqrt{4x^2-12x+8}\text{.}$

Exercise19
Hint

After substituting, use the identity $\tan^2 x = \sec^2 x - 1$ more than once.\\ Remember $\displaystyle\int \sec x \dee{x} = \log \big|\sec x + \tan x \big|+C\text{.}$

$\log(1+\sqrt{2})-\dfrac{1}{\sqrt{2}}$

Solution

We use the substitution $x=\tan u\text{,}$ $\dee{x}=\sec^2 u\dee{u}\text{.}$ Note $\tan 0=0$ and $\tan \frac{\pi}{4} =1\text{.}$

\begin{align*} \displaystyle\int_0^1\dfrac{x^2}{\sqrt{x^2+1}^3}\dee{x}&= \int_0^{\pi/4}\dfrac{\tan^2 u}{\sqrt{\tan^2 u +1}^3}\sec^2u\dee{u}\\ &= \int_0^{\pi/4}\dfrac{\tan^2 u}{\sqrt{\sec^2u}^3}\sec^2u\dee{u}\\ &= \int_0^{\pi/4}\frac{\tan^2 u}{\sec u}\dee{u}\\ &= \int_0^{\pi/4}\frac{\sec^2 u-1}{\sec u}\dee{u}\\ &= \int_0^{\pi/4}\big({\sec u}-\cos u\big)\dee{u}\\ &=\Big[\log\left|\sec u + \tan u \right| - \sin u\Big]_{0}^{\pi/4}\\ &=\left(\log\left|\sqrt{2} + 1 \right| -\frac{1}{\sqrt{2}}\right) - \left(\log|1+0|-0\right)\\ &=\log(1+\sqrt{2})-\frac{1}{\sqrt{2}} \end{align*}
Exercise20
Hint

There's no square root, but we can still make use of the substitution $x=\tan\theta\text{.}$

$\displaystyle\frac{1}{2}\left(\arctan x + \frac{x}{x^2+1}\right)+C$

Solution

There's no square root, but we can still make use of the substitution $x=\tan\theta\text{,}$ $\dee{x} = \sec^2\theta\dee{\theta}\text{.}$

\begin{align*} \int \frac{1}{(x^2+1)^2}\dee{x}&=\int\frac{1}{(\tan^2\theta+1)^2}\sec^2\theta\dee{\theta}\\ &=\int\frac{1}{\sec^4\theta}\sec^2\theta\dee{\theta} = \int\cos^2\theta\dee{\theta}\\ &=\frac{1}{2}\int \big(1 + \cos(2\theta)\big)\dee{\theta}\\ &=\frac{1}{2}\left(\theta + \frac{1}{2}\sin(2\theta)\right)+C\\ &=\frac{1}{2}\left(\theta + \sin\theta\cos\theta\right)+C\\ &=\frac{1}{2}\left(\arctan x + \frac{x}{x^2+1}\right)+C \end{align*}

Since $x = \tan\theta\text{,}$ we can draw a right triangle with angle $\theta\text{,}$ opposite side $x\text{,}$ and adjacent side $1\text{.}$ Then by the Pythagorean Theorem, its hypotenuse has length $\sqrt{x^2+1}\text{,}$ which allows us to find $\sin\theta$ and $\cos\theta\text{.}$

Exercise21
Hint

You'll probably want to use the identity $\tan^2\theta+1=\sec^2\theta$ more than once.

$\dfrac{3+x}{2}\sqrt{x^2-2x+2}+ \dfrac{1}{2}\log\left|\sqrt{x^2-2x+2}+x-1\right|+C$

Solution

We complete the square to find $x^2-2x+2 = (x-1)^2+1\text{.}$

\begin{align*} \int \dfrac{x^2}{\sqrt{x^2-2x+2}}\dee{x}&= \int \dfrac{x^2}{\sqrt{(x-1)^2+1}}\dee{x}\\ \end{align*}

We use the substitution $x-1=\tan\theta\text{,}$ which implies $\dee{x}=\sec^2\theta\dee{\theta}$ and $x=\tan\theta+1$

\begin{align*} &=\int \dfrac{(\tan\theta+1)^2}{\sqrt{(\tan\theta)^2+1}} \sec^2\theta\dee{\theta}\\ &=\int\frac{\textcolor{red}{\tan^2\theta}+2\tan\theta+\textcolor{red}1}{\sec\theta}\sec^2\theta\dee{\theta}\\ &=\int(\textcolor{red}{\sec^2\theta}+2\tan\theta)\sec\theta\dee{\theta}\\ &=\int\big( \sec^3\theta+2\tan\theta\sec\theta\big)\dee{\theta}\\ &=\frac{1}{2}\sec\theta\tan\theta + \frac{1}{2}\log|\sec\theta+\tan\theta|+2\sec\theta+C\\ &=\frac{1}{2}\sqrt{x^2-2x+2}(x-1) + \frac{1}{2}\log\left|\sqrt{x^2-2x+2}+x-1\right|+2\sqrt{x^2-2x+2}+C\\ &=\frac{3+x}{2}\sqrt{x^2-2x+2}+ \frac{1}{2}\log\left|\sqrt{x^2-2x+2}+x-1\right|+C \end{align*}

From our substitution, we know $\tan\theta = x-1\text{.}$ To find $\sec\theta\text{,}$ we can notice that in our work we already simplified $\sqrt{x^2-2x+1}=\sec\theta\text{.}$

Alternately, we can draw a right triangle with angle $\theta\text{,}$ opposite side $x-1\text{,}$ adjacent side $1\text{,}$ and use the Pythagorean Theorem to find the hypotenuse.

Exercise22
Hint

Complete the square — refer to Question 2 if you want a refresher. The constants aren't pretty, but don't let them scare you.

$\displaystyle\left| \left(\frac{6}{5}x+1\right)+\frac{2}{5}\sqrt{9x^2+15x} \right|+C$

Solution

First, we complete the square. The constants aren't integers, but we can still use the same method as in Question 2. The quadratic function under the square root is $3x^2+5x\text{.}$ We match the non-constant terms to those of a perfect square.

\begin{align*} (ax+b)^2&=a^2x^2+2abx+b^2\\ \textcolor{red}{3x^2}+\textcolor{blue}{5x}&=\textcolor{red}{a^2x^2} + \textcolor{blue}{2abx} +b^2 + c \quad\mbox{for some constant $c$} \end{align*}
• Looking at the leading term tells us $a=\sqrt{3}\text{.}$
• Then the second term tells us $5=2ab=2\sqrt{3}b\text{,}$ so $b=\frac{5}{2\sqrt3}\text{.}$
• Finally, the constant terms give us $0=b^2+c=\frac{25}{12}+c\text{,}$ so $c=-\frac{25}{12}\text{.}$

So, $3x^2+5x=\left(\sqrt{3}x+\frac{5}{2\sqrt{3}}\right)^2-\frac{25}{12}\text{.}$

\begin{align*} \int \dfrac{1}{\sqrt{3x^2+5x}}\dee{x}&=\int \dfrac{1}{\sqrt{\left(\sqrt{3}x+\frac{5}{2\sqrt3}\right)^2-\frac{25}{12}}}\dee{x}\\ \end{align*}

We use the substitution $\sqrt{3}x + \frac{5}{2\sqrt3}=\frac{5}{2\sqrt{3}}\sec\theta\text{,}$ which leads to $\sqrt{3}\dee{x} = \frac{5}{2\sqrt{3}}\sec\theta\tan\theta\dee{\theta}\text{,}$ i.e. $\dee{x} = \frac{5}{6}\sec\theta\tan\theta\dee{\theta}\text{.}$

\begin{align*} &=\int\frac{1}{\sqrt{\left(\frac{5}{2\sqrt3}\sec\theta\right)^2-\frac{25}{12}}}\cdot\frac{5}{6}\sec\theta\tan\theta\dee{\theta}\\ &=\int\frac{1}{\sqrt{\frac{25}{12}\sec^2\theta-\frac{25}{12}}}\cdot\frac{5}{6}\sec\theta\tan\theta\dee{\theta}\\ &=\int\frac{1}{\sqrt{\frac{25}{12}\tan^2\theta}}\cdot\frac{5}{6}\sec\theta\tan\theta\dee{\theta}\\ &=\int\frac{1}{{\frac{5}{2\sqrt{3}}\tan\theta}}\cdot\frac{5}{6}\sec\theta\tan\theta\dee{\theta}\\ &=\frac{1}{\sqrt3}\int\sec\theta\dee{\theta}\\ &=\frac{1}{\sqrt3}\log\left|\sec\theta+\tan\theta\right|+C\\ &=\frac{1}{\sqrt3}\log\left| \left(\frac{6}{5}x+1\right)+\frac{2}{5}\sqrt{9x^2+15x} \right|+C \end{align*}

Since we used the substitution $\sqrt3x+\frac{5}{2\sqrt{3}}=\frac{5}{2\sqrt3}\sec\theta\text{,}$ we have $\sec\theta = \frac{6}{5}x+1 = \frac{6x+5}{5}\text{.}$ To find $\tan\theta$ in terms of $x\text{,}$ we have two options. We can make a right triangle with angle $\theta\text{,}$ hypotenuse $6x+5\text{,}$ and adjacent side $5\text{,}$ then use the Pythagorean Theorem to find the opposite side. Or, we can look through our work and see that $\sqrt{3x^2+5}=\frac{5}{2\sqrt3}\tan\theta\text{,}$ so $\tan\theta = \frac{2\sqrt3}{5}\sqrt{3x^2+5}=\frac{2}{5}\sqrt{9x^2+15}\text{.}$

Remark: in applications, often the numbers involved are messier than they are in textbooks. The ideas of this problem are similar to other problems in this section, but it's good practice to apply them in a slightly messy context.

Exercise23
Hint

After substituting, use the identity $\sec^2 u = \tan^2 u +1\text{.}$ It might help to break the integral into a few pieces.

$\dfrac{1}{3}\sqrt{1+x^2}(4+x^2)+\log\left|\dfrac{1-\sqrt{1+x^2}}{x} \right|+C$

Solution

We use the substitution $x=\tan u\text{,}$ $\dee{x}=\sec^2 u\dee{u}\text{.}$

\begin{align*} \int\dfrac{\sqrt{1+x^2}^3}{x}\dee{x}&=\int\frac{\sqrt{1+\tan^2 u}^3}{\tan u}\sec^2 u\dee{u}\\ &=\int\frac{\sec^3u}{\tan u}\sec^2 u\dee{u}\\ &=\int\frac{(\sec^2u)^2}{\tan u}\sec u\dee{u}\\ &=\int\frac{(\tan^2u+1)^2}{\tan u}\sec u\dee{u}\\ &=\int\frac{\tan^4u+2\tan^2 u + 1}{\tan u}\sec u\dee{u}\\ &=\int \tan^3 u \sec u \dee{u} + \int 2\sec u \tan u\dee u + \int \frac{\sec u}{\tan u}\dee{u}\\ \end{align*}

For the first integral, we use the substitution $w=\sec u\text{.}$ The second is the antiderivative of $2\sec u\text{.}$ The third we simplify as $\frac{\sec u}{\tan u} = \frac{1}{\cos u}\cdot \frac{\cos u}{\sin u} = \csc u$ .

\begin{align*} &=\int\bigg((\sec^2 u -1)\sec u \tan u\bigg)\dee{u} + 2\sec u + \log|\cot u - \csc u |+C\\ &=\int (w^2-1)\dee{w}+2\sec u + \log|\cot u - \csc u|+C\\ &=\frac{1}{3}w^3-w+2\sec u + \log|\cot u - \csc u|+C\\ &=\frac{1}{3}\sec^3u-\sec u+2\sec u + \log|\cot u - \csc u|+C\\ &=\frac{1}{3}\sec^3u+\sec u + \log|\cot u - \csc u|+C\\ \end{align*}

We began with the substitution $x=\tan u\text{.}$ Then $\cot u = \frac{1}{x}\text{.}$ To find $\csc u$ and $\sec u\text{,}$ we draw a right triangle with angle $u\text{,}$ opposite side $x\text{,}$ and adjacent side $1\text{.}$ The Pythagorean Theorem gives us the hypotenuse.

\begin{align*} &=\frac{1}{3}\sqrt{1+x^2}^3+\sqrt{1+x^2}+\log\left| \frac{1}{x}-\frac{\sqrt{1+x^2}}{x} \right|+C\\ &=\frac{1}{3}\sqrt{1+x^2}(4+x^2)+\log\left|\frac{1-\sqrt{1+x^2}}{x} \right|+C \end{align*}
Exercise24
Hint

Make use of symmetry, and integrate with respect to $y$ (rather than $x$). The limits of integration should be reference angles.

$\dfrac{8\pi}{3}+4\sqrt{3}$

Solution

The half of the ellipse to the right of the $y$-axis is given by the equation

\begin{align*} x=f(y)&=4\sqrt{1-\left(\frac{y}{2}\right)^2}\\ \end{align*}

The area we want is the twice the area between the right-hand side of the curve and the $y$-axis, from $y=-1$ to $y=1\text{.}$ In other words,

\begin{align*} \mbox{Area}&=2\int_{-1}^1 4\sqrt{1-\left(\frac{y}{2}\right)^2}\dee{y}\\ \end{align*}

Making use of symmetry,

\begin{align*} &=16\int_{0}^1 \sqrt{1-\left(\frac{y}{2}\right)^2}\dee{y}\\ \end{align*}

We use the substitution $\frac{y}{2} = \sin \theta\text{,}$ $\frac{1}{2}\dee{y}=\cos\theta\dee{\theta}\text{.}$ Notice $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\sin 0 =0\text{.}$

\begin{align*} &=16\int_{0}^{\pi/6} \sqrt{1-\left(\sin\theta\right)^2} \cdot 2\cos\theta\dee{\theta}\\ &=32\int_{0}^{\pi/6} \sqrt{\cos^2\theta} \cos\theta\dee{\theta}\\ &=32\int_{0}^{\pi/6}\cos^2\theta\dee{\theta}\\ &=16\int_{0}^{\pi/6}\left(1+\cos(2\theta)\right)\dee{\theta}\\ &=16\left[\theta +\frac{1}{2}\sin(2\theta)\right]_{0}^{\pi/6}\\ &=16\left(\frac{\pi}{6}+\frac{1}{2}\cdot \frac{\sqrt{3}}{2} \right)\\ &=\frac{8\pi}{3}+4\sqrt{3} \end{align*}

Remark: we also investigated areas of ellipses in Question 1.2.3.16, Section 1.2.

Exercise25
Hint

Use the symmetry of the function to re-write your integrals without an absolute value.

Area: $\dfrac{4}{3} - \sqrt[4]{\dfrac{4}{3}}$

Volume: $\dfrac{\pi^2}{6} - \dfrac{\sqrt{3}\pi}{4}$

Solution

Note that $f(x)$ is an even function, nonnegative over its entire domain.\\ (a) To find the area of $R\text{,}$ we evaluate

\begin{align*} \mbox{Area}&=\int _{-1/2}^{1/2} \dfrac{|x|}{\sqrt[4]{1-x^2}}\dee{x} = 2\int _{0}^{1/2} \dfrac{x}{\sqrt[4]{1-x^2}}\dee{x}\\ \end{align*}

We use the substitution $u=1-x^2\text{,}$ $\dee{u}=-2x\dee{x}\text{.}$

\begin{align*} &=-\int_{1}^{3/4} \frac{1}{u^{1/4}}\dee{u}\\ &=-\left[\frac{4}{3}u^{3/4}\right]_1^{3/4} = -\frac{4}{3}\left(\left(\frac{3}{4}\right)^{3/4}-1\right)\\ &=\frac{4}{3} - \sqrt[4]{\frac{4}{3}} \end{align*}

(b) We slice the solid of rotation into circular disks of width $\dee{x}$ and radius $\dfrac{|x|}{\sqrt[4]{1-x^2}}\text{.}$

\begin{align*} \mbox{Volume}&=\int_{-1/2}^{1/2} \pi\left(\dfrac{|x|}{\sqrt[4]{1-x^2}}\right)^2\dee{x}\\ &=2\pi\int_{0}^{1/2} \dfrac{x^2}{\sqrt{1-x^2}}\dee{x}\\ \end{align*}

We use the substitution $x=\sin \theta\text{,}$ $\dee{x} = \cos\theta\dee{\theta}\text{,}$ so $\sqrt{1-x^2} = \sqrt{1-\sin^2\theta}=\cos \theta\text{.}$ Note $\sin 0 =0$ and $\sin\frac{\pi}{6}=\frac{1}{2}.$

\begin{align*} &=2\pi\int_{0}^{\pi/6} \frac{\sin^2 \theta}{\cos \theta}\cos \theta\dee{\theta}\\ &=2\pi\int_{0}^{\pi/6} \sin^2 \theta\dee{\theta}\\ &=\pi\int_{0}^{\pi/6}\big(1- \cos(2 \theta)\big)\dee{\theta}\\ &=\pi\left[\theta - \frac{1}{2}\sin(2\theta)\right]_0^{\pi/6}\\ &=\pi\left(\frac{\pi}{6} - \frac{1}{2}\cdot \frac{\sqrt{3}}{2}\right)\\ &=\frac{\pi^2}{6} - \frac{\sqrt{3}\pi}{4} \end{align*}
Exercise26
Hint

Think of $e^x$ as $\left(e^{x/2}\right)^2\text{,}$ and use a trig substitution. Then, use the identity $\sec^2 \theta = \tan^2 \theta +1\text{.}$

$2\sqrt{1+e^x}+2\log\left| 1-\sqrt{1+e^x} \right|-x+C$

Solution

If we think of $e^x$ as $\left(e^{x/2}\right)^2\text{,}$ the function under the square root suggests the substitution $e^{x/2}=\tan \theta\text{.}$ Then $\frac{1}{2}e^{x/2}\dee{x}=\sec^2\theta\dee{\theta}\text{,}$ so $\dee{x} = \frac{2}{e^{x/2}}\sec^2\theta\dee{\theta} = \frac{2}{\tan\theta}\sec\theta\dee{\theta}\text{.}$

\begin{align*} \int \sqrt{1+e^x}\dee{x}&=\int\frac{2\sqrt{1+\tan^2 \theta}}{\tan\theta}\sec^2\theta\dee{\theta}\\ &=2\int \frac{\sec^3\theta}{\tan\theta}\dee{\theta}\\ &=2\int \frac{\sec\theta(\tan^2\theta+1)}{\tan\theta}\dee{\theta}\\ &=2\int \left(\sec\theta \tan\theta + \frac{\sec \theta}{\tan\theta}\right)\dee{\theta}\\ &=2\int\big( \sec\theta \tan\theta + \csc\theta\big)\dee{\theta}\\ &=2\sec\theta + 2\log|\cot\theta-\csc\theta|+C\\ &=2\sqrt{1+e^x}+2\log\left| \frac{1}{e^{x/2}} - \frac{\sqrt{1+e^x}}{e^{x/2}} \right|+C\\ &=2\sqrt{1+e^x}+2\log\left| 1-\sqrt{1+e^x} \right|-2\log(e^{x/2})+C\\ &=2\sqrt{1+e^x}+2\log\left| 1-\sqrt{1+e^x} \right|-x+C \end{align*}

We used the substitution $e^{x/2}=\tan\theta\text{,}$ so $\cot\theta = \frac{1}{e^{x/2}}\text{.}$ To find $\sec \theta$ and $\csc\theta\text{,}$ we draw a right triangle with opposite side $e^{x/2}$ and adjacent side 1. They by the Pythagorean Theorem, the hypotenuse has length $\sqrt{1+e^x}\text{.}$

Remark: if we use the substitution $u=\sqrt{1+e^x}\text{,}$ then we can change the integral to $\displaystyle\int \dfrac{2u^2}{u^2-1}\dee{u}\text{.}$ We can integrate this using the method of partial fractions, which we'll learn in the next section. You can explore this option in Question 1.10.4.26, Section 1.10.

Exercise27
Hint

1. Use logarithm rules to simplify first.
3. What went wrong in part (b)? At what point in the work was that problem introduced?

There is a subtle but important point mentioned in the introductory text to Section 1.9 that may help you make sense of things.

1. $\dfrac{1}{1-x^2}$
2. False
3. The work in the question is not correct. The most salient problem is that when we make the substitution $x=\sin\theta\text{,}$ we restrict the possible values of $x$ to $[-1,1]\text{,}$ since this is the range of the sine function. However, the original integral had no such restriction.

How can we be sure we avoid this problem in the future? In the introductory text to Section 1.9 (before Example 1.9.1), the notes tell us that we are allowed to write our old variable as a function of a new variable (say $x=s(u)$) as long as that function is invertible to recover our original variable $x\text{.}$ There is one very obvious reason why invertibility is necessary: after we antidifferentiate using our new variable $u\text{,}$ we need to get it back in terms of our original variable, so we need to be able to recover $x\text{.}$ Moreover, invertibility reconciles potential problems with domains: if an inverse function $u=s^{-1}(x)$ exists, then for any $x\text{,}$ there exists a $u$ with $s(u)=x\text{.}$ (This was not the case in the work for the question, because we chose $x=\sin \theta\text{,}$ but if $x=2\text{,}$ there is no corresponding $\theta\text{.}$ Note, however, that $x=\sin\theta$ is invertible over $[-1,1]\text{,}$ so the work is correct if we restrict $x$ to those values.)

Solution

1. We can save ourselves some trouble by applying logarithm rules before we differentiate. \begin{align*} \log\left| \dfrac{1+x}{\sqrt{1-x^2}}\right|&=\log|1+x| - \log|\sqrt{1-x^2}|\\ &=\log|1+x| - \frac{1}{2}\log|1-x^2|\\ &=\log|1+x| - \frac{1}{2}\log|(1+x)(1-x)|\\ &=\log|1+x| - \frac{1}{2}\log|1+x|- \frac{1}{2}\log|1-x|\\ \diff{}{x}\left\{\log\left| \dfrac{1+x}{\sqrt{1-x^2}}\right|\right\}&= \diff{}{x}\left\{\log|1+x| - \frac{1}{2}\log|1+x|- \frac{1}{2}\log|1-x|\right\}\\ &=\frac{1}{1+x} - \frac{1/2}{1+x}+ \frac{1/2}{1-x}\\ &= \frac{1/2}{1+x}+ \frac{1/2}{1-x}\\ &=\frac{1}{1-x^2} \end{align*} Notice this is the integrand from our work in blue.
2. False: $\displaystyle\int_{2}^{3} \frac{1}{1-x^2}\dee{x}$ is a number, because it is the area under a finite portion of a continuous curve. (We note that the integrand is continuous over the interval $[2,3]\text{,}$ although it is not continuous everywhere.) However, $\left[\log\left| \dfrac{1+x}{\sqrt{1-x^2}}\right|\right]_{x=2}^{x=3}$ is not defined, since the denominator takes the square root of a negative number. So, these two expressions are not the same.
3. The work in the question is not correct. The most salient problem is that when we make the substitution $x=\sin\theta\text{,}$ we restrict the possible values of $x$ to $[-1,1]\text{,}$ since this is the range of the sine function. However, the original integral had no such restriction.

How can we be sure we avoid this problem in the future? In the introductory text to Section 1.9 (before Example 1.9.1), the notes tell us that we are allowed to write our old variable as a function of a new variable (say $x=s(u)$) as long as that function is invertible to recover our original variable $x\text{.}$ There is one very obvious reason why invertibility is necessary: after we antidifferentiate using our new variable $u\text{,}$ we need to get it back in terms of our original variable, so we need to be able to recover $x\text{.}$ Moreover, invertibility reconciles potential problems with domains: if an inverse function $u=s^{-1}(x)$ exists, then for any $x\text{,}$ there exists a $u$ with $s(u)=x\text{.}$ (This was not the case in the work for the question, because we chose $x=\sin \theta\text{,}$ but if $x=2\text{,}$ there is no corresponding $\theta\text{.}$ Note, however, that $x=\sin\theta$ is invertible over $[-1,1]\text{,}$ so the work is correct if we restrict $x$ to those values.)

Remark: in the next section, you will learn to use partial fractions to find $\displaystyle\int \dfrac{1}{1-x^2}\dee{x} = \log|1+x|-\dfrac{1}{2}\log|1-x|\text{.}$ When $-1 \lt x \lt 1\text{,}$ this is equivalent to $\log\left| \dfrac{1+x}{\sqrt{1-x^2}}\right|\text{.}$

Exercise28
Hint

Consider the ranges of the inverse trigonometric functions. For (c), also consider the domain of $\sqrt{x^2-a^2}\text{.}$

(a), (b): None.

(c): $x \lt -a$

Solution

Remember that for any value $X\text{,}$

\begin{equation*} |X| = \left\{\begin{array}{rl} X&\mbox{if }X \ge 0\\ -X&\mbox{if }X \le 0 \end{array}\right. \end{equation*}

So, $|X| \neq X$ precisely when $X \lt 0\text{.}$

(a) The range of arcsine is $\big[-\frac{\pi}{2},\frac{\pi}{2}\big]\text{.}$ So, since $u=\arcsin(x/a)\text{,}$ $u$ is in the range $\big[-\frac{\pi}{2},\frac{\pi}{2}\big]\text{.}$ Therefore $\cos u \geq 0\text{.}$ Since $a$ is positive, $a\cos u \ge 0\text{,}$ so $a\cos u = |a\cos u|\text{.}$ That is,

\begin{equation*} \sqrt{a^2-x^2}=|a\cos u|=a\cos u \end{equation*}

all the time.

(b) The range of arctangent is $\big(-\frac{\pi}{2},\frac{\pi}{2}\big)\text{.}$ So, since $u=\arctan(x/a)\text{,}$ $u$ is in the range $\big(-\frac{\pi}{2},\frac{\pi}{2}\big)\text{.}$ Therefore $\sec u = \frac{1}{\cos u} \gt 0 \text{.}$ Since $a$ is positive, $a\sec u \gt 0\text{,}$ so $a\sec u = |a\sec u|\text{.}$That is,

\begin{equation*} \sqrt{a^2+x^2}=|a\sec u|=a\sec u \end{equation*}

all the time.

(c) The range of arccosine is $\big[0,\pi \big]\text{.}$ So, since $u=\arcsec(x/a) = \arccos(a/x)\text{,}$ $u$ is in the range $\big[0,\pi\big]\text{.}$ (Actually, it's in the range $[0,\frac{\pi}{2}) \cup (\frac{\pi}{2},\pi]\text{,}$ since secant is undefined at $\pi/2\text{.}$) If $|a\tan u| \neq a\tan u\text{,}$ then $\tan u \lt 0\text{,}$ which happens when $u$ is in the range $\big (\frac{\pi}{2},\pi)\text{.}$ This is the same range over which $-1 \lt \cos u \lt 0\text{,}$ and so $-1 \lt \frac{a}{x} \lt 0\text{.}$ Since $\frac{a}{x} \lt 0\text{,}$ $a$ and $x$ have different signs, so $x \lt 0\text{.}$ Then since $-1 \lt \frac{a}{x}\text{,}$ also $x \lt -a\text{.}$

So,

\begin{equation*} \sqrt{x^2-a^2} = |a\tan u| = -a\tan u \neq a\tan u \end{equation*}

happens precisely when when $x \lt -a\text{.}$

Exercises1.10.4Exercises

Exercise1
Hint

If a quadratic function can be factored as $(ax+b)(cx+d)$ for some constants $a,b,c,d\text{,}$ then it has roots $-\frac{b}{a}$ and $-\frac{d}{c}\text{.}$

(a) (iii)

(b) (ii)

(c) (ii)

(d) (i)

Solution

If a quadratic function can be factored as $(ax+b)(cx+d)$ for some constants $a,b,c,d\text{,}$ then it has roots $-\frac{b}{a}$ and $-\frac{d}{c}\text{.}$ So, if a quadratic function has no roots, it is irreducible: this is the case for the function in graph (d).

If a quadratic function has two different roots, then $(ax+b) \neq \alpha(cx+d)$ for any constant $\alpha\text{.}$ That is, the quadratic function is the product of distinct linear factors. This is the case for the functions graphed in (b) and (c), since these each have two distinct places where they cross the $x$-axis.

Finally, if a quadratic function has precisely one root, then $\frac{b}{a}=\frac{d}{c}\text{,}$ so:

\begin{align*} (ax+b)(cx+d)&=a(x+\tfrac{b}{a})(cx+d) = a(x+\tfrac{d}{c})(cx+d) = \tfrac{a}{c}(cx+d)(cx+d) \end{align*}

That is, the quadratic function is the product of a repeated linear factor, and a constant $\frac{a}{c}$ (which might simply be $\frac{a}{c}=1$).

Exercise2(*)
Hint

Review Equations 1.10.7 through 1.10.11. Be careful to fully factor the denominator.

$\displaystyle\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{Ex+F}{x^2+1}$

Solution

Our first step is to fully factor the denominator:

\begin{equation*} (x^2-1)^2(x^2+1) = (x-1)^2(x+1)^2(x^2+1) \end{equation*}

Once a term is linear, it can't be factored further; for quadratic terms, we should check that they are irreducible. Since $x^2+1$ has no real roots (we are familiar with its graph, which is entirely above the $x$-axis), it is irreducible, so now our denominator is fully factored.

\begin{align*} \frac{x^3+3}{(x^2-1)^2(x^2+1)} &=\frac{x^3+3}{(x-1)^2(x+1)^2(x^2+1)}\\ &=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{Ex+F}{x^2+1} \end{align*}

Notice $(x-1)$ and $(x+1)$ are (repeated) linear factors, while $(x^2+1)$ is an irreducible quadratic factor. This accounts for the difference in the numerators of their corresponding terms.

Exercise3(*)
Hint

Review Example 1.10.1. Is the “Algebraic Method” or the “Sneaky Method” going to be easier?

$3$

Solution

The partial fraction decomposition has the form

\begin{gather*} \frac{3x^3-2x^2+11}{x^2(x-1)(x^2+3)} = \frac A{x-1} + \text{various terms} \end{gather*}

When we multiply through by the original denominator, this becomes

\begin{gather*} {3x^3-2x^2+11} = {x^2(x^2+3)}A + (x-1)(\text{other terms}). \end{gather*}

Evaluating both sides at $x=1$ yields $3\cdot1^3-2\cdot1^2+11=1^2(1^2+3)A+0\text{,}$ or $A=3\text{.}$

Exercise4
Hint

For each part, use long division as in Example 1.10.4.

(a) $\displaystyle\frac{x^3+2x+2}{x^2+1} = x+\frac{x+2}{x^2+1}$

(b) $\displaystyle\dfrac{15x^4+6x^3+34x^2+4x+20}{5x^2+2x+8} = 3x^2+2+\frac{4}{5x^2+2x+8}$

(c) $\displaystyle\dfrac{2x^5+9x^3+12x^2+10x+30}{2x^2+5}=x^3+2x+6$

Solution

1. We start by dividing. The leading term of the numerator is $x$ times the leading term of the denominator. The remainder is $x+2\text{.}$

That is, $x^3+2x+2 = x(x^2+1)+(x+2)\text{.}$ So,

\begin{equation*} \frac{x^3+2x+2}{x^2+1} = x+\frac{x+2}{x^2+1} \end{equation*}
2. We start by dividing. The leading term of the numerator is $3x^2$ times the leading term of the denominator.

Then $5x^2$ goes into $10x^2$ twice, so

Our remainder is 4. That is,

\begin{equation*} \dfrac{15x^4+6x^3+34x^2+4x+20}{5x^2+2x+8} = 3x^2+2+\frac{4}{5x^2+2x+8}. \end{equation*}
3. We start by dividing. The leading term of the numerator is $x^3$ times the leading term of the denominator.

Then $2x^2(2x)$ gives us $4x^3\text{.}$

Finally, $2x^2$ goes into $12x^2$ six times.

Since there is no remainder,

\begin{equation*} \dfrac{2x^5+9x^3+12x^2+10x+30}{2x^2+5}=x^3+2x+6 \end{equation*}

Remark: if we wanted to be pedantic about the question statement, we could write our final answer as $x^3+2x+6+\frac{0}{x}\text{,}$ so that we are indeed adding a polynomial to a rational function whose numerator has degree strictly smaller than its denominator.

Exercise5
Hint

(a) Look for a pattern you can exploit to factor out a linear term.

(b) If you set $y=x^2\text{,}$ this is quadratic. Remember $(x^2-a)= (x+\sqrt{a})(x-\sqrt{a})$ as long as $a$ is positive

(c),(d) Look for integer roots, then use long division.

(a) $5x^3-3x^2-10x+6=(x+\sqrt{2})(x-\sqrt{2})(5x-3)$

(b) $x^4-3x^2-5=\displaystyle\left(x+\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x-\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x^2+\frac{\sqrt{29}-3}{2}\right)$

(a) $x^4-4x^3-10x^2-11x-6 = (x+1)(x-6)(x^2+x+1)$

(b) $2x^4+12x^3-x^2-52x+15= (x+3)(x+5)\left(x-(2+\sqrt2)\right)\left(x-(2-\sqrt2)\right)$

Solution

1. The polynomial $5x^3-3x^2-10x+6$ has a repeated pattern: the ratio of the first two coefficients is the same as the ratio of the last two coefficients. We can use this to factor. \begin{align*} 5x^3-3x^2-10x+6&=x^2(5x-3)-2(5x-3) = (x^2-2)(5x-3)\\ &=(x+\sqrt{2})(x-\sqrt{2})(5x-3) \end{align*}
2. The polynomial $x^4-3x^2-5$ has only even powers of $x\text{,}$ so we can (temporarily) replace them with $x^2=y$ to turn our quartic polynomial into a quadratic. \begin{align*} x^4-3x^2-5&=y^2-3y-5\\ \end{align*}

There's no obvious factoring here, but we can find its roots, if any, using the quadratic equation.

We'd like to use the difference of two squares to factor these quadratic expressions. For this to work, the constants must be positive (so their square roots are real). Since $\sqrt{29} \gt 3\text{,}$ only the first quadratic is factorable. The other is irreducible — it's always positive, so it had no roots.

\begin{align*} x^4-3x^2-5&=\left(x+\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x-\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x^2+\frac{\sqrt{29}-3}{2}\right) \end{align*}
3. Without seeing any obvious patterns, we start hunting for roots. Since we have all integer coefficients, if there are any integer roots, they will divide our constant term, $-6\text{.}$ So, our candidates for roots are $\pm1,\,\pm2,\,\pm3,$ and $\pm6\text{.}$ To save time, we don't need to know exactly the value of our polynomial at these points: only whether or not it is 0. Write $f(x) = x^4 - 4x^3 - 10x^2 - 11x - 6\text{.}$

\begin{align*} \color{red}{f(-1)}&\color{red}{=0}& f(-2)&\neq 0& f(-3)&\neq 0& f(-6)&\neq 0\\ f(1)&\neq 0& f(2)&\neq 0& f(3)&\neq 0& \color{red}{f(6)}&\color{red}{= 0} \end{align*}

Since $x=-1$ and $x=6$ are roots of our polynomial, it has factors $(x+1)$ and $(x-6)\text{.}$ Note $(x+1)(x-6) = x^2-5x-6\text{.}$ We use long division to figure out what else is lurking in our polynomial.

So, $x^4-4x^3-10x^2-11x-6 = (x+1)(x-6)(x^2+x+1)\text{.}$

We should check whether $x^2+x+1$ is reducible or not. If we try to find its roots with the quadratic equation, we get $\dfrac{-1\pm\sqrt{-3}}{2}\text{,}$ which are not real numbers. So, we're at the end of our factoring.

4. Without seeing any obvious patterns, we start hunting for roots. Since we have all integer coefficients, if there are any integer roots, they will divide our constant term, $-15\text{.}$ So, our candidates for roots are $\pm1,\,\pm3,\,\pm5,$ and $\pm15\text{.}$ Write $f(x) = 2x^4 + 12x^3 - x^2 - 52x + 15\text{.}$

\begin{align*} f(-1)&\neq 0& \color{red}{f(-3)}&\color{red}{=0}& \color{red}{f(-5)}&\color{red}{=0}& f(-5)&\neq 0\\ f(1)&\neq 0& f(3)&\neq 0& f(5)&\neq 0& f(15)&\neq 0& \end{align*}

Since $x=-3$ and $x=-5$ are roots of our polynomial, it has factors $(x+3)$ and $(x+5)\text{.}$ Note $(x+3)(x+5) = x^2+8x+15\text{.}$ We use long division to move forward.

So, $2x^4+12x^3-x^2-52x+15= (x+3)(x+5)(2x^2-4x+1)\text{.}$

We should check whether $2x^2-4x+1$ is reducible or not. There's not an obvious way to factor it, but we can use the quadratic equation. This gives us roots $\dfrac{4\pm\sqrt{16-8}}{2}=2\pm\sqrt2\text{.}$ So, we have two more linear factors.

Specifically: $2x^4+12x^3-x^2-52x+15= (x+3)(x+5)\left(x-(2+\sqrt2)\right)\left(x-(2-\sqrt2)\right)\text{.}$

Exercise6
Hint

Why do we do partial fraction decomposition at all?

The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution $u=(ax+b)\text{.}$ It's less clear how to find the antiderivative of (2).

Solution

The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution $u=(ax+b)\text{.}$ It's less clear how to find the antiderivative of (2).

Exercise7(*)
Hint

What is the title of this section?

$\displaystyle\log\frac{4}{3}$

Solution

The integrand is a rational function, so it's a candidate for partial fraction. We quickly rule out any obvious substitution or integration by parts, so we go ahead with the decomposition.

We start by expressing the integrand, i.e. the fraction $\frac{1}{x+x^2}=\frac{1}{x(1+x)}\text{,}$ as a linear combination of the simpler fractions $\frac{1}{x}$ and $\frac{1}{x+1}$ (which we already know how to integrate). We will have

\begin{gather*} \frac{1}{x+x^2}=\frac{1}{x(1+x)} = \frac{a}{x} + \frac{b}{x+1} =\frac{a(x+1) +bx}{x(1+x)} \end{gather*}

The fraction on the left hand side is the same as the fraction on the right hand side if and only if the numerator on the left hand side, which is $1 = 0x + 1\text{,}$ is equal to the numerator on the right hand side, which is $a(x+1)+bx = (a+b)x +a\text{.}$ This in turn is the case if and only of $a=1$ (i.e. the constant terms are the same in the two numerators) and $a+b=0$ (i.e. the coefficients of $x$ are the same in the two numerators). So $a=1$ and $b=-1\text{.}$ Now we can easily evaluate the integral

\begin{align*} \int_1^2 \frac{\dee{x}}{x+x^2} &=\int_1^2 \frac{\dee{x}}{x(x+1)} =\int_1^2 \Big(\frac{1}{x}-\frac{1}{x+1}\Big)\,\dee{x} =\Big[\log x-\log(x+1)\Big]_1^2\\ &=\log 2-\log\frac{3}{2} =\log\frac{4}{3} \end{align*}
Exercise8(*)
Hint

You can save yourself some work in developing your partial fraction decomposition by renaming $x^2$ to $y$ and comparing the result with Question 7.

$-\dfrac{1}{x}-\arctan x+C$

Solution

We'll first do a partial fraction decomposition. The sneaky way is to temporarily rename $x^2$ to $y\text{.}$ Then $x^4+x^2=y^2+y$ and

\begin{equation*} \frac{1}{x^4+x^2} = \frac{1}{y(y+1)} =\frac{1}{y}-\frac{1}{y+1} \end{equation*}

as we found in Question 7. Now we restore $y$ to $x^2\text{.}$

\begin{gather*} \int \frac1{x^4+x^2}\,\dee{x} =\int \Big(\frac{1}{x^2}-\frac{1}{x^2+1}\Big)\,\dee{x} =-\frac{1}{x}-\arctan x+C \end{gather*}
Exercise9(*)
Hint

Review Steps 3 (particularly the “Sneaky Method”) and 4 of Example 1.10.3.

$4 \log |x-3| - 2 \log (x^2 + 1) + C$

Solution

The integrand is of the form $N(x)/D(x)$ with $D(x)$ already factored and $N(x)$ of lower degree. We immediately look for a partial fraction decomposition:

\begin{equation*} \frac{12x+4}{(x-3)(x^2+1)} = \frac{A}{x-3} + \frac{Bx+C}{x^2+1}. \end{equation*}

Multiplying through by the denominator yields \begin{align} 12x+4 &= A(x^2+1) + (Bx+C)(x-3) \tag{$*$} \end{align} Setting $x=3$ we find:

\begin{gather*} 36 +4 = A(9 + 1) + 0 \implies 40 =10A \implies \color{red}{A = 4} \end{gather*}

Substituting $A=4$ in $(*)$ gives

\begin{align*} 12x+4 = 4(x^2+1) + (Bx+C)(x-3) &\implies -4x^2+12 x =(x-3)(Bx + C) \\ &\implies (-4x)(x-3)=(Bx + C)(x-3) \\ & \implies \color{red}{B=-4, C=0} \end{align*}

So we have found that $A=4\text{,}$ $B=-4\text{,}$ and $C=0\text{.}$ Therefore

\begin{align*} \int \frac{12x+4}{(x-3)(x^2+1)} \,dx &= \int \bigg( \frac{4}{x-3} - \frac{4x}{x^2+1} \bigg) \,dx \\ &= 4 \log |x-3| - 2 \log (x^2 + 1) + C \end{align*}

The second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution $u=x^2+1\text{,}$ $\dee{u}=2x\,\dee{x}\text{.}$

Exercise10(*)
Hint

Review Steps 3 (particularly the “Sneaky Method”) and 4 of Example 1.10.3. Remember $\diff{}{x}\{\arctan x\} = \frac{1}{1+x^2}\text{.}$

$F(x) = \log |x-2| + \log |x^2+4| + 2\arctan (x/2) + D$

Solution

The integrand is of the form $N(x)/D(x)$ with $D(x)$ already factored and $N(x)$ of lower degree. With no obvious substitution available, we look for a partial fraction decomposition.

\begin{gather*} \frac{3x^2 -4}{(x-2)(x^2+4)} = \frac{A}{x-2} + \frac{Bx + C}{x^2+4} \end{gather*}

Multiplying through by the denominator gives \begin{align} 3x^2 -4 = A(x^2+4) + (Bx + C)(x-2) \tag{$*$} \end{align} Setting $x=2$ we find:

\begin{gather*} 12 -4 = A(4 + 4) + 0 \implies 8 =8A \implies \color{red}{A = 1} \end{gather*}

Substituting $A=1$ in $(*)$ gives

\begin{align*} 3x^2 -4 = (x^2+4) + (x-2)(Bx + C) &\implies 2x^2-8 =(x-2)(Bx + C) \\ &\implies (x-2)(2x+4)=(x-2)(Bx + C) \\ & \implies \color{red}{ B=2, C=4} \end{align*}

Thus, we have:

\begin{gather*} \frac{3x^2 -4}{(x-2)(x^2+4)} = \frac{1}{x-2} + \frac{2x + 4}{x^2+4} = \frac{1}{x-2} + \frac{2x }{x^2+4} + \frac{4}{x^2+4 } \end{gather*}

The first two of these are directly integrable:

\begin{gather*} F(x) = \log |x-2| + \log |x^2+4| + \int \frac{4}{x^2+4}\,\dee{x} \end{gather*}

(The second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution $u=x^2+4\text{,}$ $\dee{u}=2x\,\dee{x}\text{.}$) For the final integral, we substitute: $x = 2y\text{,}$ $\dee{x} = 2\dee{y}\text{,}$ and see that:

\begin{gather*} \int \frac{4}{x^2+4}\,\dee{x} = 2 \int \frac{1}{y^2 + 1}\,dy = 2 \arctan y + D = 2\arctan (x/2) + D \end{gather*}

for any constant $D\text{.}$ All together we have:

\begin{gather*} F(x) = \log |x-2| + \log |x^2+4| + 2\arctan (x/2) + D \end{gather*}
Exercise11(*)
Hint

Fill in the blank: the integrand is a $\underbar{\ \ \ \ \ \ \ \ }$ function.

$-2\log|x-3|+3\log|x+2|+C$

Solution

This integrand is a rational function, with no obvious substitution. This sure looks like a partial fraction problem. Let's go through our protocol.

• The degree of the numerator $x-13$ is one, which is strictly smaller than the dergee of the denominator $x^2-x-6\text{,}$ which is two. So we don't need long division to pull out a polynomial.
• Next we factor the denominator. \begin{gather*} x^2-x-6 = (x-3)(x+2) \end{gather*}
• Next we find the partial fraction decomposition of the integrand. It is of the form \begin{gather*} \frac{x-13}{(x-3)(x+2)} =\frac{A}{x-3} + \frac{B}{x+2} \end{gather*} To find $A$ and $B\text{,}$ using the sneaky method, we cross multiply by the denominator. \begin{gather*} x-13 = A(x+2) + B(x-3) \end{gather*} Now we can find $A$ by evaluating at $x=3$ \begin{gather*} 3-13 = A(3+2) + B(3-3) \implies \color{red}{ A=-2} \end{gather*} and find $B$ by evaluating at $x=-2\text{.}$ \begin{gather*} -2-13 = A(-2+2) + B(-2-3) \implies \color{red}{ B=3} \end{gather*} (Hmmm. $A$ and $B$ are nice round numbers. Sure looks like a rigged exam or homework question.) Our partial fraction decomposition is \begin{gather*} \frac{x-13}{(x-3)(x+2)} =\frac{-2}{x-3} + \frac{3}{x+2} \end{gather*} As a check, we recombine the right hand side and make sure that it matches the left hand side. \begin{gather*} \frac{-2}{x-3} + \frac{3}{x+2} =\frac{-2(x+2)+3(x-3)}{(x-3)(x+2)} =\frac{x-13}{(x-3)(x+2)} \end{gather*}
• Finally, we evaluate the integral. \begin{gather*} \int \frac{x-13}{x^2-x-6}\dee{x} =\int\bigg(\frac{-2}{x-3} + \frac{3}{x+2}\bigg)\dee{x} =-2\log|x-3|+3\log|x+2|+C \end{gather*}
Exercise12(*)
Hint

The integrand is yet another $\underbar{\ \ \ \ \ \ \ \ }$ function.

$-9\log|x+2|+14\log|x+3| +C$

Solution

Again, this sure looks like a partial fraction problem. So let's go through our protocol.

• The degree of the numerator $5x+1$ is one, which is strictly smaller than the dergee of the denominator $x^2+5x+6\text{,}$ which is two. So we do not long divide to pull out a polynomial.
• Next we factor the denominator. \begin{gather*} x^2+5x+6 = (x+2)(x+3) \end{gather*}
• Next we find the partial fraction decomposition of the integrand. It is of the form \begin{gather*} \frac{5x+1}{(x+2)(x+3)} =\frac{A}{x+2} + \frac{B}{x+3} \end{gather*} To find $A$ and $B\text{,}$ using the sneaky method, we cross multiply by the denominator. \begin{gather*} 5x+1= A(x+3) + B(x+2) \end{gather*} Now we can find $A$ by evaluating at $x=-2$ \begin{gather*} -10+1 = A(-2+3) + B(-2+2) \implies\color{red}{ A=-9} \end{gather*} and find $B$ by evaluating at $x=-3\text{.}$ \begin{gather*} -15+1 = A(-3+3) + B(-3+2) \implies\color{red}{ B=14} \end{gather*} So our partial fraction decomposition is \begin{gather*} \frac{5x+1}{(x+2)(x+3)} =\frac{-9}{x+2} + \frac{14}{x+3} \end{gather*} As a check, we recombine the right hand side and make sure that it matches the left hand side \begin{gather*} \frac{-9}{x+2} + \frac{14}{x+3} =\frac{-9(x+3)+14(x+2)}{(x+2)(x+3)} =\frac{5x+1}{(x+2)(x+3)} \end{gather*}
• Finally, we evaluate the integral \begin{gather*} \int \frac{5x+1}{x^2+5x+6}\dee{x} =\int\bigg(\frac{-9}{x+2} + \frac{14}{x+3}\bigg)\dee{x} =-9\log|x+2|+14\log|x+3|+C \end{gather*}
Exercise13
Hint

Since the degree of the numerator is the same as the degree of the denominator, we can't do our partial fraction decomposition before we simplify the integrand.

$\displaystyle5x+\frac{1}{2}\log|x-1| - \frac{7}{2}\log|x+1|+C$

Solution

We have a rational function with no obvious substitution, so let's use partial fraction decomposition.

• Since the degree of the numerator is the same as the degree of the denominator, we need to pull out a polynomial.

That is,

\begin{equation*} \int \frac{5x^2-3x-1}{x^2-1}\dee{x} = \int\left(5+ \frac{-3x+4}{x^2-1}\right)\dee{x} =5x+ \int \frac{-3x+4}{x^2-1}\dee{x} \end{equation*}
• Again, there's no obvious substitution for the new integrand, so we want to use partial fraction. The denominator factors as $(x-1)(x+1)\text{,}$ so our decomposition has this form: \begin{align*} \frac{-3x+4}{x^2-1}&=\frac{-3x+4}{(x-1)(x+1)} = \frac{A}{x-1}+\frac{B}{x+1} = \frac{(A+B)x+(A-B)}{(x-1)(x+1)} \end{align*} So, (1) $A+B=-3$ and (2) $A-B=4\text{.}$
• We solve (2) for $A$ in terms of $B\text{,}$ namely $A=4+B\text{.}$ Plugging this into (1), we see $(4+B)+B=-3\text{.}$ So, $\textcolor{red}{B=-\frac{7}{2}}\text{,}$ and $\textcolor{red}{A=\frac{1}{2}}\text{.}$
• Now we can write our integral in a friendlier form and evaluate. \begin{align*} \int \frac{5x^2-3x-1}{x^2-1}\dee{x} =&=5x+ \int \frac{-3x+4}{x^2-1}\dee{x} = 5x+\int \frac{1/2}{x-1} - \frac{7/2}{x+1} \dee{x}\\ &=5x+\frac{1}{2}\log|x-1| - \frac{7}{2}\log|x+1|+C \end{align*}
Exercise14
Hint

The degree of the numerator is not smaller than the degree of the denominator.\\ Your final answer will have an arctangent in it.

$\displaystyle x-\frac{2}{x}+\frac{5}{2}\arctan (2x) +C$

Solution

The integrand is a rational function with no obvious substitution, so we use partial fraction decomposition.

• The degree of the numerator is the same as the degree of the denominator. Since it's not smaller, we need to re-write our integrand. We could do this using long division, but this case is simple enough to do more informally. \begin{align*} \frac{4x^4+14x^2+2}{4x^4+x^2}&=\frac{4x^4+x^2+13x^2+2}{4x^4+x^2}\\ &=\frac{4x^4+x^2}{4x^4+x^2}+\frac{13x^2+2}{4x^4+x^2}\\ &=1+\frac{13x^2+2}{4x^4+x^2} \end{align*}
• The denominator factors as $x^2(4x^2+1)\text{.}$
• We want to find the partial fraction decomposition of the fractional part of our simplified integrand.

\begin{align*} \frac{13x^2+2}{4x^4+x^2}&= \frac{13x^2+2}{x^2(4x^2+1)} = \frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{4x^2+1}\\ \end{align*}

Multiply through by the original denominator.

\begin{align*} 13x^2+2&=Ax(4x^2+1)+B(4x^2+1)+(Cx+D)x^2 \tag{1}\\ \end{align*}

Setting $x=0$ gives us:

\begin{align*} \textcolor{red}{2}&\color{red}{=B}\\ \end{align*}

We use $B=2$ to simplify Equation (1).

\begin{align*} 13x^2+2&=Ax(4x^2+1)+\textcolor{red}{2}(4x^2+1)+(Cx+D)x^2\\ 5x^2&=Ax(4x^2+1)+(Cx+D)x^2\\ 5x&=A(4x^2+1)+(Cx+D)x\tag{2}\\ \end{align*}

Again, let $x=0\text{.}$

\begin{align*} \color{red}{0}&\color{red}{=A}\\ \end{align*}

Using $A=0\text{,}$ simplify Equation (2).

\begin{align*} 5x&=(Cx+D)x\\ 5&=Cx+D\\ \color{red}{C}& \textcolor{red}{=0},\qquad\color{red}{D=5} \end{align*}
• Now we can write our integral in pieces.

\begin{align*} \int \frac{4x^4+14x^2+2}{4x^4+x^2} \dee{x}&= \int \left(1+\frac{13x^2+2}{4x^4+x^2}\right) \dee{x}\\ &=\int\left(1+ \frac{\textcolor{red}2}{x^2}+\frac{\textcolor{red}5}{4x^2+1}\right) \dee{x} \\ &=x -\frac{2}{x} + \int \frac{5}{(2x)^2+1} \dee{x} \\ \end{align*}

Substitute $u=2x\text{,}$ $\dee{u}=2\dee{x}\text{.}$

\begin{align*} & =x -\frac{2}{x} + \int \frac{5/2}{u^2+1} \dee{u} \\ &=x-\frac{2}{x}+\frac{5}{2}\arctan u +C\\ &=x-\frac{2}{x}+\frac{5}{2}\arctan (2x) +C \end{align*}
Exercise15
Hint

In the partial fraction decomposition, several constants turn out to be 0.

$\displaystyle\frac{1}{x}-\frac{2}{x-1}+C$

Solution

The integrand is a rational function with no obvious substitution, so we'll use a partial fraction decomposition.

• Since the numerator has strictly smaller degree than the denominator, we don't need to start off with a long division.
• We do, however, need to factor the denominator. We can immediately pull out $x^2\text{;}$ the remaining part is $x^2-2x+1 = (x-1)^2\text{.}$
• Now we can perform our partial fraction decomposition. \begin{align*} \frac{x^2+2x-1}{x^4-2x^3+x^2}&= \frac{x^2+2x-1}{x^2(x-1)^2} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}\\ \end{align*}

Multiply both sides by the original denominator.

\begin{align*} x^2+2x-1&=Ax(x-1)^2+B(x-1)^2+Cx^2(x-1)+Dx^2\tag{1}\\ \end{align*}

To be sneaky, we set $x=0\text{,}$ and find:

\begin{align*} \color{red}{-1}&\color{red}{=B}\\ \end{align*}

We also set $x=1\text{,}$ and find:

\begin{align*} \color{red}{2}&\color{red}{=D}\\ \end{align*}

We use $B$ and $D$ to simplify Equation (1).

\begin{align*} x^2+2x-1&=Ax(x-1)^2\textcolor{red}{-1}(x-1)^2+Cx^2(x-1)+\textcolor{red}{2}x^2\\ 0&=Ax(x-1)^2+Cx^2(x-1)\\ &=x(x-1)[(A+C)x-A]\\ \mbox{So,}\qquad 0&=(A+C)x-A \end{align*} That is, $\textcolor{red}{A=C=0}\text{.}$
• Now we can evaluate our integral. \begin{align*} \int \frac{x^2+2x-1}{x^4-2x^3+x^2} \dee{x}&=\int \left(\frac{-1}{x^2}+\frac{2}{(x-1)^2}\right) \dee{x}\\ &=\frac{1}{x}-\frac{2}{x-1}+C \end{align*}
Exercise16
Hint

Factor $(2x-1)$ out of the denominator to get started. You don't need long division for this step.

$\displaystyle-\frac{1}{2}\log|x-2| + \frac{1}{2}\log|x+2| + \frac{3}{2}\log|2x-1|+C$

Solution

Our integrand is a rational function with no obvious substitution, so we'll use the method of partial fractions.

• The degree of the numerator is less than the degree of the denominator.
• We need to factor the denominator. The first two terms have the same ratio as the last two terms. \begin{align*} 2x^3-x^2-8x+4&=x^2(2x-1)-4(2x-1) \\ &= (x^2-4)(2x-1)\\ &=(x-2)(x+2)(2x-1) \end{align*}
• Now we find our partial fraction decomposition. \begin{align*} \frac{ 3x^2-4x-10}{2x^3-x^2-8x+4}&=\frac{ 3x^2-4x-10}{(x-2)(x+2)(2x-1)}= \frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{2x-1}\\ \end{align*}

Multiply both sides by the original denominator.

\begin{align*} 3x^2-4x-10&=A(x+2)(2x-1)+B(x-2)(2x-1)+C(x-2)(x+2)\\ \end{align*}

Distinct linear factors is the best possible scenario for the sneaky method. First, let's set $x=2\text{.}$

\begin{align*} 3(4)-4(2)-10&=A(4)(3)+B(0)+C(0)\\ \color{red}{A}&\color{red}{=-\frac{1}{2}}\\ \end{align*}

Now, let $x=-2\text{.}$

\begin{align*} 3(4)-4(-2)-10&=A(0)+B(-4)(-5)+C(0)\\ \color{red}{B}&\color{red}{=\frac{1}{2}}\\ \end{align*}

Finally, let $x=\frac{1}{2}\text{.}$

\begin{align*} \frac{3}{4}-2-10&=A(0)+B(0)+C\left(-\frac{3}{2}\right)\left(\frac{5}{2}\right)\\ \color{red}{C}&\color{red}{=3} \end{align*}
• Now we can evaluate our integral in its new form. \begin{align*} \int \frac{ 3x^2-4x-10}{2x^3-x^2-8x+4} \dee{x}&=\int \left(\frac{-1/2}{x-2}+\frac{1/2}{x+2}+\frac{3}{2x-1}\right) \dee{x}\\ &=-\frac{1}{2}\log|x-2| + \frac{1}{2}\log|x+2| + \frac{3}{2}\log|2x-1|+C\\ &=\frac{1}{2}\log\left| \frac{x+2}{x-2} \right| + \frac{3}{2}\log|2x-1|+C \end{align*}
Exercise17
Hint

When it comes time to integrate, look for a convenient substitution.

$\displaystyle\log \left(\frac{4\cdot 6^3}{5^3}\right)$

Solution

The integrand is a rational function with no obvious substitution, so we use the method of partial fractions.

• The numerator has smaller degree than the denominator.
• We need to factor the denominator. In the absence of any clues, we look for an integer root. The constant term is 5, so the possible integer roots are $\pm 1$ and $\pm 5\text{.}$ Name $f(x) = 2x^3+11x^2+6x+5\text{.}$

So, $(x+5)$ is a factor of the denominator.

• We use long division to pull out the factor of $(x+5)\text{.}$

That is, our denominator is $(x+5)(2x^2+x+1)\text{.}$

• The quadratic function $2x^2+x+1$ is irreducible: we can see this by using the quadratic equation, and finding no real roots. So, we are ready to find our partial fraction decomposition. \begin{align*} \frac{10x^2+24x+8}{2x^3+11x^2+6x+5}&=\frac{10x^2+24x+8}{(x+5)(2x^2+x+1)}=\frac{A}{x+5}+\frac{Bx+C}{2x^2+x+1}\\ \end{align*}

Multiply through by the original denominator.

\begin{align*} 10x^2+24x+8&=A(2x^2+x+1)+(Bx+C)(x+5)\tag{1}\\ \end{align*}

Set $x=-5\text{.}$

\begin{align*} 10(25)-24(5)+8&=A(2(25)-5+1) + (B(-5)+C)(0)\\ \color{red}{A}&\color{red}{=3}\\ \end{align*}

Using our value of $A\text{,}$ we simplify Equation (1).

\begin{align*} 10x^2+24x+8&=\textcolor{red}{3}(2x^2+x+1)+(Bx+C)(x+5)\\ 4x^2+21x+5&=(Bx+C)(x+5)\\ \end{align*}

We factor the left side. We know $(x+5)$ must be one of its factors.

\begin{align*} (4x+1)(x+5)&=(Bx+C)(x+5)\\ 4x+1&=Bx+C \end{align*} So, $\textcolor{red}{B=4}$ and $\textcolor{red}{C=1}\text{.}$
• Now we can write our integral in smaller pieces. \begin{align*} \int_0^1 \frac{10x^2+24x+8}{2x^3+11x^2+6x+5} \dee{x}&= \int_0^1 \left( \frac{3}{x+5} + \frac{4x+1}{2x^2+x+1}\right) \dee{x}\\ \end{align*}

The antiderivative of the left fraction is $3\log|x+5|\text{.}$ For the right fraction, we use the substitution $u=2x^2+x+1\text{,}$ $\dee{u}=(4x+1)\dee{x}$ to antidifferentiate.

\begin{align*} &=\big[3\log|x+5| + \log|2x^2+x+1|\big]_0^1\\ &=3\log 6 + \log 4 - 3\log 5 -\log 1\\ &=\log \left(\frac{4\cdot 6^3}{5^3}\right) \end{align*}
Exercise18
Hint

$\displaystyle\csc x = \frac{1}{\sin x} = \frac{\sin x}{\sin^2 x}$

$\displaystyle\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x}\right|+C$

Solution

We follow the example in the text.

\begin{align*} \int \csc x \dee{x}&=\int \frac{1}{\sin x} \dee{x}=\int \frac{\sin x}{\sin^2 x} \dee{x} = \int \frac{\sin x}{1-\cos^2 x} \dee{x} \\ \end{align*}

Let $u=\cos x\text{,}$ $\dee{u}=-\sin x \dee{x}\text{.}$

\begin{align*} &=\int\frac{-1}{1-u^2} \dee{u} = \int\frac{-1}{(1+u)(1-u)} \dee{u}\\ \end{align*}

We see an opportunity for partial fraction.

\begin{align*} \frac{-1}{(1+u)(1-u)}&=\frac{A}{1+u}+\frac{B}{1-u}\\ \end{align*}

Multiply both sides by the original denominator.

\begin{align*} -1&=A(1-u)+B(1+u)\\ \end{align*}

Let $u=1\text{.}$

\begin{align*} -1&=2B \qquad\Rightarrow\color{red}{ B = -\frac{1}{2}}\\ \end{align*}

Let $u=-1\text{.}$

\begin{align*} -1&=2A \qquad\Rightarrow \color{red}{A = -\frac{1}{2}}\\ \end{align*}

We can now re-write our integral.

\begin{align*} \int \csc x \dee{x}&= \int\frac{-1}{(1+u)(1-u)} \dee{u} =\int \left(\frac{-1/2}{1+u} + \frac{-1/2}{1-u}\right) \dee{u}\\ &=-\frac{1}{2}\log|1+u| +\frac{1}{2}\log|1-u|+C\\ &=\frac{1}{2}\log\left| \frac{1-u}{1+u}\right|+C\\ &=\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x}\right|+C \end{align*}

Remark: Elsewhere in the text, and in many tables of integrals, the antiderivative of cosecant is given as $\log|\csc x - \cot x|\text{.}$ We show that this is equivalent to our result.

\begin{align*} \log|\csc x - \cot x|&= \frac{1}{2}\log\left|\left(\csc x - \cot x\right)^2\right| = \frac{1}{2}\log\left| \csc^2 x - 2\csc x \cot x + \cot^2 x\right|\\ &=\frac{1}{2}\log\left| \frac{1}{\sin^2x} - \frac{2\cos x}{\sin^2 x}+\frac{\cos^2 x}{\sin^2x}\right|\\ &=\frac{1}{2}\log\left| \frac{1-2\cos x + \cos^2 x}{\sin^2x} \right| =\frac{1}{2}\log\left| \frac{\left(1-\cos x\right)^2}{1-\cos^2x} \right|\\ &=\frac{1}{2}\log\left| \frac{\left(1-\cos x\right)^2}{(1-\cos x)(1+\cos x)} \right| =\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x} \right| \end{align*}
Exercise19
Hint

Use the partial fraction decomposition from Queston 18 to save yourself some time.

$\displaystyle\frac{-\cos x}{2\sin^2 x} + \frac{1}{4}\log\left|\frac{1-\cos x}{1+\cos x}\right|+C$

Solution

We follow the example in the text.

\begin{align*} \displaystyle\int \csc^3 x \dee{x}&=\int \frac{1}{\sin^3x} \dee{x} =\int \frac{\sin x}{\sin^4x} \dee{x}=\int \frac{\sin x}{(1-\cos^2x)^2} \dee{x}\\ \end{align*}

Let $u=\cos x\text{,}$ $\dee{u}=-\sin x \dee{x}\text{.}$

\begin{align*} &=\int\frac{-1}{(1-u^2)^2} \dee{u}\\ \end{align*}

In Question 18 we saw $\textcolor{blue}{\frac{1}{1-u^2} = \frac{1/2}{1+u}+\frac{1/2}{1-u}}$ , so

\begin{align*} \int\frac{-1}{(1-u^2)^2} \dee{u}&=-\int\left(\textcolor{blue}{\frac{1}{1-u^2}}\right)^2 \dee{u} =-\int\left(\textcolor{blue}{\frac{1/2}{1+u}+\frac{1/2}{1-u}}\right)^2 \dee{u}\\ &=-\frac{1}{4}\int\left(\frac{1}{(1+u)^2}+\textcolor{blue}{\frac{2}{1-u^2}}+\frac{1}{(1-u)^2}\right) \dee{u}\\ &=-\frac{1}{4}\int\left(\frac{1}{(1+u)^2}+ \textcolor{blue}{\frac{1}{1+u}+\frac{1}{1-u}}+\frac{1}{(1-u)^2}\right) \dee{u}\\ &=-\frac{1}{4}\left(-\frac{1}{1+u} + \log|1+u|-\log|1-u|+\frac{1}{1-u}\right)+C\\ &=-\frac{1}{4}\left(\frac{2u}{1-u^2} + \log\left|\frac{1+u}{1-u}\right|\right)+C\\ &=\frac{-u}{2(1-u^2)} + \frac{1}{4}\log\left|\frac{1-u}{1+u}\right|+C\\ &=\frac{-\cos x}{2\sin^2 x} + \frac{1}{4}\log\left|\frac{1-\cos x}{1+\cos x}\right|+C \end{align*}

Remark: In Example 1.8.22, and in many tables of integrals, the antiderivative of $\csc^3 x$ is given as $-\frac{1}{2}\cot x \csc x + \frac{1}{2}\log|\csc x - \cot x|+C\text{.}$ This is equivalent to our result. Recall in the remark after the solution to Question 18, we saw $\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x}\right|=\log|\csc x - \cot x|\text{.}$

\begin{align*} -\frac{1}{2}\cot x \csc x + \frac{1}{2}\log|\csc x - \cot x|&= -\frac{1}{2}\cot x \csc x + \frac{1}{4}\log\left| \frac{1-\cos x}{1+\cos x}\right|\\ &=-\frac{1}{2}\left(\frac{\cos x}{\sin x}\right)\left(\frac{1}{\sin x}\right) + \frac{1}{4}\log\left| \frac{1-\cos x}{1+\cos x}\right|\\ &=\frac{-\cos x}{2\sin^2 x }+ \frac{1}{4}\log\left| \frac{1-\cos x}{1+\cos x}\right| \end{align*}
Exercise20
Hint

In the final integration, complete the square to make a piece of the integrand look more like the derivative of arctangent.

$\displaystyle3\log 2 + \frac{1}{2}+\frac{2}{\sqrt{15}}\left(\arctan\left(\frac{7}{\sqrt{15}}\right)-\arctan\left(\frac{9}{\sqrt{15}}\right)\right)$

Solution

This is a rational function, and there's no obvious substitution, so we'll use partial fraction decomposition.

• First, we check that the numerator has strictly smaller degree than the denominator, so we don't have to use long division.
• Second, we factor the denominator. We can immediately pull out a factor of $x^2\text{;}$ then we're left with the quadratic polynomial $x^2+5x+10\text{.}$ Using the quadratic equation, we check that this has no real roots, so it is irreducible.
• Once we know the factorization of the denominator, we can set up our decomposition. \begin{align*} \frac{3x^3+15x^2+35x+10}{x^4+5x^3+10x^2}&= \frac{3x^3+15x^2+35x+10}{x^2(x^2+5x+10)}\\ &=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+5x+10}\\ \end{align*}

We multiply both sides by the original denominator.

\begin{align*} 3x^3+15x^2+35x+10&=Ax(x^2+5x+10) + B(x^2+5x+10)+(Cx+D)x^2 \tag{1}\\ \end{align*}

Following the “Sneaky Method,” we plug in $x=0\text{.}$

\begin{align*} 0+10&=A(0)+B(10)+(C(0)+D)(0)\\ \color{red}{B}&\color{red}{=1} \end{align*}
• Knowing $B$ allows us to simplify our Equation (1). \begin{align*} 3x^3+15x^2+35x+10&=Ax(x^2+5x+10) + \textcolor{red}{1}(x^2+5x+10)+(Cx+D)x^2\\ 3x^3+14x^2+30x&=Ax(x^2+5x+10) +(Cx+D)x^2\\ \end{align*}

We can factor $x$ out of both sides of the equation.

\begin{align*} 3x^2+14x+30&=A(x^2+5x+10) +(Cx+D)x \tag{2} \end{align*}
• Again, we set $x=0\text{.}$ \begin{align*} 0+30&=A(10) +(C(0+D)(0)\\ \color{red}{A}&\color{red}{=3} \end{align*}
• We simplify Equation (2), using $A=3\text{.}$ \begin{align*} 3x^2+14x+30&=\textcolor{red}{3}(x^2+5x+10) +(Cx+D)x\\ -x&=Cx^2+Dx\\ \color{red}{C}&\color{red}{=0,\quad D=-1} \end{align*}
• Now that we have our coefficients, we can re-write our integral in a friendlier form. \begin{align*} \int_1^2 \frac{3x^3+15x^2+35x+10}{x^4+5x+10x^2} \dee{x}&= \int_1^2 \left(\frac{3}{x}+\frac{1}{x^2}-\frac{1}{x^2+5x+10} \right) \dee{x}\\ &=\left[3\log|x|-\frac{1}{x}\right]_1^2-\int_1^2\frac{1}{x^2+5x+10} \dee{x}\\ &=3\log 2+\frac{1}{2}-\int_1^2\frac{1}{x^2+5x+10} \dee{x}\\ \end{align*}

The remaining integral is the reciprocal of a quadratic polynomial, much like $\dfrac{1}{1+x^2}\text{,}$ whose antiderivative is arctangent. We complete the square and use the substitution $u=\left(\frac{2x+5}{\sqrt{15}}\right)\text{,}$ $\dee{u}=\frac{2}{\sqrt{15}} \dee{x}\text{.}$

\begin{align*} \int_1^2 \frac{1}{x^2+5x+10} \dee{x}&=\int_1^2 \frac{1}{\left(x+\frac{5}{2}\right)^2+\frac{15}{4}} \dee{x}\\ &=\frac{4}{15}\int_1^2 \frac{1}{\left(\frac{2x+5}{\sqrt{15}}\right)^2+1} \dee{x}\\ &=\frac{2}{\sqrt{15}}\int_{7/\sqrt{15}}^{9/\sqrt{15}} \frac{1}{u^2+1} \dee{u}\\ &=\frac{2}{\sqrt{15}}\Big[\arctan u\Big]_{7/\sqrt{15}}^{9/\sqrt{15}}\\ &=\frac{2}{\sqrt{15}}\left(\arctan\left(\frac{9}{\sqrt{15}}\right)-\arctan\left(\frac{7}{\sqrt{15}}\right)\right) \end{align*} So, all together, \begin{equation*} \int_1^2 \frac{3x^3+15x^2+35x+10}{x^4+5x^3+10x^2} \dee{x}=3\log 2 + \frac{1}{2}-\frac{2}{\sqrt{15}}\left(\arctan\left(\frac{9}{\sqrt{15}}\right)-\arctan\left(\frac{7}{\sqrt{15}}\right)\right) \end{equation*}
Exercise21
Hint

Review Question 1.9.2.20 in Section 1.9 for antidifferentiation tips.

$\displaystyle=\frac{9}{4\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{2+3x}{4(x^2+2)} +C$

Solution

Our integrand is already in the nice form that would come out of a partial fractions decomposition. Let's consider its different pieces.

• First piece: $\int \frac{3}{x^2+2} \dee{x}\text{.}$ The fraction looks somewhat like the derivative of arctangent, so we can massage it to find an appropriate substitution.

\begin{align*} \int \frac{3}{x^2+2} \dee{x}&=\frac{3}{2}\int \frac{1}{\left(\frac{x}{\sqrt2}\right)^2+1} \dee{x}\\ \end{align*}

Use the substitution $u=\frac{x}{\sqrt2}\text{,}$ $\dee{u} = \frac{1}{\sqrt{2}} \dee{x}.$

\begin{align*} &=\frac{3}{\sqrt2}\int\frac{1}{u^2+1} \dee{u}\\ &=\frac{3}{\sqrt2}\arctan u+C\\ &=\frac{3}{\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)+C \end{align*}
• The next piece is $\int \frac{x-3}{(x^2+2)^2} \dee{x}\text{.}$ If the numerator were only $x$ (and no constant), we could use the substitution $u=x^2+2\text{,}$ $\dee{u} = 2x \dee{x}\text{.}$ So, to that end, we can break up that fraction into $\frac{x}{(x^2+2)^2}-\frac{3}{(x^2+2)^2}\text{.}$ For now, we only evaluate the first half.

\begin{align*} \int \frac{x}{(x^2+2)^2} \dee{x}&=\frac{1}{2}\int\frac{1}{u^2} \dee{u} = -\frac{1}{2u}+C\\ &=-\frac{1}{2x^2+4}+C \end{align*}
• That leaves us with the final piece, $\frac{3}{(x^2+2)^2}\text{,}$ which is the hardest. We saw something similar in Question 1.9.2.20 in Section 1.9: we can use the substitution $x=\sqrt{2}\tan\theta\text{,}$ $\dee{x} = \sqrt{2}\sec^2\theta \dee{\theta}\text{.}$

\begin{align*} \int\frac{3}{(x^2+2)^2} \dee{x}&=\int\frac{3}{(2\tan^2\theta+2)^2}\sqrt{2}\sec^2\theta \dee{\theta}\\ &=\int\frac{3}{4\sec^4\theta}\sqrt{2}\sec^2\theta \dee{\theta}\\ &=\frac{3}{2\sqrt{2}}\int\cos^2\theta \dee{\theta}\\ &=\frac{3}{4\sqrt{2}}\int\big(1+\cos(2\theta)\big) \dee{\theta}\\ &=\frac{3}{4\sqrt{2}}\left(\theta+\frac{1}{2}\sin(2\theta)\right)+C\\ &=\frac{3}{4\sqrt{2}}\left(\theta+\sin\theta\cos\theta\right)+C\\ &=\frac{3}{4\sqrt{2}}\left(\arctan\left(\frac{x}{\sqrt2}\right)+\frac{x\sqrt{2}}{x^2+2}\right)+C \end{align*}

From our substitution, $\tan\theta = \frac{x}{\sqrt2}\text{.}$ So, we can draw a right triangle with angle $\theta\text{,}$ opposite side $x\text{,}$ and adjacent side $\sqrt2\text{.}$ Then by the Pythagorean Theorem, the hypotenuse has length $\sqrt{x^2+2}\text{,}$ and this gives us $\sin\theta$ and $\cos\theta\text{.}$

Now we have our integral.

\begin{align*} \int\bigg(\frac{3}{x^2+2}&+\frac{x-3}{(x^2+2)^2}\bigg) \dee{x}= \int\frac{3}{x^2+2} \dee{x}+\int\frac{x}{(x^2+2)^2} \dee{x}-\int\frac{3}{(x^2+2)^2} \dee{x}\\ &=\frac{3}{\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{1}{2x^2+4} -\frac{3}{4\sqrt{2}}\left(\arctan\left(\frac{x}{\sqrt2}\right)+\frac{x\sqrt{2}}{x^2+2}\right)+C\\ &=\frac{9}{4\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{1}{2(x^2+2)} -\frac{3x}{4(x^2+2)}+C\\ &=\frac{9}{4\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{2+3x}{4(x^2+2)} +C \end{align*}
Exercise22
Hint

Partial fraction decomposition won't simplify this any more. Use a trig substitution.

$\displaystyle\frac{3}{8}\arctan x + \frac{3x^3+5x}{8(1+x^2)^2}+C$

Solution

This is already as simplified as we can make it using partial fraction. Indeed, this is the kind of term that could likely come out of the partial fraction decomposition of a scarier rational function. So, we need to know how to integrate it. Similar to the last piece we integrated in Question 21, we can use the substitution $x=\tan\theta\text{,}$ $\dee{x} = \sec^2\theta \dee{\theta}.$

\begin{align*} \int\frac{1}{(1+x^2)^3} \dee{x}&=\int \frac{\sec^2\theta}{(1+\tan^2\theta)^3} \dee{\theta} =\int \frac{\sec^2\theta}{(\sec^2\theta)^3} \dee{\theta}\\ &=\int \cos^4\theta \dee{\theta} =\int \left[\frac{1+\cos(2\theta)}{2}\right]^2 \dee{\theta} \\ &= \frac{1}{4}\int (1+\cos(2\theta))^2 \dee{\theta}\\ &=\frac{1}{4}\int\left(1+2\cos(2\theta)+\cos^2(2\theta)\right) \dee{\theta}\\ &=\frac{1}{4}\int\left(1+2\cos(2\theta)+\frac{1}{2}(1+\cos(4\theta))\right) \dee{\theta}\\ &=\frac{1}{4}\int\left(\frac{3}{2}+2\cos(2\theta)+\frac{1}{2}\cos(4\theta))\right) \dee{\theta}\\ &=\frac{1}{4}\left(\frac{3}{2}\theta + \sin(2\theta) +\frac{1}{8}\sin(4\theta)\right)+C\\ &=\frac{3}{8}\theta + \frac{1}{4}\sin(2\theta) +\frac{1}{32}\sin(4\theta)+C\\ &=\frac{3}{8}\theta + \frac{1}{2}\sin\theta\cos\theta +\frac{1}{16}\sin(2\theta)\cos(2\theta)+C\\ &=\frac{3}{8}\theta + \frac{1}{2}\sin\theta\cos\theta +\frac{1}{8}\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)+C\\ &=\frac{3}{8}\arctan x + \frac{x}{2(1+x^2)}+\frac{1}{8}\left(\frac{x}{1+x^2}\right)\left(\frac{1-x^2}{1+x^2}\right)+C\\ &=\frac{3}{8}\arctan x + \frac{3x^3+5x}{8(1+x^2)^2}+C \end{align*}

To change our variables from $\theta$ to $x\text{,}$ recall we used the substitution $x=\tan\theta\text{.}$ So, we draw a right triangle with angle $\theta\text{,}$ opposite side length $x\text{,}$ and adjacent side length 1. By the Pythagorean Theorem, the hypotenuse has length $\sqrt{1+x^2}\text{.}$ This allows us to find $\sin\theta$ and $\cos\theta\text{.}$

Exercise23
Hint

To evaluate the antiderivative, break one of the fractions into two fractions.

$\displaystyle\frac{3}{2}x^2+\frac{1}{\sqrt{5}}\arctan \left(\frac{x}{\sqrt5}\right)+ \frac{3}{2}\log|x^2+5|-\frac{3}{2x^2+10}+C$

Solution

Our integrand is already as simplified as the method of partial fractions can make it. The first term is easy to antidifferentiate. The second term would be easier if it were broken into two pieces: one where the numerator is a constant, and one where the numerator is a multiple of $x\text{.}$

\begin{align*} \int \left(3x+\frac{3x+1}{x^2+5}+\frac{3x}{(x^2+5)^2}\right) \dee{x} &= \frac{3}{2}x^2+\int\left(\frac{1}{x^2+5}+\frac{3x}{x^2+5}+\frac{3x}{(x^2+5)^2}\right) \dee{x}\\ &= \frac{3}{2}x^2+\int\frac{1}{x^2+5}\dee{x}+\int\left(\frac{3x}{x^2+5}+\frac{3x}{(x^2+5)^2}\right) \dee{x}\\ \end{align*}

The first integral looks similar to the derivative of arctangent. For the second integral, we use the substitution $u=x^2+5\text{,}$ $\dee{u}=2x \dee{x}\text{.}$

\begin{align*} &= \frac{3}{2}x^2+\frac{1}{5}\int\frac{1}{\left(\frac{x}{\sqrt5}\right)^2+1}\dee{x}+\int\left(\frac{3/2}{u}+\frac{3/2}{u^2}\right) \dee{u}\\ \end{align*}

For the first integral, use the substitution $w=\frac{x}{\sqrt5}\text{,}$ $\dee{w} = \frac{1}{\sqrt5} \dee{x}\text{.}$

\begin{align*} &= \frac{3}{2}x^2+\frac{1}{\sqrt{5}}\int\frac{1}{w^2+1}\dee{w}+ \frac{3}{2}\log|u|-\frac{3}{2u}\\ &= \frac{3}{2}x^2+\frac{1}{\sqrt{5}}\arctan w+ \frac{3}{2}\log|x^2+5|-\frac{3}{2x^2+10}+C\\ &= \frac{3}{2}x^2+\frac{1}{\sqrt{5}}\arctan \left(\frac{x}{\sqrt5}\right)+ \frac{3}{2}\log|x^2+5|-\frac{3}{2x^2+10}+C \end{align*}
Exercise24
Hint

$\cos^2\theta = 1-\sin^2\theta$

$\displaystyle\log\left| \frac{\sin\theta-1}{\sin\theta-2}\right|+C$

Solution

If our denominator were all sines, we could use the substitution $x=\sin\theta\text{.}$ To that end, we apply the identity $\cos^2\theta =1- \sin^2\theta\text{.}$

\begin{align*} \int \frac{\cos\theta}{3\sin\theta+\cos^2\theta-3} \dee{\theta}&= \int \frac{\cos\theta}{3\sin\theta+1-\sin^2\theta-3} \dee{\theta}= \int \frac{\cos\theta}{3\sin\theta-\sin^2\theta-2} \dee{\theta}\\ \end{align*}

We use the substitution $x=\sin \theta\text{,}$ $\dee{x}=\cos\theta \dee{\theta}\text{.}$

\begin{align*} &=\int \frac{1}{3x-x^2-2} \dee{x}=\int \frac{-1}{x^2-3x+2} \dee{x} =\int \frac{-1}{(x-1)(x-2)} \dee{x}\\ \end{align*}

Now we can find a partial fraction decomposition.

\begin{align*} \frac{-1}{(x-1)(x-2)}&=\frac{A}{x-1}+\frac{B}{x-2}\\ -1&=A(x-2)+B(x-1)\\ \end{align*}

Setting $x=1$ and $x=2\text{,}$ we see

Now, we can evaluate our integral.

\begin{align*} \int \frac{\cos\theta}{3\sin\theta+\cos^2\theta-3} \dee{\theta}&=\int \frac{-1}{(x-1)(x-2)} \dee{x} =\int \left(\frac{1}{x-1} - \frac{1}{x-2}\right) \dee{x}\\ &=\log|x-1| - \log|x-2|+C=\log\left| \frac{x-1}{x-2}\right|+C\\ &=\log\left| \frac{\sin\theta-1}{\sin\theta-2}\right|+C \end{align*}
Exercise25
Hint

If you're having a hard time making the substitution, multiply the numerator and the denominator by $e^x\text{.}$

$\displaystyle t - \frac{1}{2}\log|e^{2t}+e^t+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2e^t+1}{\sqrt3}\right)+C$

Solution

This looks a lot like a rational function, but with the function $e^t$ in place of the variable. So, we would like to make the substitution $x=e^t\text{,}$ $\dee{x}=e^t\dee{t}\text{.}$ Then $\dee{t} = \frac{1}{e^t}\,\dee{x} = \frac{1}{x}\,\dee{x}\text{.}$

\begin{align*} \int\frac{1}{e^{2t}+e^t+1} \dee{t}&=\int\frac{1}{x\left(x^2+x+1\right)}\dee{x}\\ \end{align*}

The factor $x^2+x+1$ is an irreducible quadratic, so the denominator is completely factored. Now we can use partial fraction decomposition.

\begin{align*} \frac{1}{x\left(x^2+x+1\right)}&=\frac{A}{x}+\frac{Bx+C}{x^2+x+1}\\ 1&=A(x^2+x+1)+(Bx+C)x\\ 1&=(A+B)x^2+(A+C)x+A\\ \end{align*}

The constant terms tell us $\textcolor{red}{A=1}\text{;}$ then the coefficient of $x$ tells us $\textcolor{red}{C=-A=-1}\text{.}$ Finally, the coefficient of $x^2$ tells us $\textcolor{red}{B=-A=-1}\text{.}$ Now we can evaluate our integral.

\begin{align*} \int\frac{1}{e^{2t}+e^t+1} \dee{t}&=\int\frac{1}{x\left(x^2+x+1\right)}\dee{x}\\ &=\int\left(\frac{1}{x} - \frac{x+1}{x^2+x+1}\right) \dee{x}\\ &=\int\left(\frac{1}{x} - \frac{x+1/2+1/2}{x^2+x+1}\right) \dee{x}\tag{$*$}\\ &=\int\frac{1}{x} \dee{x}-\int \frac{x+1/2}{x^2+x+1} \dee{x}-\int \frac{1/2}{x^2+x+1} \dee{x}\\ &=\log|x| - \frac{1}{2}\log|x^2+x+1|-\int \frac{1/2}{x^2+x+1} \dee{x}\\ \end{align*}

In step ($*$), we set ourselves up so that we could evaluate the second integral with the substitution $u=x^2+x+1\text{.}$ For the remaining integral, we complete the square, so that the integrand looks something like the derivative of arctangent.

\begin{align*} &=\log|x| - \frac{1}{2}\log|x^2+x+1|-\int \frac{1/2}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} \dee{x}\\ &=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{2}{3}\int \frac{1}{\left(\frac{2x+1}{\sqrt3}\right)^2+1} \dee{x}\\ \end{align*}

We use the substitution $u=\frac{2x+1}{\sqrt{3}}\text{,}$ $\dee{u} = \frac{2}{\sqrt3}\text{.}$

\begin{align*} &=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{1}{\sqrt3}\int \frac{1}{u^2+1} \dee{u}\\ &=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{1}{\sqrt3}\arctan u+C\\ &=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2x+1}{\sqrt3}\right)+C\\ &=\log|e^t| - \frac{1}{2}\log|e^{2t}+e^t+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2e^t+1}{\sqrt3}\right)+C\\ &=t - \frac{1}{2}\log|e^{2t}+e^t+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2e^t+1}{\sqrt3}\right)+C \end{align*}
Exercise26
Hint

Try the substitution $u=\sqrt{1+e^x}\text{.}$ You'll need to do long division before you can use partial fraction decomposition.

$\displaystyle2\sqrt{1+e^x}+\log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C$

Solution

• Solution 1: We use the substitution $u=\sqrt{1+e^x}\text{.}$ \\ Then $\dee{u} = \dfrac{e^x}{2\sqrt{1+e^x}}\dee{x}\text{,}$ so $\dee{x}=\dfrac{2u}{u^2-1}\dee{u}\text{.}$ \begin{align*} \int\sqrt{1+e^x} \dee{x}&=\int u\cdot\dfrac{2u}{u^2-1}\dee{u} = \int \dfrac{2u^2}{u^2-1}\dee{u} \\ &= \int \dfrac{2(u^2-1)+2}{u^2-1}\dee{u} = \int\left(2+ \dfrac{2}{u^2-1}\right)\dee{u} \\ \end{align*}

We use a partial fraction decomposition on the fractional part of the integrand.

\begin{align*} &\frac{2}{u^2-1} = \frac{2}{(u-1)(u+1)}=\frac{A}{u-1}+\frac{B}{u+1}=\frac{(A+B)u+(A-B)}{(u-1)(u+1)}\\ & A+B=0,\quad A-B=2\\ &\color{red}{A=1,\quad B=-1}\\ \int\sqrt{1+e^x} \dee{x}&=\int\left(2+ \dfrac{2}{u^2-1}\right)\dee{u} =\int\left( 2 + \frac{1}{u-1} - \frac{1}{u+1}\right) \dee{u}\\ &=2u+\log|u-1|-\log|u+1|+C=2u+\log\left| \frac{u-1}{u+1}\right|+C\\ &=2\sqrt{1+e^x}+\log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C \end{align*}
• Solution 2: It might not occur to us right away to use the fruitful substitution in Solution 1. More realistically, we might start with the “inside function,” $u=1+e^x\text{.}$ Then $\dee{u}=e^x\,\dee{x}\text{,}$ so $\dee{x} = \frac{1}{u-1}\dee{u}\text{.}$ \begin{align*} \int\sqrt{1+e^x} \dee{x}&=\int\frac{ \sqrt{u}}{u-1}\dee{u}\\ \end{align*}

This isn't quite a rational function, because we have a square root on top. If we could turn it into a rational function, we could use partial fraction. To that end, let $w=\sqrt{u}\text{,}$ $\dee{w} = \frac{1}{2\sqrt{u}}\dee{u}\text{,}$ so $\dee{u}=2w\dee{w}\text{.}$

\begin{align*} &=\int \frac{w}{w^2-1}2w\dee{w}=\int \frac{2w^2}{w^2-1}\dee{w}\\ &=\int \frac{2(w^2-1)+2}{w^2-1}\dee{w} = \int 2 + \frac{2}{w^2-1}\dee{w}\\ \end{align*}

Now we can use partial fraction decomposition.

\begin{align*} & \frac{2}{w^2-1} = \frac{2}{(w-1)(w+1)} = \frac{A}{w-1}+\frac{B}{w+1} = \frac{(A+B)w+(A-B)}{(w-1)(w+1)}\\ & A+B=0,\quad A-B=2\\ &\color{red}{ A=1,\quad B=-1}\\ \end{align*}

This allows us to antidifferentiate.

\begin{align*} \int\sqrt{1+e^x} \dee{x}&= \int \left(2 + \frac{2}{w^2-1}\right)\dee{w}=\int \left(2 + \frac{1}{w-1}-\frac{1}{w+1}\right) \dee{w}\\ &= 2w + \log|w-1| - \log|w+1|+C\\ &=2w+\log\left| \frac{w-1}{w+1}\right|+C\\ &=2\sqrt{u}+\log\left| \frac{\sqrt{u}-1}{\sqrt{u}+1}\right|+C\\ &=2\sqrt{1+e^x}+\log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C \end{align*}

Remark: we also evaluated this integral using trigonometric substitution in Section 1.9, Question 1.9.2.26. In that question, we found the antiderivative to be $2\sqrt{1+e^x}+2\log\left| 1-\sqrt{1+e^x} \right|-x+C\text{.}$ These expressions are equivalent:

\begin{align*} \log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|&= \log\left| \sqrt{1+e^x}-1\right|+\log\left| \frac{1}{\sqrt{1+e^x}+1}\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left| \left(\frac{1}{\sqrt{1+e^x}+1}\right)\left(\frac{1-\sqrt{1+e^x}}{1-\sqrt{1+e^x}}\right)\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left|\frac{1-\sqrt{1+e^x}}{1-(1+e^x)}\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left|\frac{1-\sqrt{1+e^x}}{-e^x}\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left|1-\sqrt{1+e^x}\right|-\log|-e^x|\\ &=2\log\left| \sqrt{1+e^x}-1\right|-x \end{align*}
Exercise27(*)
Hint

The mechanically easiest way to answer part (c) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what $R$ looks like. So look at the sketch in part (a) very carefully when identifying the left endpoints of your horizontal strips.

(a) The region $R$ is

(b) $\displaystyle10\pi\log\frac{9}{4}=20\pi\log\frac{3}{2}$ \qquad (c) $20\pi$

Solution

(a) Let's graph $y=\dfrac{10}{\sqrt{25-x^2}}\text{.}$ We start with the endpoints: $(3,\frac{5}{2})$ and $(4,\frac{10}{3})\text{.}$ Then we consider the first derivative:

\begin{equation*} \diff{}{x}\left\{\frac{10}{\sqrt{25-x^2}}\right\} = \frac{10x}{\sqrt{25-x^2}^3} \end{equation*}

Over the interval $[3,4]\text{,}$ this is always positive, so our function is increasing over the entire interval. The second derivative,

\begin{equation*} \ddiff{2}{}{x}\left\{\frac{10}{\sqrt{25-x^2}}\right\}=\diff{}{x}\left\{\frac{10x}{\sqrt{25-x^2}^3}\right\} = \frac{10(2x^2+25)}{\sqrt{25-x^2}^5}, \end{equation*}

is always positive, so our function is concave up over the entire interval. So, the region $R$ is:

(b) Let $\cV_1$ be the solid obtained by revolving $R$ about the $x$-axis. The portion of $\cV_1$ with $x$-coordinate between $x$ and $x+\dee{x}$ is obtained by rotating the red vertical strip in the figure on the left below about the $x$-axis. That portion is a disk of radius $\frac{10}{\sqrt{25-x^2}}$ and thickness $\dee{x}\text{.}$ The volume of this disk is $\pi\big(\frac{10}{\sqrt{25-x^2}}\big)^2\,\dee{x}\text{.}$ So the total volume of $\cV_1$ is

\begin{align*} \int_3^4\pi{\Big(\frac{10}{\sqrt{25-x^2}}\Big)}^2\dee{x} &=100\pi\int_3^4\frac{1}{25-x^2}\dee{x} =100\pi\int_3^4\frac{1}{(5-x)(5+x)}\dee{x} \\ &=10\pi\int_3^4\Big(\frac{1}{5-x}+\frac{1}{5+x}\Big)\dee{x} =10\pi\Big[-\log(5-x)+\log(5+x)\Big]_3^4 \\ &=10\pi\Big[-\log1+\log9+\log2-\log8\Big] =10\pi\log\frac{9}{4}=20\pi\log\frac{3}{2} \end{align*}

(c) We'll use horizontal washers as in Example 1.6.5.

• We cut $\cR$ into thin horizontal strips of width $\dee{y}$ as in the figure on the right above.
• When we rotate $\cR$ about the $y$-axis, each strip sweeps out a thin washer

• whose outer radius is $r_{out}=4\text{,}$ and
• whose inner radius is $r_{in}= \sqrt{25-\frac{100}{y^2}}$ when $y\ge \frac{10}{\sqrt{25-3^2}} = \frac{10}{4} =\frac{5}{2}$ (see the red strip in the figure on the right above), and whose inner radius is $r_{in}= 3$ when $y\le \frac{5}{2}$ (see the blue strip in the figure on the right above) and
• whose thickness is $\dee{y}$ and hence
• whose volume is $\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big(\frac{100}{y^2}-9\big)\dee{y}$ when $y\ge \frac{5}{2}$ and whose volume is $\pi(r_{out}^2 - r_{in}^2)\dee{y} = 7 \pi\,\dee{y}$ when $y\le \frac{5}{2}$ and
• As our bottommost strip is at $y=0$ and our topmost strip is at $y=\frac{10}{3}$ (since at the top $x=4$ and $y= \frac{10}{\sqrt{25-x^2}} =\frac{10}{\sqrt{25-4^2}} =\frac{10}{3}$), the volume is \begin{align*} &\int _{5/2}^{10/3} \pi\Big(\frac{100}{y^2}-9\Big)\dee{y} +\int _ 0^{5/2}7 \pi\,\dee{y} \\ &=\pi{\Big[-\frac{100}{y}-9y\Big]}_{5/2}^{10/3} +\frac{35}{2}\pi\\ &=\pi \Big[-30+40-30+\frac{45}{2}\Big] +\frac{35}{2}\pi\\ &=20\pi \end{align*}
Exercise28
Hint

You'll need to use two regions, because the curves cross.

$\displaystyle2\log\frac53+\frac{4}{\sqrt3}\arctan\frac{1}{4\sqrt3}$

Solution

In order to find the area between the curves, we need to know which one is on top, and which on the bottom. Let's start by finding where they meet.

\begin{align*} \dfrac{4}{3+x^2}&=\frac{2}{x(x+1)}\\ 2x^2+2x&=3+x^2\\ x^2+2x-3&=0\\ (x-1)(x+3)&=0 \end{align*}

In the interval $[\frac14,3]\text{,}$ the curves only meet at $x=1\text{.}$ So, to find which is on top and on bottom in the intervals $[\frac14,1)$ and $(1,3]\text{,}$ it suffices to check some point in each interval.

 $x$ $\frac{4}{3+x^2}$ $\frac{2}{x(x+1)}$ Top: $1/2$ $16/13$ $8/3$ $\frac{2}{x(x+1)}$ $2$ $4/7$ $1/3$ $\frac{4}{3+x^2}$

So, $\frac{2}{x(x+1)}$ is the top function when $\frac{1}{4}\leq x \lt 1\text{,}$ and $\frac{4}{3+x^2}$ is the top function when $1 \lt x \leq 3\text{.}$ Then the area we want to find is:

\begin{equation*} \mbox{Area}=\int_{\frac14}^1 \left( \frac{2}{x(x+1)} - \frac{4}{3+x^2}\right) \dee{x}+ \int_{1}^3 \left( \frac{4}{3+x^2} -\frac{2}{x(x+1)} \right) \dee{x} \end{equation*}

We'll need to antidifferentiate both these functions. We can antidifferentiate $\dfrac{2}{x(x+1)}$ using partial fraction decomposition.

\begin{align*} \frac{2}{x(x+1)}&=\frac{A}{x}+\frac{B}{x+1} = \frac{(A+B)x+A}{x(x+1)}\\ \color{red}{A}&\color{red}{=2,\quad B=-2}\\ \int\frac{2}{x(x+1)} \dee{x}&=\int\left(\frac{2}{x}-\frac{2}{x+1}\right) \dee{x}=2\log|x|-2\log|x+1|+C\\ &=2\log\left| \frac{x}{x+1}\right|+C \end{align*}

We can antidifferentiate $\dfrac{4}{3+x^2}$ using the substitution $u=\frac{x}{\sqrt3}\text{,}$ $\dee{u}=\frac{1}{\sqrt3} \dee{x}\text{.}$

\begin{align*} \int\frac{4}{3+x^2} \dee{x}&= \int\frac{4}{3\left(1+\left(\frac{x}{\sqrt3}\right)^2\right)} \dee{x}= \int\frac{4\sqrt3}{3\left(1+u^2\right)} \dee{u}\\ &=\frac{4}{\sqrt3}\arctan u +C=\frac{4}{\sqrt3}\arctan \left(\frac{x}{\sqrt3}\right) +C \end{align*}

Now, we can find our area.

\begin{align*} \mbox{Area}&=\int_{\frac14}^1 \left( \frac{2}{x(x+1)} - \frac{4}{3+x^2}\right) \dee{x}+ \int_{1}^3 \left( \frac{4}{3+x^2} -\frac{2}{x(x+1)} \right) \dee{x}\\ &=\left[2\log\left|\frac{x}{x+1}\right| - \frac{4}{\sqrt3}\arctan\left(\frac{x}{\sqrt3}\right)\right] _{1/4}^1+ \left[ \frac{4}{\sqrt3}\arctan\left(\frac{x}{\sqrt3}\right)-2\log\left|\frac{x}{x+1}\right|\right]_1^3\\ &=\left(2\log\frac{1}{2}- \frac{4}{\sqrt3}\cdot\frac{\pi}{6} - 2\log\frac{1}{5}+\frac{4}{\sqrt3}\arctan\frac{1}{4\sqrt3}\right)+\\ &\hphantom{=} \left(\frac{4}{\sqrt3}\cdot\frac{\pi}{3} - 2\log\frac34 - \frac{4}{\sqrt3}\cdot \frac{\pi}{6}+2\log\frac{1}{2}\right)\\ &=2\log\frac53+\frac{4}{\sqrt3}\arctan\frac{1}{4\sqrt3} \end{align*}
Exercise29
Hint

For (b), use the Fundamental Theorem of Calculus Part 1.

(a) $\displaystyle\frac{1}{6}\left(\log\left| 2\cdot\frac{x-3}{x+3} \right|\right)$ \qquad (b) $F'(x) = \frac{1}{x^2-9}$

Solution

(a) To antidifferentiate $\dfrac{1}{t^2-9}\text{,}$ we use a partial fraction decomposition.

\begin{align*} \frac{1}{t^2-9}&=\frac{1}{(t-3)(t+3)} = \frac{A}{t-3}+\frac{B}{t+3}=\frac{(A+B)t+3(A-B)}{(t-3)(t+3)}\\ A+B&=0,\quad A-B = \frac{1}{3}\\ \color{red}{A}&\color{red}{=\frac{1}{6},\quad B = -\frac{1}{6}}\\ F(x)&=\int_1^x\frac{1}{t^2-9} \dee{x} = \int_1^x \left(\frac{1/6}{t-3}-\frac{1/6}{t+3}\right) \dee{x}\\ &=\left[\frac{1}{6}\log|t-3| - \frac{1}{6}\log|t+3|\right]_1^x\\ &=\left(\frac{1}{6}\log|x-3| - \frac{1}{6}\log|x+3| -\frac{1}{6}\log2 +\frac{1}{6}\log4\right)\\ &=\frac{1}{6}\left(\log\left| 2\cdot\frac{x-3}{x+3} \right|\right) \end{align*}

(b) Rather than differentiate our answer from (a), we use the Fundamental Theorem of Calculus Part 1 to conclude

\begin{equation*} F'(x) = \diff{}{x}\left\{ \displaystyle\int_1^x \frac{1}{t^2-9} \dee{t}\right\} = \frac{1}{x^2-9} \end{equation*}

Exercises1.11.6Exercises

Exercise1
Hint

The absolute error is the difference of the two values; the relative error is the absolute error divided by the exact value; the percent error is one hundred times the relative error.

Relative error: $\approx 0.08147\text{;}$ absolute error: $0.113\text{;}$ percent error: $\approx 8.147\%\text{.}$

Solution

The absolute error is the difference between the two values:

\begin{equation*} |1.387-1.5| = 0.113 \end{equation*}

The relative error is the absolute error divided by the exact value:

\begin{equation*} \frac{0.113}{1.387}\approx 0.08147 \end{equation*}

The percent error is 100 times the relative error:

\begin{equation*} \approx 8.147\% \end{equation*}
Exercise2
Hint

You should have four rectangles in one drawing, and four trapezoids in another.

Midpoint rule:

Trapezoidal rule:

Solution

Midpoint rule:

Trapezoidal rule:

Exercise3
Hint

$M=6.25\text{,}$ $L=2$

Solution

1. Differentiating, we find $f''(x) = -x^2+7x-6\text{.}$ Since $f''(x)$ is quadratic, we have a pretty good idea of what it looks like.

• It factors as $f(x) = -(x-6)(x-1)\text{,}$ so its two roots are at $x=6$ and $x=1\text{.}$
• The “flat part” of the parabola is at $x=3.5$ (since this is exactly half way between $x=1$ and $x=6\text{;}$ alternately, we can check that $f'''(3.5)=0$).
• Since the coefficient of $x^2$ is negative, $f(x)$ is increasing from $-\infty$ to $3.5\text{,}$ then decreasing from $3.5$ to $\infty\text{.}$

Therefore, over the interval $[1,6]\text{,}$ the largest positive value of $f''(x)$ occurs when $x=3.5\text{,}$ and this is $f''(3.5) = -(3.5-6)(3.5-1)=6.25\text{.}$

So, we take $M=6.25\text{.}$

2. We differentiate further to find $f^{(4)}(x)=-2\text{.}$ This is constant everywhere, so we take $L=|-2|=2\text{.}$
Exercise4
Hint

You don't have to find the actual, exact maximum the second derivative achieves--you only have to give a reasonable “ceiling” that it never breaks through.

One reasonable answer is $M=3\text{.}$

Solution

Let's start by differentiating.

\begin{align*} f(x)&=x\sin x +2\cos x\\ f'(x)&=x\cos x + \sin x - 2\sin x =x\cos x - \sin x\\ f''(x)&=-x\sin x + \cos x -\cos x = -x\sin x \end{align*}

For any value of $x\text{,}$ $|\sin x| \leq 1\text{.}$ When $-3 \leq x \leq 2\text{,}$ then $|x| \leq 3\text{.}$ So, it is true (and not unreasonably sloppy) that

\begin{equation*} f''(x) \leq 3 \end{equation*}

whenever $x$ is in the interval $[-3,2]\text{.}$ So, we can take $M=3\text{.}$

Note that $|f''(x)|$ is actually smaller than 3 whenever $x$ is in the interval $[-3,2]\text{,}$ because when $x=-3\text{,}$ $\sin x \neq 1\text{.}$ In fact, since 3 is pretty close to $\pi\text{,}$ $\sin 3$ is pretty small. (The actual maximum value of $|f''(x)|$ when $-3\leq x\leq 2$ is about 1.8.) However, we find parameters like $M$ for the purpose of computing error bounds. There is often not much to be gained from taking the time to find the actual maximum of a function, so we content ourselves with reasonable upper bounds. Question 31 has a further investigation of “sloppy” bounds like this.

Exercise5
Hint

To compute the upper bound on the error, find an upper bound on the fourth derivative of cosine, then use Theorem 1.11.13 in the text.

To find the actual error, you need to find the actual value of $A\text{.}$

(a) $\dfrac{\pi^5}{180\cdot8}$ \qquad (b) $0$ \qquad (c) $0$

Solution

1. Let $f(x) = \cos x\text{.}$ Then $f^{(4)}(x)=\cos x\text{,}$ so $|f^{(4)}(x)| \leq 1$ when $-\pi \leq x \leq \pi\text{.}$ So, using $L=1\text{,}$ we find the upper bound of the error using Simpson's rule with $n=4$ is:

\begin{equation*} \frac{L(b-a)^5}{180 n^4} = \frac{(2\pi)^5}{180\cdot 4^4} = \frac{\pi^5}{180\cdot8}\approx 0.2 \end{equation*}

The error bound comes from Theorem 1.11.13 in the text. We used a calculator to find the approximate decimal value.

2. We use the general form of Simpson's rule (Equation 1.11.9 in the text) with $\Delta x = \frac{b-a}{n}=\frac{2\pi}{4} = \frac{\pi}{2}\text{.}$ \begin{align*} A& \approx \frac{\Delta x}{3}\left(f(x_0) + 4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)\right)\\ &=\frac{\pi/2}{3}\left(f(-\pi) + 4f(\tfrac{-\pi}{2})+2f(0)+4f(\tfrac{\pi}{2})+f(\pi)\right)\\ &= \frac{\pi}{6}\left(-1 + 4(0)+2(1)+4(0)-1\right)=0 \end{align*}
3. To find the actual error in our approximation, we compare the approximation from (b) to the exact value of $A\text{.}$ In fact, $A=0\text{:}$ this is a fact you've probably seen before by considering the symmetry of cosine, but it's easy enough to calculate: \begin{equation*} A = \int_{-\pi}^{\pi} \cos x \dee{x}= \sin \pi - \sin (-\pi) = 0 \end{equation*} So, our approximation was exactly the same as our exact value. The absolute error is 0.

Remark: the purpose of this question was to remind you that the error bounds we calculate are not (usually) the same as the actual error. Often our approximations are better than we give them credit for. In normal circumstances, we would be approximating an integral precisely to avoid evaluating it exactly, so we wouldn't find our exact error. The bound is a quick way of ensuring that our approximation is not too far off.

Exercise6
Hint

Find a function with $f''(x)=3$ for all $x$ in $[0,1]\text{.}$

Possible answers: $f(x) = \dfrac{3}{2}x^2+Cx+D$ for any constants $C\text{,}$ $D\text{.}$

Solution

Using Theorem 1.11.13 in the text, the error using the trapezoidal rule as described is at most

\begin{equation*} \dfrac{M(b-a)^3}{12\cdot n^2} = \dfrac{M}{48} \leq \frac{3}{48}=\frac{1}{16}. \end{equation*}

So, we're really being asked to find a function with the maximum possible error using the trapezoidal rule, given its second derivative.

With that in mind, our function should have the largest second derivative possible: let's set $f''(x)=3$ for every $x\text{.}$ Then:

\begin{align*} &&f''(x)&=3\\ &&f'(x)&=3x+C\\ &&f(x)&=\frac{3}{2}x^2+Cx+D\\ \end{align*}

for some constants $C$ and $D\text{.}$ Now we can find the exact and approximate values of $\displaystyle\int_0^1 f(x) \dee{x}\text{.}$

\begin{align*} &\mbox{Exact:}&\int_0^1 f(x) \dee{x}&=\int_0^1 \left(\frac{3}{2}x^2+Cx+D\right) \dee{x}\\ &&&=\left[\frac{1}{2}x^3+\frac{C}{2}x^2+Dx\right]_0^1\\ &&&=\textcolor{blue}{\frac{1}{2}+\frac{C}{2}+D}\\ &\mbox{Approximate:}&\int_0^1 f(x) \dee{x}&\approx \Delta x \left[\frac{1}{2}f(0)+f(\tfrac{1}{2})+\frac{1}{2}f(1)\right]\\ &&&=\frac{1}{2}\left[\frac{1}{2}(D)+\left(\frac{3}{8}+\frac{C}{2}+D\right)+\frac{1}{2}\left(\frac{3}{2}+C+D\right)\right]\\ &&&=\frac{1}{2}\left[\frac{9}{8}+C+2D\right]\\ &&&=\textcolor{red}{\frac{9}{16}+\frac{C}{2}+D} \end{align*}

So, the absolute error associated with the trapezoidal approximation is:

\begin{equation*} \left|\left(\textcolor{blue}{\frac{1}{2}+\frac{C}{2}+D}\right) - \left(\textcolor{red}{\frac{9}{16}+\frac{C}{2}+D}\right) \right| = \frac{1}{16} \end{equation*}

So, for any constants $C$ and $D\text{,}$ $f(x) = \frac{3}{2}x^2+Cx+D$ has the desired error.

Remark: contrast this question with Question 5. In this problem, our absolute error was exactly as bad as the bound predicted, but sometimes it is much better. The thing to remember is that, in general, we don't know our absolute error. We only guarantee that it's not any worse than some worst-case-scenario bound.

Exercise7
Hint

You're allowed to use common sense for this one.

my mother

Solution

Under any reasonable assumptions  16 Anyone caught trying to come up with a scenario in which I am older than my mother will be sent to maximum security grad school. , my mother is older than I am.

Exercise8
Hint

For part (b), consider Question 7.

Solution

(a) Since both expressions are positive, and $\frac{1}{24} \leq \frac{1}{12}\text{,}$ the inequality is true.

(b) False. The reasoning is the same as in Question 7. The error bound given by Theorem 1.11.13 is always better for the trapezoid rule, but this doesn't necessarily mean the error is better.

To see how the trapezoid approximation could be better than the corresponding midpoint approximation in some cases, consider the function $f(x)$ sketched below.

The trapezoidal approximation of $\displaystyle\int_a^b f(x) \dee{x}$ with $n=1$ misses the thin spike, and gives a mild underapproximation. By contrast, the midpoint approximation with $n=1$ takes the spike as the height of the entire region, giving a vast overapproximation.

Exercise9(*)
Hint

Draw a sketch.

True. Because $f(x)$ is positive and concave up, the graph of $f(x)$ is always below the top edges of the trapezoids used in the trapezoidal rule.

Solution

True. Because $f(x)$ is positive and concave up, the graph of $f(x)$ is always below the top edges of the trapezoids used in the trapezoidal rule.

Exercise10
Hint

The error bound for the approximation is given in Theorem 1.11.13 in the text. You want this bound to be zero.

Any polynomial of degree at most 3 will do. For example, $f(x)=5x^3-27\text{,}$ or $f(x)=x^2\text{.}$

Solution

According to Theorem 1.11.13 in the text, the error associated with the Simpson's rule approximation is no more than $\dfrac{L}{180}\dfrac{(b-a)^5}{n^4}\text{,}$ where $L$ is a constant such that $|f^{(4)}(x)| \leq L$ for all $x$ in $[a,b]\text{.}$ If $L=0\text{,}$ then the error is no more than 0 regardless of $a\text{,}$ $b\text{,}$ or $n$--that is, the approximation is exact.

Any polynomial $f(x)$ of degree at most 3 has $f^{(4)}(x)=0$ for all $x\text{.}$ So, any polynomial of degree at most 3 is an acceptable answer. For example, $f(x)=5x^3-27\text{,}$ or $f(x)=x^2\text{.}$

Exercise11
Hint

Follow the formulas in Equations 1.11.2, 1.11.6, and 1.11.9 in the text.

Midpoint:

\begin{equation*} \displaystyle \int_0^{30} \frac{1}{x^3+1}\,\dee{x}\approx\left[\tfrac{1}{\left(2.5\right)^3+1} +\tfrac{1}{\left(7.5\right)^3+1} +\tfrac{1}{\left(12.5\right)^3+1} +\tfrac{1}{\left(17.5\right)^3+1} +\tfrac{1}{\left(22.5\right)^3+1} +\tfrac{1}{\left(27.5\right)^3+1} \right]5 \end{equation*}

Trapezoidal:

\begin{equation*} \displaystyle\int_0^{30} \frac{1}{x^3+1}\,\dee{x} \approx\left[ \frac{1/2}{0^3+1}+ \frac{1}{5^3+1}+ \frac{1}{10^3+1}+ \frac{1}{15^3+1}+ \frac{1}{20^3+1}+ \frac{1}{25^3+1}+ \frac{1/2}{30^3+1} \right]5 \end{equation*}

Simpson's:

\begin{equation*} \displaystyle\int_0^{30} \frac{1}{x^3+1}\,\dee{x}\approx \Big[\frac{1}{{0}^3+2}\!+\frac{4}{{5}^3+1}\!+\frac{2}{{10}^3+1}\!+\frac{4}{{15}^3+1}\!+\frac{2}{{20}^3+1}\!+\frac{4}{{25}^3+1}\!+ \frac{1}{{30}^3+1}\Big]\frac{5}{3} \end{equation*}
Solution

• For all three approximations, $\Delta x = \dfrac{b-a}{n}=\dfrac{30-0}{6}=5\text{.}$
• For the trapezoidal rule and Simpson's rule, the $x$-values where we evaluate $\dfrac{1}{x^3+1}$ start at $x=a=0$ and move up by $\Delta x = 5\text{:}$ $x_0=0\text{,}$ $x_1=5\text{,}$ $x_2=10\text{,}$ $x_3=15\text{,}$ $x_4=20\text{,}$ $x_5=25\text{,}$ and $x_6=30\text{.}$

• For the midpoint rule, the $x$-values where we evaluate $\dfrac{1}{x^3+1}$ start at $x=2.5 = \frac{x_0+x_1}{2}$ and move up by $\Delta x = 5\text{:}$ $\bar x_1=2.5\text{,}$ $\bar x_2=7.5\text{,}$ $\bar x_3=12.5\text{,}$ $\bar x_4=17.5\text{,}$ $\bar x_5=22.5\text{,}$ and $\bar x_6=27.5\text{.}$

• Following Equation 1.11.2, the midpoint rule approximation is: \begin{align*} \int_0^{30} \frac{1}{x^3+1}\,\dee{x}&\approx\Big[f(\bar x_1)+f(\bar x_2)+\cdots +f(\bar x_n)\Big]\De x\\ &=\left[\tfrac{1}{\left(2.5\right)^3+1} +\tfrac{1}{\left(7.5\right)^3+1} +\tfrac{1}{\left(12.5\right)^3+1} +\tfrac{1}{\left(17.5\right)^3+1} +\tfrac{1}{\left(22.5\right)^3+1} +\tfrac{1}{\left(27.5\right)^3+1} \right]5 \end{align*}
• Following Equation 1.11.6, the trapezoidal rule approximation is: \begin{align*} \int_0^{30} \frac{1}{x^3+1}\,\dee{x} &\approx\Big[\half f(x_0)+f(x_1)+f(x_2)+\cdots+ f(x_{n-1})+\half f(x_n)\Big]\De x\\ &=\left[ \frac{1/2}{0^3+1}+ \frac{1}{5^3+1}+ \frac{1}{10^3+1}+ \frac{1}{15^3+1}+ \frac{1}{20^3+1}+ \frac{1}{25^3+1}+ \frac{1/2}{30^3+1} \right]5 \end{align*}
• Following Equation 1.11.9, the Simpson's rule approximation is: \begin{align*} \int_0^{30} \frac{1}{x^3+1}\,\dee{x} &\approx\Big[f(x_0)\!+4f(x_1)\!+2f(x_2)\!+4f(x_3)\!+2f(x_4)\!+4f(x_{5})\!+ f(x_6)\Big]\tfrac{\De x}{3}\\ &=\Big[\frac{1}{{0}^3+2}\!+\frac{4}{{5}^3+1}\!+\frac{2}{{10}^3+1}\!+\frac{4}{{15}^3+1}\!+\frac{2}{{20}^3+1}\!+\frac{4}{{25}^3+1}\!+ \frac{1}{{30}^3+1}\Big]\frac{5}{3} \end{align*}
Exercise12(*)
Hint

See Section 1.11.1. You should be able to simplify your answer to an exact value (in terms of $\pi$).

$\dfrac{2\pi}{3}$

Solution

By Equation 1.11.2, the midpoint rule approximation to $\int_a^b f(x)\dee{x}$ with $n=3$ is

\begin{gather*} \int_a^b f(x)\,\dee{x}\approx\big[f(\bar x_1)+f(\bar x_2)+f(\bar x_3)\big]\De x \end{gather*}

where $\De x = \tfrac{b-a}{3}$ and

For this problem, $a=0\text{,}$ $b=\pi$ and $f(x) = \sin x\text{,}$ so that $\De x = \tfrac{\pi}{3}$ and

Therefore,

\begin{align*} \int_0^\pi \sin x\dee{x} &\approx\left[\sin\frac{\pi}{6} +\sin \frac{\pi}{2} +\sin \frac{5\pi}{6}\right]\frac{\pi}{3} =\left[\frac{1}{2} +1 +\frac{1}{2}\right]\frac{\pi}{3} =\frac{2\pi}{3} \end{align*}
Exercise13(*)
Hint

See Section 1.11.2. To set up the volume integral, see Example 1.6.6. Note the dimensions given for the cross sections are diameters, not radii.

$1720\pi\approx 5403.5\ {\rm cm}^3$

Solution

Let $f(x)$ denote the diameter at height $x\text{.}$ As in Example 1.6.6, we slice $V$ into thin horizontal “pancakes”, which in this case are circular.

• We are told that the pancake at height $x$ is a circular disk of diameter $f(x)$ and so
• has cross-sectional area $\pi\big(\frac{f(x)}{2}\big)^2$ and thickness $\dee{x}$ and hence
• has volume $\pi\big(\frac{f(x)}{2}\big)^2\dee{x}\text{.}$

Hence the volume of $V$ is

\begin{align*} \int_0^{40}\pi\Big{[\frac{f(x)}{2}\Big]}^2\,\dee{x} &\approx \frac{\pi}{4}10\Big[\half f(0)^2+f(10)^2+f(20)^2+f(30)^2+\half f(40)^2\Big]\\ &=\frac{\pi}{4}10\Big[\half 24^2+16^2+10^2+6^2+\half 4^2\Big]\\ &=688\times 2.5\pi =1720\pi\approx 5403.5 \end{align*}

where we have approximated the integral using the trapezoidal rule with $\De x=10\text{,}$ and used a calculator to get a decimal approximation.

Exercise14(*)
Hint

See Section 1.11.3 and compare to Question 1.11.6.13. Note the table gives diameters, not radii.

$\displaystyle\frac{\pi}{12}(16.72)\approx4.377\ {\rm m}^3$

Solution

Let $f(x)$ be the diameter a distance $x$ from the left end of the log. If we slice our log into thin disks, the disks $x$ metres from the left end of the log has

• radius $\frac{f(x)}{2}\text{,}$
• width $\dee{x}\text{,}$ and so
• volume $\pi\left(\frac{f(x)}{2}\right)^2 \dee{x}=\frac{\pi}{4}f(x)^2 \dee{x}\text{.}$

Using Simpson's Rule with $\De x=1\text{,}$ the volume of the log is:

\begin{align*} V=\int_0^6 \frac{\pi}{4}f(x)^2\,\dee{x} &\approx\frac{\pi}{4}\frac{1}{3} \Big[f(0)^2+4f(1)^2+2f(2)^2+4f(3)^2+2f(4)^2+4f(5)^2+f(6)^2\Big]\\ &=\frac{\pi}{12} \Big[1.2^2+4(1)^2+2(0.8)^2+4(0.8)^2+2(1)^2+4(1)^2+1.2^2\Big]\\ &=\frac{\pi}{12}(16.72)\\ &\approx4.377\ {\rm m}^3 \end{align*}

where we used a calculator to approximate the decimal value.

Exercise15(*)
Hint

See §1.11.3. To set up the volume integral, see Example 1.6.6, or Question 14.

Note that the table gives the circumference, not radius, of the tree at a given height.

$\dfrac{12.94}{6\pi} \approx0.6865\ {\rm m}^3$

Solution

At height $x$ metres, let the circumference of the tree be $c(x)\text{.}$ The corresponding radius is $\dfrac{c(x)}{2\pi}\text{,}$ so the corresponding cross--sectional area is $\pi\left(\dfrac{c(x)}{2\pi}\right)^2=\dfrac{c(x)^2}{4\pi}\text{.}$

The height of a very thin cross--sectional disk is $\dee{x}\text{,}$ so the volume of a cross-sectional disk is $\dfrac{c(x)^2}{4\pi} \dee{x}\text{.}$ Therefore, total volume of the tree is:

\begin{align*} \int_0^8 \frac{c(x)^2}{4\pi}\,\dee{x} &\approx \frac{1}{4\pi}\frac{2}{3}\Big[c(0)^2+4c(2)^2+2c(4)^2+4c(6)^2+c(8)^2\Big]\cr &=\frac{1}{6\pi}\Big[1.2^2+4(1.1)^2+2(1.3)^2+4(0.9)^2+0.2^2\Big]\\ &=\frac{12.94}{6\pi} \approx0.6865 \end{align*}

where we used Simpson's rule with $\De x = 2$ and $n=4$ to approximate the value of the integral based on the values of $c(x)$ given in the table.

Exercise16(*)

Solution

For both approximations, $\De x = 10$ and $n=6\text{.}$

(a) The Trapezoidal Rule gives

\begin{align*} V&=\int_0^{60} A(h)\,\dee{h} \approx 10\Big[\half A(0)+A(10)+A(20)+A(30)+A(40)+A(50)+\half A(60)\Big] \\ &=\text{363,500} \end{align*}

(b) Simpson's Rule gives

\begin{align*} V&=\int_0^{60} A(h)\,\dee{h} \approx \frac{10}{3}\Big[A(0)+4A(10)+2A(20)+4A(30)+2A(40)+4A(50)+A(60)\Big]\\ &=\text{367,000} \end{align*}
Exercise17(*)

(a) $\dfrac{49}{2}$ \qquad (b) $\dfrac{77}{3}$

Solution

Call the curve in the graph $y=f(x)\text{.}$ It looks like

We're estimating $\int_2^6 f(x) \dee{x}$ with $n=4\text{,}$ so $\De x = \frac{6-2}{4}=1\text{.}$

(a) The trapezoidal rule gives

\begin{gather*} T_4=\left[\frac{3}{2}+ 8+7+ 6+\frac{4}{2}\right]\times 1=\frac{49}{2} \end{gather*}

(b) Simpson's rule gives

\begin{gather*} S_4=\frac{1}{3}\big[3+4\times 8+2\times 7+4\times 6+4\big]\times 1 =\frac{77}{3} \end{gather*}
Exercise18(*)
Hint

The main step is to find an appropriate value of $M\text{.}$ It is not necessary to find the smallest possible $M\text{.}$

Let $f(x) = \sin(x^2)\text{.}$ Then $f'(x) = 2x \cos(x^2)$ and

\begin{equation*} f''(x) = 2\cos(x^2) - 4x^2\sin(x^2). \end{equation*}

Since $|x^2|\le1$ when $|x|\leq 1\text{,}$ and $\left|\sin\theta\right|\le1$ and $\left|\cos\theta\right|\leq 1$ for all $\theta\text{,}$ we have

\begin{gather*} \left|2\cos(x^2) - 4x^2\sin(x^2)\right| \le 2|\cos(x^2)| + 4x^2|\sin(x^2)| \le 2\times 1 +4\times 1\times 1 = 2+4 = 6 \end{gather*}

We can therefore choose $M=6\text{,}$ and it follows that the error is at most

\begin{gather*} \frac{M[b-a]^3}{24n^2} \le \frac{6\cdot [1-(-1)]^3}{24 \cdot 1000^2} = \frac{2}{10^6} = 2\cdot 10^{-6} \end{gather*}
Solution

Let $f(x) = \sin(x^2)\text{.}$ Then $f'(x) = 2x \cos(x^2)$ and

\begin{equation*} f''(x) = 2\cos(x^2) - 4x^2\sin(x^2). \end{equation*}

Since $|x^2|\le1$ when $|x|\leq 1\text{,}$ and $\left|\sin\theta\right|\le1$ and $\left|\cos\theta\right|\leq 1$ for all $\theta\text{,}$ we have

\begin{gather*} \left|2\cos(x^2) - 4x^2\sin(x^2)\right| \le 2|\cos(x^2)| + 4x^2|\sin(x^2)| \le 2\times 1 +4\times 1\times 1 = 2+4 = 6 \end{gather*}

We can therefore choose $M=6\text{,}$ and it follows that the error is at most

\begin{gather*} \frac{M[b-a]^3}{24n^2} \le \frac{6\cdot [1-(-1)]^3}{24 \cdot 1000^2} = \frac{2}{10^6} = 2\cdot 10^{-6} \end{gather*}
Exercise19(*)
Hint

The main step is to find $M\text{.}$ This question is unusual in that its wording requires you to find the smallest possible allowed $M\text{.}$

$\dfrac{3}{100}$

Solution

Setting $f(x) = 2 x^4$ and $b-a = 1-(-2)=3\text{,}$ we compute $f''(x) = 24x^2\text{.}$ The largest value of $24x^2$ on the interval $[-2,1]$ occurs at $x=-2\text{,}$ so we can take $M = 24\cdot(-2)^2=96\text{.}$ Thus the total error for the midpoint rule with $n=60$ points is bounded by

\begin{gather*} \frac{M (b-a)^3}{24n^2} = \frac{96 \times 3^3}{24 \times 60 \times 60} = \frac{3}{100} \end{gather*}

That is: we are guaranteed our absolute error is certainly no more  17 This is what the error bound always tells us. than $\frac{3}{100}\text{,}$ and using the bound stated in the problem we cannot give a better guarantee. (The second part of the previous sentence comes from the fact that we used the smallest possible $M\text{:}$ if we had used a larger value of $M\text{,}$ we would still have some true statement about the error, for example “the error is no more than $\frac{5}{100}\text{,}$” but it would not be the best true statement we could make.)

Exercise20(*)
Hint

The main steps in part (b) are to find the smallest possible values of $M$ and $L\text{.}$

(a) $\dfrac{1/3}3 \Big( (-3)^5 + 4\Big( \frac13-3 \Big)^5 + 2\Big( \frac23-3 \Big)^5 + 4(-2)^5 + 2\Big( \frac43-3 \Big)^5 + 4\Big( \frac53-3 \Big)^5 + (-1)^5 \Big)$

(b) Simpson's Rule results in a smaller error bound.

Solution

(a) Since $a=0\text{,}$ $b=2$ and $n=6\text{,}$ we have $\Delta x=\frac{b-a}{n}=\frac{2-0}6 = \frac{1}{3}\text{,}$ and so $x_0=0\text{,}$ $x_1=\frac{1}{3}\text{,}$ $x_2=\frac{2}{3}\text{,}$ $x_3=1\text{,}$ $x_4=\frac{4}{3}\text{,}$ $x_5=\frac{5}{3}\text{,}$ and $x_6=2\text{.}$ Since Simpson's Rule with $n=6$ in general is

\begin{gather*} \frac{\Delta x}3 \big[ f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + 4f(x_5) + f(x_6) \big], \end{gather*}

the desired approximation is

\begin{gather*} \frac{1/3}3 \bigg( (-3)^5 + 4\Big( \frac13-3 \Big)^5 + 2\Big( \frac23-3 \Big)^5 + 4(-2)^5 + 2\Big( \frac43-3 \Big)^5 + 4\Big( \frac53-3 \Big)^5 + (-1)^5 \bigg) \end{gather*}

(b) Here $f(x) = (x-3)^5\text{,}$ which has derivatives

\begin{align*} f'(x) &= 5(x-3)^4 & f''(x) &= 20(x-3)^3 \\ f^{(3)}(x) &= 60(x-3)^2 & f^{(4)}(x) &= 120(x-3). \end{align*}

For $0\le x\le 2\text{,}$ $(x-3)$ runs from $-3$ to $-1\text{,}$ so the maximum absolute values are found at $x=0\text{,}$ giving $M= 20\cdot|0-3|^3=540$ and $L=120\cdot|0-3|=360\text{.}$ Consequently, for the Midpoint Rule with $n=100\text{,}$

\begin{gather*} |E_M| \le \frac{M(b-a)^3}{24n^2} = \frac{540 \times 2^3}{24 \times 10^4} = \frac{180}{10^4}; \end{gather*}

whereas for Simpson's Rule with $n=10\text{,}$

\begin{gather*} |E_S| \le \frac{360 \times 2^5 }{180 \times 10^4} = \frac{64}{10^4}. \end{gather*}

Since $64 \lt 180\text{,}$ Simpson's Rule results in a smaller error bound.

Exercise21(*)
Hint

As usual, the biggest part of this problem is finding $L\text{.}$ Don't be thrown off by the error bound being given slightly differently from Theorem 1.11.13 in the text: these expressions are equivalent, since $\De x = \frac{b-a}{n}\text{.}$

$\dfrac{8}{15}$

Solution

In general the error in approximating $\int_a^b f(x)\dee{x}$ using Simpson's rule with $n$ steps is bounded by $\frac{L(b-a)}{180}(\De x)^4$ where $\De x=\frac{b-a}{n}$ and $L\ge |f^{(4)}(x)|$ for all $a\le x\le b\text{.}$ In this case, $a=1\text{,}$ $b=5\text{,}$ $n=4$ and $f(x)=\frac{1}{x}\text{.}$ We need to find $L\text{,}$ so we differentiate.

and

\begin{gather*} \big|f^{(4)}(x)\big|\le 24\text{ for all }x\ge 1 \end{gather*}

So we may take $L=24$ and $\De x=\frac{5-1}{4}=1\text{,}$ which leads to

\begin{gather*} |\text{Error }|\le \frac{24(5-1)}{180}(1)^4=\frac{24}{45}=\frac{8}{15} \end{gather*}
Exercise22(*)
Hint

The function $e^{-2x} = \dfrac{1}{e^{2x}}$ is positive and decreasing, so its maximum occurs when $x$ is as small as possible.

$\displaystyle\frac{1}{180\times 3^4} =\frac{1}{14580}$

Solution

In general, the error in approximating $\int_a^b f(x)\dee{x}$ using Simpson's rule with $n$ steps is bounded by $\displaystyle\frac{L(b-a)}{180}(\De x)^4$ where $\De x=\dfrac{b-a}{n}$ and $L\ge |f^{(4)}(x)|$ for all $a\le x\le b\text{.}$ In this case, $a=0\text{,}$ $b=1\text{,}$ $n=6$ and $f(x)=e^{-2x}+3x^3\text{.}$ We need to find $L\text{,}$ so we differentiate.

Since $e^{-2x} = \dfrac{1}{e^{2x}}\text{,}$ we see $f^{(4)}(x)$ is a positive, decreasing function. So, its maximum occurs when $x$ is as small as possible. In the interval $[0,1]\text{,}$ that means $x=0\text{.}$

\begin{gather*} \big|f^{(4)}(x)\big|\le f(0)=16\text{ for all }x\ge 0 \end{gather*}

So, we take $L=16$ and $\De x=\frac{1-0}{6}=\frac{1}{6}\text{.}$

\begin{gather*} |\text{Error }|\le \frac{L(b-a)}{180}(\De x)^4=\frac{16(1-0)}{180}(1/6)^4 =\frac{16}{180\times 6^4} =\frac{1}{180\times 3^4} =\frac{1}{14580} \end{gather*}
Exercise23(*)
Hint

Since $\dfrac{1}{x^5}$ is a decreasing function when $x \gt 0\text{,}$ look for its maximum value when $x$ is as small as possible.

(a) $\displaystyle T_4 =\frac{1}{4}\left[\left(\frac{1}{2}\times 1\right)+\frac{4}{5}+\frac{2}{3}+ \frac{4}{7}+\left(\frac{1}{2}\times\frac{1}{2}\right)\right]\text{,}$

(b) $\displaystyle S_4 =\frac{1}{12}\left[1+\left(4\times\frac{4}{5}\right)+\left(2\times \frac{2}{3}\right)+\left(4\times \frac{4}{7}\right)+\frac{1}{2}\right]$

(c) $\displaystyle\Big|I -S_4\Big| \le \frac{24}{180\times 4^4}=\frac{3}{5760}$

Solution

For both approximations, $a=1\text{,}$ $b=2\text{,}$ $n=4\text{,}$ $f(x)=\frac{1}{x}$ and $\De x=\frac{b-a}{n}=\frac{1}{4}\text{.}$

Then $x_0 = 1\text{,}$ $x_1=\frac{5}{4}\text{,}$ $x_2 = \frac{3}{2}\text{,}$ $x_3=\frac{7}{4}\text{,}$ and $x_4=2\text{.}$

(a)

\begin{align*} T_4&=&\De x&\left[\frac{1}{2}f(x_0)+f(x_1)+f(x_2)+f(x_3)+\frac{1}{2}f(x_4)\right]\\ &=&\De x&\left[\frac{1}{2}f(1)+f(5/4)+f(3/2)+f(7/4)+\frac{1}{2}f(2)\right]\\ &=&\frac{1}{4}&\left[\left(\frac{1}{2}\times 1\right)+\frac{4}{5}+\frac{2}{3}+ \frac{4}{7}+\left(\frac{1}{2}\times\frac{1}{2}\right)\right] \end{align*}

(b)

\begin{align*} S_4&=\frac{\De x}{3}\big[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+f(x_4)\big]\\ &=\frac{\De x}{3}\big[f(1)+4f(5/4)+2f(3/2)+4f(7/4)+f(2)\big]\\ &=\frac{1}{12}\left[1+\left(4\times\frac{4}{5}\right)+\left(2\times \frac{2}{3}\right)+\left(4\times \frac{4}{7}\right)+\frac{1}{2}\right] \end{align*}

(c) In this case, $a=1\text{,}$ $b=2\text{,}$ $n=4$ and $f(x)=\frac{1}{x}\text{.}$ We need to find $L\text{,}$ so we differentiate.

So,

\begin{gather*} \big|f^{(4)}(x)\big|\le 24\text{ for all }x\text{ in the interval } [1,2] \end{gather*}

We take $L=24\text{.}$

\begin{gather*} |\text{Error }|\le \frac{L(b-a)^5}{180\times n^4} \le \frac{24(2-1)^5}{180\times 4^4}=\frac{24}{180\times 4^4}=\frac{3}{5760} \end{gather*}
Exercise24(*)
Hint

The “best ... approximations that you can” means using the maximum number of intervals, given the information available.

The final sentence in part (b) is just a re-statement of the error bounds we're familiar with from Theorem 1.11.13 in the text. The information $\big|s^{(k)}(x)\big|\le \dfrac{k}{1000}$ gives you values of $M$ and $L$ when you set $k=2$ and $k=4\text{,}$ respectively.

(a) $T_4=8.03515\text{,}$ $S_4\approx 8.03509$

(b) $\displaystyle\Big|\int_a^b f(x)\ dx -T_n\Big| \le \frac{2}{1000}\frac{8^3}{12(4)^2} \le0.00533\text{,}$\quad $\displaystyle\Big|\int_a^b f(x)\ dx -S_n\Big| \le \frac{4}{1000}\frac{8^5}{180(4)^4}\le0.00284$

Solution

Set $a=0$ and $b=8\text{.}$ Since we have information about $s(x)$ when $x$ is 0, 2, 4, 6, and 8, we set $\De x=\frac{b-a}{n}=2\text{,}$ so $n=4\text{.}$ (Recall with the trapezoid rule and Simpson's rule, $n=4$ intervals actually uses the value of the function at 5 points.)

We could perform the trapezoidal approximations with fewer intervals, for example $n=2\text{,}$ but this would involve ignoring some of the points we're given. Since the question asks for the best estimation we can give, we use $n=4$ intervals and no fewer.

1. \begin{align*} T_4&={\De x}\left[\frac{1}{2}s(0)+s(2)+s(4)+s(6)+\frac{1}{2}s(8)\right]\\ &=2\left[\frac{1.00664}{2}+1.00543+ 1.00435+1.00331+ \frac{1.00233}{2}\right]\\ &=8.03515\\ S_4&=\frac{\De x}{3}\big[s(0)+4s(2)+2s(4)+4s(6)+s(8)\big]\\ &=\frac{2}{3}\big[1.00664+4\times 1.00543+2\times 1.00435+4\times1.00331+ 1.00233\big]\\ &\approx 8.03509 \end{align*}
2. The information $\big|s^{(k)}(x)\big|\le \dfrac{k}{1000}\text{,}$ with $k=2\text{,}$ tells us $|s''(x)|\leq \frac{2}{1000}$ for all $x$ in the interval $[0,8]\text{.}$ So, we take $K_2$ (also called $M$ in your text) to be $\frac{2}{1000}\text{.}$

Then the absolute error associated with our trapezoid rule approximation is at most

\begin{align*} \bigg|\int_a^b f(x)\ dx -T_n\bigg|&\le \frac{K_2(b-a)^3}{12n^2} &&\le \frac{2}{1000}\cdot\frac{8^3}{12(4)^2} &\le 0.00533 \end{align*}

For $k=4\text{,}$ we see $|s^{(4)}(x)|\leq \frac{4}{1000}$ for all $x$ in the interval $[0,8]\text{.}$ So, we take $K_4$ (also called $L$ in your text) to be $\frac{4}{1000}\text{.}$

Then the absolute error associated with our Simpson's rule approximation is at most

\begin{align*} \bigg|\int_a^b f(x)\ dx -S_n\bigg|&\le \frac{K_4(b-a)^5}{180n^4} &&\le \frac{4}{1000}\cdot\frac{8^5}{180(4)^4} &\le0.00284 \end{align*}
Exercise25(*)
Hint

Set the error bound to be less than $0.001\text{,}$ then solve for $n\text{.}$

Any $n\ge 68$ works.

Solution

In this case, $a=1\text{,}$ $b=4\text{.}$ Since $-2 \leq f''(x) \leq 0$ over the relevant interval, we take $M=2\text{.}$ (Remember $M$ is an upper bound on $|f''(x)|\text{,}$ not $f''(x)\text{.}$) So we need $n$ to obey

\begin{gather*} \frac{ 2(4 - 1)^3}{12n^2} \le 0.001 \iff n^2\ge \frac{2(3)^3}{12} 1000 =\frac{27000}{6}=\frac{9000}{2}=4500 \end{gather*}

One obvious allowed $n$ is $100\text{.}$ Since $\sqrt{4500} \approx 67.01\text{,}$ and $n$ has to be a whole number, any $n\ge 68$ works.

Exercise26(*)
Hint

See Section 1.11.3. To set up the volume integral, see Example 1.6.2.

Since the cross-sections of the pool are semi-circular disks, a section that is $d$ metres across will have area $\frac{1}{2}\pi\left(\frac{d}{2}\right)^2$ square feet. Based on the drawing, you may assume the very ends of the pool have distance 0 feet across.

$\dfrac{472}{3}\approx 494 \ {\rm ft}^3$

Solution

Denote by $f(x)$ the width of the pool $x$ feet from the left-hand end. From the sketch, $f(0)=0\text{,}$ $f(2)=10\text{,}$ $f(4)=12\text{,}$ $f(6)=10\text{,}$ $f(8)=8\text{,}$ $f(10)=6\text{,}$ $f(12)=8\text{,}$ $f(14)=10$ and $f(16)=0\text{.}$

A cross-section of the pool $x$ feet from the left end is half of a circular disk with diameter $f(x)$ (so, radius $\frac{f(x)}{2}$) and thickness $\dee{x}\text{.}$ So, the volume of the part of the pool with $x$--coordinate running from $x$ to ($x+\dee{x}$) is $\half\pi\big(\frac{f(x)}{2}\big)^2\,\dee{x} = \frac{\pi}{8}[f(x)]^2\,\dee{x}\text{.}$

The total volume is given by the following integral.

\begin{align*} V&=\frac{\pi}{8}\int_0^{16}f(x)^2\,\dee{x}\cr &\approx\frac{\pi}{8}\cdot\frac{\De x}{3} \Big[f(0)^2+4f(2)^2+2f(4)^2+4f(6)^2+2f(8)^2 +4f(10)^2+2f(12)^2+4f(14)^2\!+f(16)^2\Big]\cr &=\frac{\pi}{8}\cdot\frac{2}{3} \Big[0+4(10)^2+2(12)^2+4(10)^2+2(8)^2 +4(6)^2+2(8)^2+4(10)^2+0\Big]\cr &=\frac{472}{3}\pi\approx 494 \ {\rm ft}^3 \end{align*}
Exercise27(*)
Hint

See Example 1.11.15.

Don't get caught up in the interpretation of the integral. It's nice to see how integrals can be used, but for this problem, you're still just approximating the integral given, and bounding the error.

When you find the second derivative to bound your error, pay attention to the difference between the integrand and $g(r)\text{.}$

(a) $0.025635$ \qquad (b) $1.8\times 10^{-5}$

Solution

(a) The Trapezoidal Rule with $n=4\text{,}$ $a=0\text{,}$ $b=1\text{,}$ and $\De x = \frac{1}{4}$ gives:

\begin{align*} W=2\pi 10^{-6}\int_0^1 rg(r)\,\dee{r} &\approx 2\pi 10^{-6}\De x\left[\frac{1}{2}x_0g(x_0)+x_1g(x_1)+x_2g(x_2)+x_3g(x_3)+ \frac{1}{2}x_4g(x_4)\right]\\ &=2\pi 10^{-6}\ \frac{1}{4} \Big[\frac{1}{2} 0g(0)+\frac{1}{4}g\Big(\frac{1}{4}\Big) +\frac{1}{2}g\Big(\frac{1}{2}\Big)+\frac{3}{4}g\Big(\frac{3}{4}\Big) +\frac{1}{2} g(1)\Big] \\ &=\pi 10^{-6}\ \frac{1}{2} \Big[\frac{8100}{4} +\frac{8144}{2}+\frac{3\cdot 8170}{4} +\frac{8190}{2} \Big] \\ & = \frac{32639\pi}{4\cdot 10^6}\approx 0.025635 \end{align*}

(b) Using the product rule, the integrand $f(r)=2\pi 10^{-6} rg(r)$ obeys

\begin{gather*} f''(r)=2\pi 10^{-6} \diff{}{r}\big[g(r)+ rg'(r)\big] =2\pi 10^{-6} \big[2g'(r)+ rg''(r)\big] \end{gather*}

and hence, for $0\le r\le 1\text{,}$

\begin{gather*} \big|f''(r)\big|\le 2\pi 10^{-6} \big[2\times200 + 1\times 150\big] =1.1\pi 10^{-3} \end{gather*}

So,

\begin{gather*} |\text{Error}|\le \frac{1.1\pi 10^{-3}(1-0)^3}{12(4)^2} \le1.8\times 10^{-5} \end{gather*}
Exercise28(*)
Hint

See Example 1.11.16. You'll want to use a calculator for the approximation in (a), and for finding the appropriate number of intervals in (b). Remember that Simpson's rule requires an even number of intervals.

(a) $\approx 0.6931698$ \qquad (b) $n\ge 12$ with $n$ even

Solution

(a) Let $f(x)=\frac{1}{x}\text{,}$ $a=1\text{,}$ $b=2$ and $\De x=\frac{b-a}{6}=\frac{1}{6}\text{.}$ Using Simpson's rule:

\begin{align*} \int_1^2\frac{1}{x}\dee{x} &\approx \frac{\De x}{3}\Big[ f(1)+4f\Big(\frac{7}{6}\Big)+2f\Big(\frac{8}{6}\Big) +4f\Big(\frac{9}{6}\Big)+2f\Big(\frac{10}{6}\Big) +4f\Big(\frac{11}{6}\Big)+f(2)\Big]\\ &= \frac{1}{18}\Big[ 1+\frac{24}{7}+\frac{12}{8}+\frac{24}{9}+\frac{12}{10}+\frac{24}{11} +\frac{1}{2}\Big]\approx 0.6931698 \end{align*}

(b) The integrand is $f(x)=\frac{1}{x}\text{.}$ The first four derivatives of $f(x)$ are:

On the interval $1\le x\le 2\text{,}$ the fourth derivative is never bigger in magnitude than $L=24\text{.}$

\begin{align*} |E_n|&\le\frac{L(b-a)^5}{180n^4} =\frac{24(2-1)^5}{180n^4} =\frac{4}{30n^4}\\ \end{align*}

So, we want an even number $n$ such that

\begin{align*} \frac{4}{30n^4} &\leq 0.00001 = \frac{1}{10^5}\\ n^4 & \geq \frac{40000}{3}\\ n&\geq \sqrt[4]{\frac{40000}{3}}\approx 10.7 \end{align*}

So, any even number greater than or equal to 12 will do.

Exercise29(*)
Hint

See Example 1.11.16.

Rather than calculating the fourth derivative of the integrand, use the graph to find the largest absolute value it attains over our interval.

(a) $0.01345$ \qquad (b) $n\ge 28$ with $n$ even

Solution

(a) From the figure, we see that the magnitude of $|f''''(x)|$ never exceeds 310 for $0\le x\le 2\text{.}$ So, the absolute error is bounded by

\begin{gather*} \frac{310(2-0)^5}{180\times 8^4}\le 0.01345 \end{gather*}

(b) We want to choose $n$ such that:

\begin{align*} \frac{310(2-0)^5}{180\times n^4}&\le 10^{-4}\\ n^4&\ge \frac{310\times 2^5}{180}10^4\\ n&\ge 10\root 4 \of {\frac{310\times 32}{180}}\approx27.2 \end{align*}

For Simpson's rule, $n$ must be even, so any even integer obeying $n\ge28$ will guarantee us the requisite accuracy.

Exercise30(*)
Hint

See Example 1.11.15.

You'll have to differentiate $f(x)\text{.}$ To that end, you may also want to review the fundamental theorem of calculus and, in particular, Example 1.3.5.

You don't have to find the best possible value for $M\text{.}$ A reasonable upper bound on $|f''(x)|$ will do.

To have five decimal places of accuracy, your error must be less than 0.000005. This ensures that, if you round your approximation to five decimal places, they will all be correct.

$n\ge 259$

Solution

Let $g(x)=\displaystyle\int_0^x\sin(\sqrt{t})\,\dee{t}\text{.}$ By the Fundamental Theorem of Calculus Part 1, $g'(x)= \sin(\sqrt{x})\text{.}$ By its definition, $f(x)=g(x^2)\text{,}$ so we use the chain rule to differentiate $f(x)\text{.}$

Since $|\sin x|,|\cos x|\le 1\text{,}$ we have $|f''(x)|\le 2+2|x|$ and, for $0\le t\le 1\text{,}$ $|f''(t)|\le 4\text{.}$ When the trapezoidal rule with $n$ subintervals is applied, the resulting error $E_n$ obeys

\begin{align*} E_n&\le\frac{4(1-0)^3}{12n^2}=\frac{1}{3n^2}\\ \end{align*}

We want an integer $n$ such that

\begin{align*} \frac{1}{3n^2}&\le 0.000005\\ n^2&\ge \frac{4}{12\times 0.000005}\\ n&\ge\sqrt{ \frac{1}{3\times 0.000005}} \approx258.2 \end{align*}

Any integer $n \geq 259$ will do.

Exercise31
Hint

To find the maximum value of $|f''(x)|\text{,}$ check its critical points and endpoints.

(a) When $0 \leq x \leq1\text{,}$ then $x^2 \leq 1$ and $x+1 \geq 1\text{,}$ so $|f''(x)| = \dfrac{x^2}{|x+1|}\leq \dfrac{1}{1}=1\text{.}$\\ (b) $\dfrac{1}{2}$\qquad (c) $n \geq 65$\qquad (d) $n \geq 46$

Solution

1. When $0 \leq x \leq1\text{,}$ then $x^2 \leq 1$ and $x+1 \geq 1\text{,}$ so $|f''(x)| = \dfrac{x^2}{|x+1|}\leq \dfrac{1}{1}=1\text{.}$
2. To find the maximum value of a function over a closed interval, we test the function's values at the endpoints of the interval and at its critical points inside the interval. The critical points are where the function's derivative is zero or does not exist.

The function we're trying to maximize is $|f''(x)| = \frac{x^2}{|x+1|} = \frac{x^2}{x+1}=f''(x)$ (since our interval only contains nonnegative numbers). So, the critical points occur when $f'''(x) = 0$ or does not exist. We find $f'''(x)$ Using the quotient rule.

The only critical point in $[0,1]$ is $x=0\text{.}$ So, the extrema of $f''(x)$ over $[0,1]$ will occur at its endpoints. Indeed, since $f'''(x) \geq 0$ for all $x$ in $[0,1]\text{,}$ $f''(x)$ is increasing over this interval, so its maximum occurs at $x=1\text{.}$ That is,

\begin{equation*} |f''(x)|\leq f''(1)=\frac{1}{2} \end{equation*}
3. The absolute error using the midpoint rule is at most $\dfrac{M(b-a)^3}{24n^2}\text{.}$ Using $M=1\text{,}$ if we want this to be no more than $10^{-5}\text{,}$ we find an acceptable value of $n$ with the following calculation: \begin{align*} \dfrac{M(b-a)^3}{24n^2}&\leq 10^{-5}\\ \dfrac{1}{24n^2}&\leq 10^{-5}&(b-a=1,\,M=1)\\ \frac{10^5}{24} & \leq n^2\\ n & \geq 65 \end{align*}
4. The absolute error using the midpoint rule is at most $\dfrac{M(b-a)^3}{24n^2}\text{.}$ Using $M=\frac{1}{2}\text{,}$ if we want this to be no more than $10^{-5}\text{,}$ we find an acceptable value of $n$ with the following calculation: \begin{align*} \dfrac{M(b-a)^3}{24n^2}&\leq 10^{-5}\\ \dfrac{1}{48n^2}&\leq 10^{-5}&(b-a=1,\,M=\frac{1}{2})\\ \frac{10^5}{48} & \leq n^2\\ n & \geq 46 \end{align*}

Remark: how accurate you want to be in these calculations depends a lot on your circumstances. Imagine, for instance, that you were finding $M$ by hand, using this to find $n$ by hand, then programming a computer to evaluate the approximation. For a simple integral like this, the difference between computing time for 65 intervals versus 46 is likely to be miniscule. So, there's not much to be gained by the extra work in (b). However, if your original sloppy $M$ gave you something like $n=1000000\text{,}$ you might want to put some time into improving it, to shorten computation time. Moreover, if you were finding the approximation by hand, the difference between adding 46 terms and adding 65 terms would be considerable, and you would probably want to put in the effort up front to find the most accurate $M$ possible.

Exercise32
Hint

In using Simpson's rule to approximate $\displaystyle\int_1^{x\vphantom{\frac{1}{2}}} \frac{1}{t} \dee{t}$ with $n$ intervals, $a=1\text{,}$ $b=x\text{,}$ and $\De x = \dfrac{x-1}{n}\text{.}$

$\displaystyle\frac{x-1}{12}\left[1+\frac{16}{x+3}+\frac{4}{x+1}+\frac{16}{3x+1}+\frac{1}{x}\right]$

Solution

Before we can take our Simpson's rule approximation of $\displaystyle\int_1^x \dfrac{1}{t} \dee{t}\text{,}$ we need to know how many intervals to use. That means we need to bound our error, which means we need to bound $\ddiff{4}{}{t}\left\{\frac{1}{t}\right\}\text{.}$

\begin{align*} \diff{}{t}\left\{\frac{1}{t}\right\}&=-\frac{1}{t^2}& \ddiff{2}{}{t}\left\{\frac{1}{t}\right\}&=\frac{2}{t^3}\\ \ddiff{3}{}{t}\left\{\frac{1}{t}\right\}&=-\frac{6}{t^4}& \ddiff{4}{}{t}\left\{\frac{1}{t}\right\}&=\frac{24}{t^5} \end{align*}

So, over the interval $[1,3]\text{,}$ $\displaystyle\left|\ddiff{4}{}{t}\left\{\frac{1}{t}\right\}\right| \leq 24\text{.}$

Now, we can find an appropriate $n$ to ensure our error will be be less than 0.1 for any $x$ in $[1,3]\text{:}$

\begin{align*} \frac{L(b-a)^5}{180 n^4}& \lt 0 .1\\ \frac{24(x-1)^5}{180 n^4}& \lt \frac{1}{10}\\ n^4 & \gt \frac{24\cdot (x-1)^5}{18}\\ \end{align*}

Because $x-1 \leq 2$ for every $x$ in $[1,3]\text{,}$ if $n^4 \gt \dfrac{24\cdot 2^5}{18}\text{,}$ then $n^4 \gt \dfrac{24\cdot (x-1)^5}{18}$ for every allowed $x\text{.}$

\begin{align*} n^4& \gt \frac{24\cdot 2^5}{18}= \frac{128}{3}\\ n& \gt \sqrt[4]{\frac{128}{3}}\approx 2.6 \end{align*}

Since $n$ must be even, $n=4$ is enough intervals to guarantee our error is not too high for any $x$ in $[1,3]\text{.}$ Now we find our Simpson's rule approximation with $n=4\text{,}$ $a=1\text{,}$ $b=x\text{,}$ and $\Delta x = \dfrac{x-1}{4}\text{.}$ The points where we evaluate $\frac{1}{t}$ are:

\begin{align*} x_0 =& 1 & x_1&=1+ \frac{x-1}{4}& x_2&=1+2\cdot\frac{x-1}{4}& x_3&=1+3\cdot\frac{x-1}{4}& x_4&=1+4\cdot\frac{x-1}{4}\\ & & &=\frac{x+3}{4}& &=\frac{x+1}{2}&&=\frac{3x+1}{4}& &=x \end{align*}

\begin{align*} \log x=\int_1^x \frac{1}{t} \dee{t}&\approx \frac{\De x}{3}\left[\frac{1}{x_0}+\frac{4}{x_1}+\frac{2}{x_2}+\frac{4}{x_3}+\frac{1}{x_4}\right]\\ &=\frac{x-1}{12}\left[1+\frac{16}{x+3}+\frac{4}{x+1}+\frac{16}{3x+1}+\frac{1}{x}\right]\\ &=f(x) \end{align*}

Below is a graph of our approximation $f(x)$ and natural logarithm on the same axes. The natural logarithm function is shown red and dashed, while our approximating function is solid blue. Our approximation appears to be quite accurate for small, positive values of $x\text{.}$

Exercise33
Hint

• $\int_1^2 \frac{1}{1+x^2} \dee{x} = \arctan(2) - \frac{\pi}{4}\text{,}$ so $\arctan(2) = \frac{\pi}{4}+\int_1^2 \frac{1}{1+x^2} \dee{x}$
• If an approximation $A$ of the integral $\int_1^2 \frac{1}{1+x^2} \dee{x}$ has error at most $\varepsilon\text{,}$ then $A-\varepsilon \leq \int_1^2 \frac{1}{1+x^2} \dee{x} \leq A+\varepsilon\text{.}$
• Looking at our target interval will tell you how small $\varepsilon$ needs to be, which in turn will tell you how many intervals you need to use.
• You can show, by considering the numerator and denominator separately, that $|f^{(4)}(x)| \leq 30.75$ for every $x$ in $[1,2]\text{.}$
• If you use Simpson's rule to approximate $\int_1^2 \frac{1}{1+x^2} \dee{x}\text{,}$ you won't need very many intervals to get the requisite accuracy.

Note: for more detail, see the solutions.

First, we use Simpson's rule with $n=4$ to approximate $\int_1^2 \frac{1}{1+x^2}\,\dee{x}\text{.}$ The choice of this method (what we're approximating, why $n=4\text{,}$ etc.) is explained in the solutions--here, we only show that it works.

\begin{equation*} \int_1^2 \frac{1}{1+x^2}\,\dee{x} \approx\frac{1}{12}\left[ \frac{1}{2} + \frac{64}{41}+\frac{8}{13}+\frac{64}{65}+\frac{1}{5} \right]\approx 0.321748 \end{equation*}

For ease of notation, define $A=0.321748\text{.}$

Now, we bound the error associated with this approximation. Define $N(x) = 24(5x^4-10x^2+1)$ and $D(x) = (x^2+1)^5\text{,}$ so $N(x)/D(x)$ gives the fourth derivative of $\frac{1}{1+x^2}\text{.}$ When $1 \le x \le 2\text{,}$ $|N(x)| \le N(2)=984$ (because $N(x)$ is increasing over that interval) and $|D(x)| \geq D(1) = 2^5$ (because $D(x)$ is also increasing over that interval), so $\left| \ddiff{4}{}{x}\left\{\frac{1}{1+x^2}\right\}\right| = \left| \frac{N(x)}{D(x)}\right| \leq \frac{984}{2^5}=30.75\text{.}$ Now we find the error bound for Simpson's rule with $L=30.75\text{,}$ $b=2\text{,}$ $a=1\text{,}$ and $n=4\text{.}$

\begin{equation*} \left| \int_1^2 \frac{1}{1+x^2}\,\dee{x} - A\right| =|\text{error}| \leq \frac{L(b-a)^5}{180\cdot n^4}\ =\frac{30.75}{180\cdot 4^4} \lt 0.00067 \end{equation*}

So,

\begin{align*} -0.00067 && \lt && \int_1^2 \frac{1}{1+x^2}\,\dee{x} - A && \lt && 0.00067\\ A-0.0067 && \lt && \int_1^2 \frac{1}{1+x^2}\,\dee{x} && \lt && A+0.00067\\ A-0.00067 && \lt && \arctan(2)-\arctan(1) && \lt && A+0.00067\\ A-0.00067 && \lt && \arctan(2)-\frac{\pi}{4} && \lt && A+0.00067\\ \frac{\pi}{4}+A-0.00067 && \lt && \arctan(2) && \lt &&\frac{\pi}{4}+ A+0.00067\\ \frac{\pi}{4}+0.321748-0.00067 && \lt && \arctan(2) && \lt &&\frac{\pi}{4}+0.321748+0.00067\\ \frac{\pi}{4}+0.321078 && \lt && \arctan(2) && \lt &&\frac{\pi}{4}+0.322418\\ \frac{\pi}{4}+0.321 && \lt && \arctan(2) && \lt &&\frac{\pi}{4}+0.323 \end{align*}

This was the desired bound.

Solution

First, we want a strategy for approximating $\arctan 2\text{.}$ Our hints are that involves integrating $\dfrac{1}{1+x^2}\text{,}$ which is the antiderivative of arctangent, and the number $\dfrac{\pi}{4}\text{,}$ which is the same as $\arctan(1)\text{.}$ With that in mind:

\begin{align*} \int_1^2 \frac{1}{1+x^2} \dee{x}& = \arctan(2) - \arctan(1) = \arctan(2) - \frac{\pi}{4}\\ \text{So, }\qquad \arctan(2) &= \frac{\pi}{4} + \int_1^2 \frac{1}{1+x^2} \dee{x}\tag{$*$}\\ \end{align*}

We won't know the value of the integral exactly, but we'll have an approximation $A$ bounded by some positive error bound $\varepsilon\text{.}$ Then,

\begin{align*} - \varepsilon &\leq \left(\int_1^2 \frac{1}{1+x^2} \dee{x} - A\right) \leq \varepsilon\\ A - \varepsilon &\leq \left(\int_1^2 \frac{1}{1+x^2} \dee{x}\right) \leq A + \varepsilon\\ \text{So, from ($*$), }\qquad \frac{\pi}{4}+A-\varepsilon& \leq \arctan(2) \leq \frac{\pi}{4}+A+\varepsilon \end{align*}

Which approximation should we use? We're given the fourth derivative of $\dfrac{1}{1+x^2}\text{,}$ which is the derivative we need for Simpson's rule. Simpson's rule is also usually quite efficient, and we're very interested in not adding up dozens of terms, so we choose Simpson's rule.

Now that we've chosen Simpson's rule, we should decide how many intervals to use. In order to bound our error, we need to find a bound for the fourth derivative. To that end, define $N(x) = 24(5x^4-10x^2+1)\text{.}$ Then $N'(x) = 24(20x^3-20x)= 480x(x^2-1)\text{,}$ which is positive over the interval $[1,2]\text{.}$ So, $N(x) \leq N(2)=24(5\cdot 2^4 - 10\cdot 2^2+1)=984$ when $1 \leq x \leq 2\text{.}$ Furthermore, let $D(x)=(x^2+1)^5\text{.}$ If $1 \leq x \leq 2\text{,}$ then $D(x) \geq 2^5\text{.}$ Now we can find a reasonable value of $L\text{:}$

\begin{align*} |f^{(4)}(x)|&=\left|\frac{25(5x^4-10x^2+1)}{(x^2+1)^5}\right| = \left|\frac{N(x)}{D(x)}\right| \leq \frac{984}{2^5} = \frac{123}{4} = 30.75 \end{align*}

So, we take $L=30.75\text{.}$

We want $\left[\dfrac{\pi}{4}+A-\varepsilon ,\, \dfrac{\pi}{4}+A+\varepsilon \right]$ to look something like $\left[\dfrac{\pi}{4}+0.321,\, \dfrac{\pi}{4}+0.323\right]\text{.}$ Note $\varepsilon$ is half the length of the first interval. Half the length of the second interval is $0.001 = \frac{1}{1000}\text{.}$ So, we want a value of $\varepsilon$ that is no larger than this. Now we can find our $n\text{:}$

\begin{align*} \dfrac{L(b-a)^5}{180\cdot n^4} &\leq \frac{1}{1000}\\ \dfrac{30.75}{180\cdot n^4} &\leq \frac{1}{1000}\\ n^4 & \geq \frac{30.75\times 1000}{180}\\ n & \geq \sqrt[4]{\frac{30750}{180}}\approx 3.62 \end{align*}

So, we choose $n=4$), and are guaranteed that the absolute error in our approximation will be no more than $\dfrac{30.75}{180\cdot 4^4} \lt 0.00067\text{.}$

Since $n=4\text{,}$ then $\De x = \dfrac{b-a}{n}=\dfrac{1}{4}\text{,}$ so:

Now we can find our Simpson's rule approximation $A\text{:}$

\begin{align*} \int_0^1 \frac{1}{1+x^2} \dee{x}&\approx \frac{\De x}{3}\big[ f(x_0) + 4f(x_1)+2f(x_2)+4f(x_3)+f(x_4) \big]\\ &=\frac{1/4}{3}\big[ f(1) + 4f(5/4)+2f(3/2)+4f(7/4)+f(2) \big]\\ &=\frac{1}{12}\left[ \frac{1}{1+1} + \frac{4}{25/16+1}+\frac{2}{9/4+1}+\frac{4}{49/16+1}+\frac{1}{4+1} \right]\\ &=\frac{1}{12}\left[ \frac{1}{2} + \frac{4\cdot 16}{25+16}+\frac{2\cdot 4}{9+4}+\frac{4\cdot 16}{49+16}+\frac{1}{5} \right]\\ &=\frac{1}{12}\left[ \frac{1}{2} + \frac{64}{41}+\frac{8}{13}+\frac{64}{65}+\frac{1}{5} \right]\\ &\approx 0.321748=A \end{align*}

As we saw before, the error associated with this approximation is at most $\dfrac{30.75}{180\cdot 4^4} \lt 0.00067=\varepsilon\text{.}$ So,

This is precisely what we wanted to show.

Exercises1.12.4Exercises

Exercise1
Hint

There are two kinds of impropreity in an integral: an infinite discontinuity in the integrand, and an infinite limit of integration.

Any real number in $[1,\infty)$ or $(-\infty,-1]\text{,}$ and $b = \pm \infty\text{.}$

Solution

If $= \pm \infty\text{,}$ then our integral is improper because one limit is not a real number.

Furthermore, our integral will be improper if its domain of integration contains either of its infinite discontinuities, $x=1$ and $x=-1\text{.}$ Since one limit of integration is 0, the integral is improper if $b \geq 1$ or if $b \leq -1\text{.}$

Below, we've graphed $\frac{1}{x^2-1}$ to make it clearer why values of $b$ in $(-1,1)$ are the only values that don't result in an improper integral when the other limit of integration is $a=0\text{.}$

Exercise2
Hint

The integrand is continuous for all $x\text{.}$

$b = \pm\infty$

Solution

Since the integrand is continuous for all real $x\text{,}$ the only kind of impropriety available to us is to set $b = \pm\infty\text{.}$

Exercise3
Hint

What matters is which function is bigger for large values of $x\text{,}$ not near the origin.

The red function is $f(x)\text{,}$ and the blue function is $g(x)\text{.}$

Solution

For large values of $x\text{,}$ $|\text{red function}|\leq \text{(blue function)}$ and $0 \leq \text{(blue function)}\text{.}$ If the blue function's integral converged, then the red function's integral would as well (by the comparison test, Theorem 1.12.17 in the text). Since one integral converges and the other diverges, the blue function is $g(x)$ and the red function is $f(x)\text{.}$

Exercise4(*)
Hint

Read both the question and Theorem 1.12.17 very carefully.

False. For example, the functions $f(x)=e^{-x}$ and $g(x)=1$ provide a counterexample.

Solution

False. The inequality goes the “wrong way” for Theorem 1.12.17: the area under the curve $f(x)$ is finite, but the area under $g(x)$ could be much larger, even infinitely larger.

For example, if $f(x)=e^{-x}$ and $g(x)=1\text{,}$ then $0 \leq f(x) \leq g(x)$ and $\displaystyle\int_{1}^{\infty} f(x) \,\dee{x}$ converges, but $\displaystyle\int_{1}^{\infty} g(x) \,\dee{x}$ diverges.

Exercise5
Hint

(a) What if $h(x)$ is negative? What if it's not?\\ (b) What if $h(x)$ is very close to $f(x)$ or $g(x)\text{,}$ rather than right in the middle?\\ (c) Note $|h(x)| \leq 2f(x)\text{.}$

1. Not enough information to decide. For example, consider $h(x) = 0$ versus $h(x) = -1\text{.}$
2. Not enough information to decide. For example, consider $h(x)= f(x)$ versus $h(x) = g(x)\text{.}$
3. $\displaystyle\int_{0\vphantom{\frac12}}^{\infty}h(x) \dee{x}$ converges by the comparison test, since $|h(x)| \leq 2f(x)$ and $\displaystyle\int_0^\infty 2f(x) \dee{x}$ converges.
Solution

1. Not enough information to decide. For example, consider $h(x) = 0$ versus $h(x) = -1\text{.}$ In both cases, $h(x) \leq f(x)\text{.}$ However, $\displaystyle\int_0^\infty 0 \dee{x}$ converges to 0, while $\displaystyle\int_0^\infty -1 \dee{x}$ diverges.

Note: if we had also specified $0 \leq h(x)\text{,}$ then we would be able to conclude that $\int_0^\infty h(x) \dee{x}$ converges by the comparison test.

2. Not enough information to decide. For example, consider $h(x)= f(x)$ versus $h(x) = g(x)\text{.}$ In both cases, $f(x) \leq h(x) \leq g(x)\text{.}$
3. $\displaystyle\int_{0\vphantom{\frac12}}^{\infty}h(x) \dee{x}$ converges.

• From the given information, $|h(x)| \leq 2f(x)\text{.}$
• We claim $\displaystyle\int_{0\vphantom{\frac12}}^{\infty} 2f(x) \dee{x}$ converges.

• We can see this by writing $\displaystyle\int_{0\vphantom{\frac12}}^{\infty} 2f(x) \dee{x}= 2\int_0^{\infty} f(x) \dee{x}$ and noting that the second integral converges.
• Alternately, we can use the limiting comparison test, Theorem 1.12.22. Since $f(x) \geq 0\text{,}$ $\displaystyle\int_0^\infty f(x) \dee{x}$ converges, and $\lim\limits_{x \to \infty}\dfrac{2f(x)}{f(x)}=2$ (the limit exists), we conclude $\displaystyle\int_0^\infty 2f(x) \dee{x}$ converges.
• So, comparing $h(x)$ to $2f(x)\text{,}$ by the comparison test (Theorem 1.12.17) $\displaystyle\int_0^{\vphantom{\frac12}\infty}h(x) \dee{x}$ converges.
Exercise6(*)
Hint

First: is the integrand unbounded, and if so, where?

Second: when evaluating integrals, always check to see if you can use a simple substitution before trying a complicated procedure like partial fractions.

The integral diverges.

Solution

The denominator is zero when $x=1\text{,}$ but the numerator is not, so the integrand has a singularity (infinite discontinuity) at $x=1\text{.}$ Let's replace the limit $x=1$ with a variable that creeps toward 1.

\begin{equation*} \int_0^1\frac{x^4}{x^5-1}\,\dee{x} =\lim_{t\rightarrow 1^-}\int_0^t\frac{x^4}{x^5-1}\,\dee{x} \end{equation*}

To evaluate this integral we use the substitution $u=x^5\text{,}$ $\dee{u}=5x^4\dee{x}\text{.}$ When $x=0$ we have $u=0\text{,}$ and when $x=t$ we have $u=t^5\text{,}$ so

\begin{align*} \int_0^1\frac{x^4}{x^5-1}\,\dee{x}&=\lim_{t \to 1^-}\int_{0}^{t}\frac{x^4}{x^5-1}\,\dee{x} =\lim_{t \to 1^-}\int_{u=0}^{u=t^5}\frac{1}{5(u-1)}\,\dee{u}\\ &=\lim_{t \to 1^-}\left( \left[\frac{1}{5}\log|u-1|\right]_0^{t^5}\right) =\lim_{t \to 1^-}\frac{1}{5}\log|t^5-1|=-\infty \end{align*}

The limit diverges, so the integral diverges as well.

Exercise7(*)
Hint

Is the integrand bounded?

The integral diverges.

Solution

The denominator of the integrand is zero when $x=-1\text{,}$ but the numerator is not. So, the integrand has a singularity (infinite discontinuity) at $x=-1\text{.}$ This is the only “source of impropriety” in this integral, so we only need to make one break in the domain of integration.

\begin{equation*} \int_{-2}^2\frac{1}{(x+1)^{4/3}}\,\dee{x} =\lim_{t\rightarrow -1^-}\int_{-2}^t \frac{1}{(x+1)^{4/3}}\,\dee{x} +\lim_{t\rightarrow -1^+}\int_t^2 \frac{1}{(x+1)^{4/3}}\,\dee{x} \end{equation*}

Let's start by considering the left limit.

\begin{align*} \lim_{t \to -1^-}\int_{-2}^t \frac{1}{(x+1)^{4/3}}\,\dee{x} &=\lim_{t \to -1^-}\left(\left[-\frac{3}{(x+1)^{1/3}}\right|_{-2}^t\right)\\ &=\lim_{t \to -1^-}\left(-\frac{3}{(t+1)^{1/3}}+\frac{3}{(-1)^{1/3}}\right)=\infty \end{align*}

Since this limit diverges, the integral diverges. (A similar argument shows that the second integral diverges. Either one of them diverging is enough to conclude that the original integral diverges.)

Exercise8(*)
Hint

See Example 1.12.21. Rather than antidifferentiating, you can find a nice comparison.

The integral does not converge.

Solution

First, let's identify all “sources of impropriety.” The integrand has a singularity when $4x^2-x=0\text{,}$ that is, when $x(4x-1)=0\text{,}$ so at $x=0$ and $x=\frac{1}{4}\text{.}$ Neither of these are in our domain of integration, so the only “source of impropriety” is the unbounded domain of integration.

We could antidifferentiate this function (it looks like a nice candidate for a trig substitution), but is seems easier to use a comparison. For large values of $x\text{,}$ the term $x^2$ will be much larger than $x\text{,}$ so we might guess that our integral behaves similarly to $\int_1^\infty \frac{1}{\sqrt{4x^2}} \dee{x}=\int_1^\infty \frac{1}{2x} \dee{x}\text{.}$

For all $x\ge 1\text{,}$ $\sqrt{4x^2-x}\le\sqrt{4x^2}=2x\text{.}$ So, $\frac{1}{\sqrt{4x^2-x}} \geq \frac{1}{2x}\text{.}$ Note $\int_1^\infty \frac{1}{2x} \dee{x}$ diverges:

\begin{gather*} \lim_{t\rightarrow\infty}\int_1^t\frac{1}{2x}\,\dee{x} =\lim_{t\rightarrow\infty}\left(\frac{1}{2} \big[\log x\big]_1^t\right) =\lim_{t\rightarrow\infty}\frac{1}{2} \log t =\infty \end{gather*}

So:

• $\frac{1}{2x}$ and $\frac{1}{\sqrt{4x^2-x}}$ are defined and continuous for all $x \geq 1\text{,}$
• $\frac{1}{2x} \geq 0$ for all $x \geq 1\text{,}$
• $\frac{1}{\sqrt{4x^2-x}} \ge\frac{1}{\sqrt{4x^2}} = \frac{1}{2x}$ for all $x \ge 1\text{,}$ and
• $\int_1^\infty \frac{1}{2x} \dee{x}$ diverges.

By the comparison test, Theorem 1.12.17 in the text, the integral does not converge.

Exercise9(*)
Hint

Which of the two terms in the denominator is more important when $x\approx 0\text{?}$ Which one is more important when $x$ is very large?

The integral converges.

Solution

The integrand is positive everywhere. So, either the integral converges to some finite number, or it is infinite. We want to generate a guess as to which it is.

When $x$ is small, $\sqrt{x} \gt x^2\text{,}$ so we might guess that our integral behaves like the integral of $\frac{1}{\sqrt{x}}$ when $x$ is near to 0. On the other hand, when $x$ is large, $\sqrt{x} \lt x^2\text{,}$ so we might guess that our integral behaves like the integral of $\frac{1}{x^2}$ as $x$ goes to infinity. This is the hunch that drives the following work:

\begin{align*} 0\le\frac{1}{x^2+\sqrt{x}} &\le \frac{1}{\sqrt{x}}\quad &&\text{and the integral }\int_0^1\frac{\dee{x}}{\sqrt{x}}\text{ converges by Example }\knowl{./knowl/eg_IMPp2.html}{\text{1.12.9}}\text{, and} \\ 0\le\frac{1}{x^2+\sqrt{x}} &\le \frac{1}{x^2} &&\text{and the integral }\int_1^\infty\frac{\dee{x}}{x^2}\text{ converges by Example }\knowl{./knowl/eg_IMPp1.html}{\text{1.12.8}} \end{align*}

Note $\frac{\dee{x}}{x^2+\sqrt{x}}$ is defined and continuous everywhere, $\frac{1}{\sqrt{x}}$ is defined and continuous for $x \gt 0\text{,}$ and $\frac{1}{x^2}$ is defined and continuous for $x \ge 1\text{.}$ So, the integral converges by the comparison test, Theorem 1.12.17 in the text, together with Remark 1.12.16.

Exercise10
Hint

Remember to break the integral into two pieces.

The integral diverges.

Solution

There are two “sources of impropriety:” the two (infinite) limits of integration. So, we break our integral into two pieces.

\begin{align*} \int_{-\infty}^\infty \cos x \dee{x}&=\int_{-\infty}^0 \cos x \dee{x}+\int_{0}^\infty \cos x \dee{x}\\ &=\lim_{a \to \infty }\left[\int_{-a}^0 \cos x \dee{x}\right]+ \lim_{b \to \infty }\left[\int_{0}^b \cos x \dee{x}\right]\\ \end{align*}

These are easy enough to antdifferentiate.

\begin{align*} &=\lim_{a \to \infty }\left[\sin 0 - \sin(- a) \right]+ \lim_{b \to \infty }\left[\sin b - \sin 0 \right]\\ &=\text{DNE} \end{align*}

Since the limits don't exist, the integral diverges. (It happens that both limits don't exist; even if only one failed to exist, the integral would still diverge.)

Exercise11
Hint

Remember to break the integral into two pieces.

The integral diverges.

Solution

There are two “sources of impropriety:” the two bounds. So, we break our integral into two pieces.

\begin{align*} \int_{-\infty}^\infty \sin x \dee{x}&=\int_{-\infty}^0 \sin x \dee{x}+\int_{0}^\infty \sin x \dee{x}\\ &=\lim_{a \to \infty }\left[\int_{-a}^0 \sin x \dee{x}\right]+ \lim_{b \to \infty }\left[\int_{0}^b \sin x \dee{x}\right]\\ &=\lim_{a \to \infty }\left[-\cos 0 + \cos(- a) \right]+ \lim_{b \to \infty }\left[-\cos b + \cos 0 \right]\\ &=\text{DNE} \end{align*}

Since the limits don't exist, the integral diverges. (It happens that both limits don't exist; even if only one failed to exist, the integral would diverge.)

Remark: it's very tempting to think that this integral should converge, because as an odd function the area to the right of the $x$-axis “cancels out” the area to the left when the limits of integration are symmetric. One justification for not using this intuition is given in Example 1.12.11 in the text. Here's another: In Question 1.12.4.10 we saw that $\int_{-\infty}^\infty \cos x \dee{x}$ diverges. Since $\sin x = \cos (x-\pi/2)\text{,}$ the area bounded by sine and the area bounded by cosine over an infinite region seem to be the same--only shifted by $\pi/2\text{.}$ So if $\int_{-\infty}^\infty \sin x \dee{x}=0\text{,}$ then we ought to also have $\int_{-\infty}^\infty \cos x \dee{x}=0\text{,}$ but we saw in Question 1.12.4.10 this is not the case.

Exercise12
Hint

The easiest test in this case is limiting comparison, Theorem 1.12.22.

The integral diverges.

Solution

First, we check that the integrand has no singularities. The denominator is always positive when $x \ge 10\text{,}$ so our only “source of impropriety” is the infinite limit of integration.

We further note that, for large values of $x\text{,}$ the integrand resembles $\dfrac{x^4}{x^5} = \dfrac{1}{x\vphantom{\frac12}}\text{.}$ So, we have a two-part hunch: that the integral diverges, and that we can show it diverges by comparing it to $\displaystyle\int_{10}^\infty \frac{1}{x} \dee{x}\text{.}$

In order to use the comparison test, we'd need to show that $\displaystyle\frac{x^4-5x^3+2x-7}{x^5+3x+8} \geq \frac{1}{\vphantom{\frac12}x}\text{.}$ If this is true, it will be difficult to prove--and it's not at all clear that it's true. So, we will use the limiting comparison test instead, Theorem 1.12.22, with $g(x)= \dfrac{1}{x}\text{,}$ $f(x)=\dfrac{x^4-5x^3+2x-7}{x^5+3x+8} \text{,}$ and $a=10\text{.}$

• Both $f(x)$ and $g(x)$ are defined and continuous for all $x \gt 0\text{,}$ so in particular they are defined and continuous for $x \geq 10\text{.}$
• $g(x) \geq 0$ for all $x \geq 10$
• $\displaystyle\int_{10}^\infty g(x) \dee{x}$ diverges.
• Using l'Hôpital's rule (5 times!), or simply dividing both the numerator and denominator by $x^5$ (the common leading term), tells us: \begin{align*} \lim_{x \to \infty} \frac{f(x)}{g(x)}&= \lim_{x \to \infty} \frac{\frac{x^4-5x^3+2x-7}{x^5+3x+8} }{ \frac{1}{x}}= \lim_{x \to \infty} x\cdot\frac{x^4-5x^3+2x-7}{x^5+3x+8} \\ &=\lim_{x \to \infty} \frac{x^5-5x^4+2x^2-7x}{x^5+3x+8} =1 \end{align*} That is, the limit exists and is nonzero.

By the limiting comparison test, we conclude $\displaystyle\int_{10}^\infty f(x) \dee{x}$ diverges.

Exercise13
Hint

Not all discontinuities cause an integral to be improper--only infinite discontinuities.

The integral diverges.

Solution

Our domain of integration is finite, so the only potential “sources of impropriety” are infinite discontinuities in the integrand. To find these, we factor.

\begin{align*} \int_0^{10} \frac{x-1}{x^2-11x+10} \dee{x}&= \int_0^{10} \frac{x-1}{(x-1)(x-10)} \dee{x}\\ \end{align*}

A removable discontinuity doesn't affect the integral.

\begin{align*} &=\int_0^{10} \frac{1}{x-10} \dee{x}\\ \end{align*}

Use the substitution $u=x-10\text{,}$ $\dee{u}=\dee{x}\text{.}$ When $x=0\text{,}$ $u=-10\text{,}$ and when $x=10\text{,}$ $u=0\text{.}$

\begin{align*} &=\int_{-10}^0 \frac{1}{u} \dee{u} \end{align*}

This is a $p$-integral with $p=1\text{.}$ From Example 1.12.9 and Theorem 1.12.20, we know it diverges.

Exercise14(*)
Hint

Which of the two terms in the denominator is more important when $x$ is very large?

The integral diverges.

Solution

You might think that, because the integrand is odd, the integral converges to $0\text{.}$ This is a common mistake-- see Example 1.12.11 in the text, or Question 11 in this section. In the absence of such a shortcut, we use our standard procedure: identifying problem spots over the domain of integration, and replacing them with limits.

There are two “sources of impropriety,” namely $x\to +\infty$ and $x\to -\infty\text{.}$ So, we split the integral in two, and treat the two halves separately. The integrals below can be evaluated with the substitution $u=x^2+1\text{,}$ $\frac{1}{2}\dee{u} = x\dee{x}\text{.}$

\begin{align*} \int_{-\infty}^{+\infty}\frac{x}{x^2+1}\dee{x} &=\textcolor{blue}{ \int_{-\infty}^0\frac{x}{x^2+1}\dee{x}} + \textcolor{red}{\int_0^{+\infty}\frac{x}{x^2+1}\dee{x}}\\ \color{blue}{\int_{-\infty}^0\frac{x}{x^2+1}\dee{x}} &=\lim_{R\rightarrow\infty}\int_{-R}^0\frac{x}{x^2+1}\dee{x} =\lim_{R\rightarrow\infty}\frac{1}{2}\log(x^2+1)\Big|_{-R}^0\\ &=\lim_{R \to \infty}\frac{1}{2}\left[\log 1-\log(R^2+1)\right]=\lim_{R\rightarrow\infty}-\frac{1}{2}\log(R^2+1) =-\infty \\ \color{red}{\int_0^{+\infty}\frac{x}{x^2+1}\dee{x}} &=\lim_{R\rightarrow\infty}\int_0^R\frac{x}{x^2+1}\dee{x} =\lim_{R\rightarrow\infty}\frac{1}{2}\log(x^2+1)\Big|_0^R\\ &=\lim_{R \to \infty}\frac{1}{2}\left[ \log(R^2+1)-\log 1\right] =\lim_{R\rightarrow\infty}\frac{1}{2}\log(R^2+1) =+\infty \end{align*}

Both halves diverge, so the whole integral diverges.

Once again: after we found that one of the limits diverged, we could have stopped and concluded that the original integrand diverges. Don't make the mistake of thinking that $\infty-\infty=0\text{.}$ That can get you into big trouble. $\infty$ is not a normal number. For example $2\infty=\infty\text{.}$ So if $\infty$ were a normal number we would have both $\infty-\infty=0$ and $\infty-\infty=2\infty-\infty =\infty\text{.}$

Exercise15(*)
Hint

Which of the two terms in the denominator is more important when $x\approx 0\text{?}$ Which one is more important when $x$ is very large?

The integral converges.

Solution

We don't want to antidifferentiate this integrand, so let's use a comparison. Note the integrand is positive when $x \gt 0\text{.}$

For any $x\text{,}$ $|\sin x| \leq 1\text{,}$ so $\dfrac{|\sin x|}{x^{3/2}+x^{1/2}} \leq \dfrac{1}{x^{3/2}+x^{1/2}} \text{.}$

Since $x=0$ and $x \to \infty$ both cause the integral to be improper, we need to break it into two pieces. Since both terms in the denominator give positive numbers when $x$ is positive, $\dfrac{1}{x^{3/2}+x^{1/2}} \leq \dfrac{1}{x^{3/2}}$ and $\dfrac{1}{x^{3/2}+x^{1/2}} \leq \dfrac{1}{x^{1/2}}\text{.}$ That gives us two options for comparison.

When $x$ is positive and close to zero, $x^{1/2} \ge x^{3/2}\text{,}$ so we guess that we should compare our integrand to $\frac{1}{x^{1/2}}$ near the limit $x=0\text{.}$ In contrast, when $x$ is very large, $x^{1/2} \le x^{3/2}\text{,}$ so we guess that we should compare our integrand to $\frac{1}{x^{3/2}}$ as $x$ goes to infinity.

\begin{align*} \frac{|\sin x|}{x^{3/2}+x^{1/2}} &\le \frac{1}{x^{1/2}} \quad&&\text{and the integral }\int_0^1\frac{\dee{x}}{x^{1/2}}\text{ converges by the $p$-test, Example }\knowl{./knowl/eg_IMPp2.html}{\text{1.12.9}} \\ \frac{|\sin x|}{x^{3/2}+x^{1/2}} &\le \frac{1}{x^{3/2}} \quad&&\text{and the integral }\int_1^\infty\frac{\dee{x}}{x^{3/2}}\text{ converges by the $p$-test, Example }\knowl{./knowl/eg_IMPp1.html}{\text{1.12.8}} \end{align*}

Now we have all the data we need to apply the comparison test, Theorem 1.12.17.

• $\dfrac{|\sin x|}{x^{3/2}+x^{1/2}}$\,, $\dfrac{1}{x^{1/2}}$\,, and $\dfrac{1}{x^{3/2}}$ are defined and continuous for $x \gt 0$
• $\dfrac{1}{x^{1/2}}$ and $\dfrac{1}{x^{3/2}}$ are nonnegative for $x \ge 0$
• $\dfrac{|\sin x|}{x^{3/2}+x^{1/2}}\le \dfrac{1}{x^{1/2}}$ for all $x \gt 0$ and $\displaystyle\int_0^1\dfrac{1}{x^{1/2}} \dee{x}$ converges, so (using Remark 1.12.16) $\displaystyle\int_0^{1\vphantom{\frac12}}\dfrac{|\sin x|}{x^{3/2}+x^{1/2}} \dee{x}$ converges.
• $\dfrac{|\sin x|}{x^{3/2}+x^{1/2}}\le \dfrac{1}{x^{3/2}}$ for all $x \ge 1$ and $\displaystyle\int_1^\infty\dfrac{1}{x^{3/2}} \dee{x}$ converges, so $\displaystyle\int_1^{\infty\vphantom{\frac12}}\dfrac{|\sin x|}{x^{3/2}+x^{1/2}} \dee{x}$ converges.

Therefore, our integral $\displaystyle\int_0^\infty \frac{|\sin x|}{x^{3/2}+x^{1/2}} \dee{x}$ converges.

Exercise16(*)
Hint

What are the “problem $x$'s” for this integral? Get a simple approximation to the integrand near each.

The integral converges.

Solution

The integrand is positive everywhere, so either the integral converges to some finite number or it is infinite. There are two potential “sources of impropriety” — a possible singularity at $x=0$ and the fact that the domain of integration extends to $\infty\text{.}$ So we split up the integral.

\begin{gather*} \int_0^\infty\frac{x+1}{x^{1/3}(x^2+x+1)}\,\dee{x} = \int_0^1\frac{x+1}{x^{1/3}(x^2+x+1)}\,\dee{x} +\int_1^\infty\frac{x+1}{x^{1/3}(x^2+x+1)}\,\dee{x} \end{gather*}

Let's develop a hunch about whether the integral converges or diverges. When $x\approx 0\text{,}$ $x^2$ and $x$ are both a lot smaller than 1, so we guess we should compare the integrand to $\frac{1}{x^{1/3}}\text{.}$

\begin{equation*} \frac{x+1}{x^{1/3}(x^2+x+1)} \approx \frac{1}{x^{1/3}(1)} = \frac{1}{x^{1/3}} \end{equation*}

Note $\int_0^1 \frac{1}{x^{1/3}} \dee{x}$ converges by Example 1.12.9 (it's a $p$-type integral), so we guess $\int_0^1 \frac{x+1}{x^{1/3}(x^2+x+1)} \dee{x}$ converges as well.

When $x$ is very large, $x^2$ is much bigger than $x\text{,}$ which is much bigger than 1, so we guess we should compare the integrand to $\frac{1}{x^{4/3}}\text{.}$

\begin{equation*} \frac{x+1}{x^{1/3}(x^2+x+1)} \approx \frac{x}{x^{1/3}(x^2)} = \frac{1}{x^{4/3}} \end{equation*}

Note $\int_1^\infty \frac{1}{x^{4/3}} \dee{x}$ converges by Example 1.12.8 (it's a $p$-type integral), so we guess $\int_1^\infty \frac{x+1}{x^{1/3}(x^2+x+1)} \dee{x}$ converges as well.

Now it's time to verify our guesses with the limiting comparison test, Theorem 1.12.22. Be careful: our “$\approx$” signs are not strong enough to use either the limiting comparison test or the comparison test, they are only enough to suggest a reasonable function to compare to.

• $\frac{x+1}{x^{1/3}(x^2+x+1)}$\,, $\frac{1}{x^{1/3}}$\,, and $\frac{1}{x^{4/3}}$ are defined and continuous for all $x \gt 0$
• $\frac{1}{x^{1/3}}$ and $\frac{1}{x^{4/3}}$ are positive for all $x \gt 0$
• $\int_0^1 \frac{1}{x^{1/3}} \dee{x}$ and $\int_1^\infty \frac{1}{x^{4/3}} \dee{x}$ both converge
• $\lim\limits_{x \to 0}\dfrac{\frac{x+1}{x^{1/3}(x^2+x+1)}}{\frac{1}{x^{1/3}}} =\lim\limits_{x \to 0}\dfrac{x+1}{x^2+x+1}=\dfrac{0+1}{0+0+1}=1 \text{;}$ in particular, this limit exists.
• Using the limiting comparison test (Theorem 1.12.22, together with Remark 1.12.16 because our impropriety is due to a singularity), $\int_0^1 \frac{x+1}{x^{1/3}(x^2+x+1)} \dee{x}$ converges.
• $\lim\limits_{x \to \infty}\dfrac{\frac{x+1}{x^{1/3}(x^2+x+1)}}{\frac{1}{x^{4/3}}} =\lim\limits_{x \to 0}\dfrac{x(x+1)}{x^2+x+1}=1 \text{;}$ in particular, this limit exists.
• Using the limiting comparison test (Theorem 1.12.22), $\int_1^\infty \frac{x+1}{x^{1/3}(x^2+x+1)} \dee{x}$ converges.

We conclude $\displaystyle\int_0^\infty \frac{x+1}{x^{1/3}(x^2+x+1)} \dee{x}$ converges.

Exercise17
Hint

To find the volume of the solid, cut it into horizontal slices, which are thin circular disks.

The true/false statement is equivalent to saying that the improper integral giving the volume of the solid when $a=0$ diverges to infinity.

false

Solution

To find the volume of the solid, we cut it into horizontal slices, which are thin circular disks. At height $y\text{,}$ the disk has radius $x=\frac{1}{y}$ and thickness $\dee{y}\text{,}$ so its volume is $\frac{\pi}{y^2}\dee{y}\text{.}$ The base of the solid is at height $y=1\text{,}$ and its top is at height $y=\frac{1}{a}\text{.}$ So, the volume of the entire solid is:

\begin{gather*} \int_1^{1/a} \frac{\pi}{y^2}\dee{y} = \left[-\frac{\pi}{y}\right]_1^{1/a} = \pi (1-a) \end{gather*}

If we imagine sliding $a$ closer and closer to 0, the volume increases, getting closer and closer to $\pi$ units, but never quite reaching it.

So, the statement is false. For example, if we set $M=4\text{,}$ no matter which $a$ we choose our solid has volume strictly less than $M\text{.}$

Remark: we've seen before that $\int_0^1 \frac{1}{x}\,\dee{x}$ diverges. If we imagine the solid that would result from choosing $a=0\text{,}$ it would have a scant volume of $\pi$ cubic units, but a silhouette (side view) of infinite area.

Exercise18(*)
Hint

Review Example 1.12.8. Remember the antiderivative of $\frac{1}{x}$ looks very different from the antiderivative of other powers of $x\text{.}$

$q=\frac{1}{5}$

Solution

Our goal is to decide when this integral diverges, and where it converges. We will leave $q$ as a variable, and antidifferentiate. In order to antidifferentiate without knowing $q\text{,}$ we'll need different cases. The integrand is $x^{-5q}\text{,}$ so when $-5q \neq -1\text{,}$ we use the power rule (that is, $\int x^n \dee{x} = \frac{x^{n+1}}{n+1}$) to antidifferentiate. Note $x^{(-5q)+1} = x^{1-5q} = \dfrac{1}{x^{5q-1}}\text{.}$

\begin{align*} \int_1^t \frac1{x^{5q}}\,\dee{x} &= \begin{cases} \left[\frac{x^{1-5q}}{1-5q} \right]_1^t \text{ with }1-5q \gt 0 &\text{if } q \lt \frac15 \\ \left[ \log x\vphantom{\frac12}\right]_1^{t}\ &\text{if } q = \frac15 \\ \left[\frac{1}{(1-5q)x^{5q-1}} \right]_1^{t}\text{ with }5q-1 \gt 0 &\text{if } q \gt \frac15 \end{cases}\\ &=\begin{cases} \frac1{1-5q}(t^{1-5q}-1) \text{ with }1-5q \gt 0 &\text{if } q \lt \frac15 \\ \log t &\text{if } q = \frac15 \\ \frac1{5q-1}(1-\frac1{t^{5q-1}})\text{ with }5q-1 \gt 0 &\text{if } q \gt \frac15. \end{cases} \end{align*}

Therefore,

\begin{align*} \int_1^\infty \frac1{x^{5q}}\,\dee{x} = \displaystyle\lim_{t\to\infty} \left( \int_1^t \frac1{x^{5q}}\,\dee{x} \right) = \begin{cases} \frac1{1-5q}\left(\displaystyle\lim_{t\to\infty} t^{1-5q}-1\right) = \infty &\text{if } q \lt \frac15 \\ \displaystyle\lim_{t\to\infty}\log t = \infty &\text{if } q = \frac15 \\ \frac1{5q-1}\left(1-\displaystyle\lim_{t\to\infty}\frac1{t^{5q-1}}\right) = \frac1{5q-1} &\text{if } q \gt \frac15. \end{cases} \end{align*}

The first two cases are divergent, and so the largest such value is $q=\frac{1}{5}\text{.}$ (Alternatively, we might recognize this as a “$p$-integral” with $p=5q\text{,}$ and recall that the $p$-integral diverges precisely when $p\le1\text{.}$)

Exercise19
Hint

Compare to Example 1.12.14 in the text. You can antidifferentiate with a $u$-substitution.

$p \gt 1$

Solution

This integrand is a nice candidate for the substitution $u=x^2+1\text{,}$ $\frac{1}{2}\dee{u} = x\dee{x}\text{.}$ Remember when we use substitution on a definite integral, we also need to adjust the limits of integration.

\begin{align*} \int_0^\infty \dfrac{x}{(x^2+1)^p} \dee{x} &= \lim_{t \to \infty}\int_0^t \frac{x}{(x^2+1)^p}\dee{x}\\ &=\lim_{t \to \infty}\frac{1}{2}\int_1^{t^2+1} \dfrac{1}{u^p} \dee{u}\\ &=\lim_{t \to \infty}\frac{1}{2}\int_1^{t^2+1}u^{-p} \dee{u}\\ &=\begin{cases} \frac12\displaystyle\lim_{t \to \infty}\left[\frac{u^{1-p}}{1-p}\right]_1^{t^2+1}&\text{ if }p \neq 1\\ \frac12\displaystyle\lim_{t \to \infty}\Big[\log|u|\Big]_1^{t^2+1}&\text{ if }p=1 \end{cases}\\ &=\begin{cases} \frac12\displaystyle\lim_{t \to \infty}\frac{1}{1-p}\left[(t^2+1)^{1-p}-1\right]&\text{ if }p \neq 1\\ \frac12\displaystyle\lim_{t \to \infty}\Big[\log(t^2+1)\Big]=\infty&\text{ if }p=1 \end{cases}\\ \end{align*}

At this point, we can see that the integral diverges when $p=1\text{.}$ When $p \neq 1\text{,}$ we have the limit

\begin{align*} \displaystyle\lim_{t \to \infty}\frac{1/2}{1-p}\left[(t^2+1)^{1-p}-1\right]&=\frac{1/2}{1-p}\left[\lim_{t \to \infty} (t^2+1)^{1-p}\right] - \frac{1/2}{1-p} \end{align*}

Since $t^2+1 \to \infty\text{,}$ this limit converges exactly when the exponent $1-p$ is negative; that is, it converges when $p \gt 1\text{,}$ and diverges when $p \lt 1\text{.}$

So, the integral in the question converges when $p \gt 1\text{.}$

Exercise20
Hint

To evaluate the integral, you can factor the denominator. \\ Recall $\displaystyle\lim_{x \to \infty}\arctan x = \frac{\pi}{2}\text{.}$ For the other limits, use logarithm rules, and beware of indeterminate forms.

$\dfrac{\log 3-\pi}{4} + \dfrac{1}{2}\arctan 2$

Solution

• First, we notice there is only one “source of impropriety:” the domain of integration is infinite. (The integrand has a singularity at $t=1\text{,}$ but this is not in the domain of integration, so it's not a problem for us.)
• We should try to get some intuition about whether the integral converges or diverges. When $t \to \infty\text{,}$ notice the integrand “looks like” the function $\frac{1}{t^4}\text{.}$ We know $\int_1^\infty \frac{1}{t^4} \dee{t}$ converges, because it's a $p$-integral with $p=4 \gt 1$ (see Example 1.12.8). So, our integral probably converges as well. If we were only asked show it converges, we could use a comparison test, but we're asked more than that.
• Since we guess the integral converges, we'll need to evaluate it. The integrand is a rational function, and there's no obvious substitution, so we use partial fractions.
\begin{align*} \frac{1}{t^4-1} &= \frac{1}{(t^2+1)(t^2-1)}= \frac{1}{(t^2+1)(t+1)(t-1)} = \frac{At+B}{t^2+1} + \frac{C}{t+1}+\frac{D}{t-1}\\ \end{align*}

Multiply by the original denominator.

\begin{align*} 1&=(At+B)(t+1)(t-1) + C(t^2+1)(t-1) +D(t^2+1)(t+1)\tag{$*$}\\ \end{align*}

Set $t=1\text{.}$

Set $t=-1\text{.}$

Simplify ($*$) using $D=\frac{1}{4}$ and $C=-\frac{1}{4}\text{.}$

\begin{align*} 1&=(At+B)(t+1)(t-1) \textcolor{red}{-\frac{1}{4}}(t^2+1)(t-1) +\textcolor{red}{\frac{1}{4}}(t^2+1)(t+1)\\ &=(At+B)(t+1)(t-1) + \frac{1}{2}(t^2+1)\\ &=At^3+\left(B+\frac{1}{2}\right)t^2-At+\left(\frac{1}{2}-B\right)\\ \end{align*}

By matching up coefficients of corresponding powers of $t\text{,}$ we find $A=0$ and $B=-\frac{1}{2}$.

\begin{align*} \int_2^\infty \frac{1}{t^4-1} \dee{t}&=\int_2^\infty \left( \frac{-1/2}{t^2+1} - \frac{1/4}{t+1}+\frac{1/4}{t-1}\right) \dee{t}\\ &=\lim_{R \to \infty}\int_2^R \left( \frac{-1/2}{t^2+1} - \frac{1/4}{t+1}+\frac{1/4}{t-1}\right) \dee{t}\\ &=\lim_{R \to \infty} \left[-\frac{1}{2}\arctan t - \frac{1}{4}\log|t+1|+\frac{1}{4}\log|{t-1}|\right]_2^R\\ &=\lim_{R \to \infty} \left[-\frac{1}{2}\arctan t + \frac{1}{4}\log\left| \frac{t-1}{t+1} \right|\right]_2^R\\ &=\lim_{R \to \infty} \left(-\frac{1}{2}\arctan R +\frac12\arctan 2 + \frac{1}{4}\log\left| \frac{R-1}{R+1} \right|-\frac{1}{4}\log\left| \frac{2-1}{2+1} \right|\right)\\ \end{align*}

We can use l'Hôpital's rule to see $\displaystyle\lim_{R \to \infty}\frac{R-1}{R+1}=1\text{.}$ Also note $-\log (1/3) = \log 3\text{.}$

\begin{align*} &=-\frac{1}{2}\left(\frac{\pi}{2}\right) +\frac{1}{2}\arctan 2 + \frac{1}{4}\log1+\frac{1}{4}\log3\\ &=\frac{\log 3-\pi}{4} + \frac{1}{2}\arctan 2 \end{align*}
Exercise21
Hint

Break up the integral. The absolute values give you a nice even function, so you can replace $|x-a|$ with $x-a$ if you're careful about the limits of integration.

The integral converges.

Solution

There are three singularities in the integrand: $x=0\text{,}$ $x=1\text{,}$ and $x=2\text{.}$ We'll need to break up the integral at each of these places.

\begin{align*} &\int_{-5}^5 \left(\frac{1}{\sqrt{|x|}} + \frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-2|}}\right)\dee{x}\\ = &\int_{-5}^0 \left(\frac{1}{\sqrt{|x|}} + \frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-2|}}\right)\dee{x}+ \int_{0}^1 \left(\frac{1}{\sqrt{|x|}} + \frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-2|}}\right)\dee{x}\\ +& \int_{1}^2 \left(\frac{1}{\sqrt{|x|}} + \frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-2|}}\right)\dee{x}+ \int_{2}^5 \left(\frac{1}{\sqrt{|x|}} + \frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-2|}}\right)\dee{x} \end{align*}

This looks rather unfortunate. Let's think again. If all of the integrals below converge, then we can write:

\begin{align*} \int_{-5}^5 \left(\frac{1}{\sqrt{|x|}} + \frac{1}{\sqrt{|x-1|}}+\frac{1}{\sqrt{|x-2|}}\right)\dee{x}&= \int_{-5}^5 \frac{1}{\sqrt{|x|}} \dee{x}+ \int_{-5}^5 \frac{1}{\sqrt{|x-1|}}\dee{x}+ \int_{-5}^5\frac{1}{\sqrt{|x-2|}}\dee{x}\\ \end{align*}

That looks a lot better. Also, we have a good reason to guess these integrals converge--they look like $p$-integrals with $p=\frac{1}{2}\text{.}$ Let's take a closer look at each one.

\begin{align*} \int_{-5}^5 \frac{1}{\sqrt {|x|}}\dee{x}&=\int_{-5}^0 \frac{1}{\sqrt {|x|}}\dee{x}+ \int_{0}^5 \frac{1}{\sqrt {|x|}}\dee{x}\\ &=2\int_{0}^5 \frac{1}{\sqrt {|x|}}\dee{x} \qquad\text{(even function)}\\ &=2\int_{0}^5 \frac{1}{\sqrt {x}}\dee{x}\\ \end{align*}

This is a $p$-integral, with $p=\frac{1}{2}\text{.}$ By Example 1.12.9 (and Theorem 1.12.20, since the upper limit of integration is not 1), it converges. The other two pieces behave similarly.

\begin{align*} \int_{-5}^5 \frac{1}{\sqrt{|x-1|}}\dee{x}&= \int_{-5}^1 \frac{1}{\sqrt{|x-1|}}\dee{x}+ \int_{1}^5 \frac{1}{\sqrt{|x-1|}}\dee{x}\\ \end{align*}

Use $u=x-1\text{,}$ $\dee{u}=\dee{x}$

\begin{align*} &= \int_{-6}^0 \frac{1}{\sqrt{|u|}}\dee{u}+ \int_{0}^4 \frac{1}{\sqrt{|u|}}\dee{x}\\ &= \int_{0}^6 \frac{1}{\sqrt{u}}\dee{u}+ \int_{0}^4 \frac{1}{\sqrt{u}}\dee{x}\\ \end{align*}

Since our function is even, we use the reasoning of Example 1.2.10 in the text to consider the area under the curve when $x \ge 0\text{,}$ rather than when $x\leq 0\text{.}$ Again, these are $p$-integrals with $p = \frac{1}{2}\text{,}$ so they both converge. Finally:

\begin{align*} \int_{-5}^5\frac{1}{\sqrt{|x-2|}}\dee{x}&= \int_{-5}^2\frac{1}{\sqrt{|x-2|}}\dee{x}+ \int_{2}^5\frac{1}{\sqrt{|x-2|}}\dee{x}\\ \end{align*}

Use $u=x-2\text{,}$ $\dee{u}=\dee{x}\text{.}$

\begin{align*} &= \int_{-7}^0\frac{1}{\sqrt{|u|}}\dee{u}+ \int_{0}^3\frac{1}{\sqrt{|u|}}\dee{u}\\ &= \int_{0}^7\frac{1}{\sqrt{u}}\dee{u}+ \int_{0}^3\frac{1}{\sqrt{u}}\dee{u} \end{align*}

Since $p = \frac{1}{2}\text{,}$ so they both converge. We conclude our original integral, as the sum of convergent integrals, converges.

Exercise22
Hint

Use integration by parts twice to find the antiderivative of $e^{-x}\sin x\text{,}$ as in Example 1.7.10. Be careful with your signs--it's easy to make a mistake with all those negatives.

If you're having a hard time taking the limit at the end, review the Squeeze Theorem (see the CLP-I text).

$\dfrac{1}{2}$

Solution

We can use integration by parts twice to find the antiderivative of $e^{-x}\sin x\text{,}$ as in Example 1.7.10. To keep our work a little simpler, we'll find the antiderivative first, then take the limit.

Let $u=e^{-x}\text{,}$ $\dee{v}=\sin x \dee{x}\text{,}$ so $\dee{u}=-e^{-x} \dee{x}$ and $v=-\cos x\text{.}$

\begin{align*} \int e^{-x}\sin x \dee{x}&=-e^{-x}\cos x - \int e^{-x}\cos x \dee{x}\\ \end{align*}

Now let $u=e^{-x}\text{,}$ $\dee{v}=\cos x \dee{x}\text{,}$ so $\dee{u}=-e^{-x} \dee{x}$ and $v=\sin x\text{.}$

\begin{align*} &=-e^{-x}\cos x-\left[e^{-x}\sin x + \int e^{-x}\sin x \dee{x} \right]\\ &=-e^{-x}\cos x-e^{-x}\sin x - \int e^{-x}\sin x \dee{x} \\ \end{align*}

All together, we found

\begin{align*} \color{red}{\int e^{-x}\sin x \dee{x}} &=-e^{-x}\cos x-e^{-x}\sin x -\color{red}{\int e^{-x}\sin x \dee{x}} +C\\ \color{red}{2\int e^{-x}\sin x \dee{x}} &=-e^{-x}\cos x-e^{-x}\sin x +C\\ \int e^{-x}\sin x \dee{x}&=-\frac{1}{2e^x}(\cos x+\sin x) +C\\ \end{align*}

(Remember, since $C$ is an arbitrary constant, we can rename $\frac{C}{2}$ to simply $C\text{.}$) Now we can evaluate our improper integral.

\begin{align*} \int_0^\infty e^{-x}\sin x \dee{x}&= \lim_{b \to \infty}\int_0^b e^{-x}\sin x \dee{x}\\ &= \lim_{b \to \infty}\left[-\frac{1}{2e^x}(\cos x+\sin x) \right]_0^b\\ &= \lim_{b \to \infty}\left(\frac{1}{2}-\frac{1}{2e^b}(\cos b+\sin b) \right)\\ \end{align*}

To find the limit, we use the Squeeze Theorem (see the CLP-I text). Since $|\sin b|,|\cos b| \leq 1$ for any $b\text{,}$ we can use the fact that $-2 \le \cos b + \sin x \le 2$ for any $b\text{.}$

\begin{align*} &\qquad\frac{-2}{2e^b} \leq \frac{1}{2e^b}(\cos b + \sin b) \leq \frac{2}{2e^b}\\ &\qquad\lim_{b \to \infty}\frac{-2}{2e^b} = 0 = \frac{2}{2e^b}\\ &\qquad\mbox{So, }\qquad\lim_{b \to \infty}\left[\frac{1}{2e^b}(\cos b + \sin b)\right]=0\\ \mbox{Therefore, }\qquad\frac{1}{2}&= \lim_{b \to \infty}\left(\frac{1}{2}-\frac{1}{2e^b}(\cos b+\sin b) \right) \end{align*}

That is, $\displaystyle\int_0^\infty e^{-x}\sin x \dee{x}= \frac{1}{2}\text{.}$

Exercise23(*)
Hint

What is the limit of the integrand when $x\rightarrow 0\text{?}$

The integral converges.

Solution

The integrand is positive everywhere. So either the integral converges to some finite number or it is infinite. There are two potential “sources of impropriety” — a possible singularity at $x=0$ and the fact that the domain of integration extends to $\infty\text{.}$ So, we split up the integral.

\begin{equation*} \int_0^\infty\frac{\sin^4 x}{x^2}\, \dee{x} =\int_0^1\frac{\sin^4 x}{x^2}\, \dee{x} + \int_1^\infty\frac{\sin^4 x}{x^2}\, \dee{x} \end{equation*}

Let's consider the first integral. By l'Hôpital's rule (see the CLP-I text),

\begin{equation*} \lim_{x\to 0} \frac{\sin x}{x} =\lim_{x\to 0} \frac{\cos x}{1} =\cos 0 =1 \end{equation*}

Consequently,

\begin{equation*} \lim_{x\to 0} \frac{\sin^4 x}{x^2} =\Big(\lim_{x\to 0} \sin^2 x \Big) \Big(\lim_{x\to 0} \frac{\sin x}{x} \Big) \Big(\lim_{x\to 0} \frac{\sin x}{x} \Big) =0\times 1\times 1 =0 \end{equation*}

and the first integral is not even improper.

Now for the second integral. Since $|\sin x|\le 1\text{,}$ we'll compare it to $\int_1^\infty \frac{1}{x^2}\text{.}$

• $\frac{\sin^4 x}{x^2}$ and $\frac{1}{x^2}$ are defined and continuous for every $x \geq 1$
• $0 \leq \frac{\sin^4 x}{x^2} \leq \frac{1^4}{x^2} = \frac{1}{x^2}$ for every $x \geq 1$
• $\int_1^\infty \frac{1}{x^2} \dee{x}$ converges by Example 1.12.8 (it's a $p$-type integral with $p \gt 1$)

By the comparison test, Theorem 1.12.17, $\displaystyle\int_1^\infty\frac{\sin^4 x}{x^2}\, \dee{x}$ converges.

Since $\displaystyle\int_0^1\frac{\sin^4 x}{x^2}\, \dee{x}$ and $\displaystyle\int_1^\infty\frac{\sin^4 x}{x^2}\, \dee{x}$ both converge, we conclude $\displaystyle\int_0^\infty\frac{\sin^4 x}{x^2}\, \dee{x}$ converges as well.

Exercise24
Hint

The only “source of impropriety” is the infinite domain of integration. Don't be afraid to be a little creative to make a comparison work.

The integral converges.

Solution

Since the denominator is positive for all $x \geq 0\text{,}$ the integrand is continuous over $[0, \infty)\text{.}$ So, the only “source of impropriety” is the infinite domain of integration.

• Solution 1: Let's try to use a direct comparison. Note $\dfrac{x}{e^x + \sqrt{x}} \geq 0$ whenever $x \geq 0\text{.}$ Also note that, for large values of $x\text{,}$ $e^x$ is much larger than $\sqrt{x}\text{.}$ That leads us to consider the following inequalty:

\begin{equation*} 0\leq \dfrac{x}{e^x + \sqrt{x}} \leq \dfrac{x}{e^x } \end{equation*}

If $\int_0^\infty \frac{x}{e^x} \dee{x}$ converges, we're in business. Let's figure it out. The integrand looks like a candidate for integration by parts: take $u=x\text{,}$ $\dee{v} = e^{-x} \dee{x}\text{,}$ so $\dee{u}=\dee{x}$ and $v=-e^{-x}\text{.}$

\begin{align*} \int_0^\infty \frac{x}{e^x} \dee{x}&=\lim_{b \to \infty}\int_0^b \frac{x}{e^x} \dee{x}= \lim_{b \to \infty}\left( \left[-\frac{x}{e^x}\right]_0^b + \int_0^b e^{-x} \dee{x} \right)\\ &= \lim_{b \to \infty}\left(-\frac{b}{e^b} +\left[-e^{-x}\right]_0^b \right)= \lim_{b \to \infty}\left(-\frac{b}{e^b} -\frac{1}{e^b}+1 \right)\\ &= \lim_{b \to \infty}\bigg(1-\underbrace{\frac{b+1}{e^b}}_{\atp{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}} \bigg)=\lim_{b \to \infty}\bigg(1-\frac{1}{e^b} \bigg)=1 \end{align*}

Using l'Hôpital's rule, we see $\int_0^\infty \frac{x}{e^x} \dee{x}$ converges. All together:

• $\frac{x}{e^x}$ and $\frac{x}{e^x+\sqrt{x}}$ are defined and continuous for all $x \geq 0\text{,}$
• $\left|\frac{x}{e^x+\sqrt{x}}\right|\leq \frac{x}{e^x}\text{,}$ and
• $\int_0^\infty \frac{x}{e^x} \dee{x}$ converges.

So, by Theorem 1.12.17, our integral $\displaystyle\int_0^\infty \frac{x}{e^x+\sqrt{x}} \dee{x}$ converges.

• Solution 2: Let's try to use a different direct comparison from Solution 1, and avoid integration by parts. We'd like to compare to something like $\dfrac{1}{e^x}\text{,}$ but the inequality goes the wrong way. So, we make a slight modification: we consider $2e^{-x/2}\text{.}$ To that end, we claim $x \lt 2e^{x/2}$ for all $x \geq 0\text{.}$ We can prove this by noting the following two facts:

• $0 \lt 2=2e^{0/2}\text{,}$ and
• $\diff{}{x}\{x\} = 1 \le e^{x/2} = \diff{}{x}\{2e^{x/2}\}\text{.}$

So, when $x=0\text{,}$ $x \lt 2e^{x/2}\text{,}$ and then as $x$ increases, $2e^{x/2}$ grows faster than $x\text{.}$

Now we can make the following comparison:

\begin{align*} 0\leq \dfrac{x}{e^x + \sqrt{x}} &\leq \dfrac{x}{e^x } \lt \frac{2e^{x/2}}{e^x} = \frac{2}{e^{x/2}} \\ \end{align*}

We have a hunch that $\int_0^\infty \frac{2}{e^{x/2}} \dee{x}$ converges, just like $\int_0^\infty \frac{1}{e^{x}} \dee{x}\text{.}$ This is easy enough to prove. We can guess an antiderivative, or use the substitution $u=x/2\text{.}$

\begin{align*} \int_0^\infty \frac{2}{e^{x/2}} \dee{x}&=\lim_{R \to \infty} \int_0^R \frac{2}{e^{x/2}} \dee{x} =\lim_{R \to \infty} \left[- \frac{4}{e^{x/2}}\right]_0^R\\ &=\lim_{R \to \infty} \left[\frac{4}{e^{0}}- \frac{4}{e^{R/2}}\right]_0^R=4 \end{align*}

Now we know:

• $0 \leq \frac{x}{e^x+\sqrt{x}} \leq \frac{2}{e^{x/2}}\text{,}$ and
• $\int_0^\infty \frac{2}{e^{x/2}} \dee{x}$ converges.
• Furthermore, $\frac{x}{e^x+\sqrt{x}}$ and $\frac{2}{e^{x/2}}$ are defined and continuous for all $x \geq 0\text{.}$

By the comparison test (Theorem 1.12.17), we conclude the integral converges.

• Solution 3: Let's use the limiting comparison test (Theorem 1.12.22). We have a hunch that our integral behaves similarly to $\int_0^\infty \frac{1}{e^x}\,\dee{x}\text{,}$ which converges (see Example 1.12.18). Unfortunately, if we choose $g(x) = \frac{1}{e^x}$ (and, of course, $f(x) = \frac{x}{e^x+\sqrt{x}}$), then

\begin{equation*} \lim_{x \to \infty}\frac{f(x)}{g(x)} = \lim_{x \to \infty}\frac{x}{e^x+\sqrt{x}}\cdot e^x = \lim_{x \to \infty}\frac{x}{1+\underbrace{\tfrac{\sqrt{x}}{e^x}}_{\to 0}} = \infty \end{equation*}

That is, the limit does not exist, so the limiting comparison test does not apply. (To find $\lim\limits_{x \to \infty}\frac{\sqrt{x}}{e^x}\text{,}$ you can use l'Hôpital's rule.)

This setback encourages us to try a slightly different angle. If $g(x)$ gave larger values, then we could decrease $\frac{f(x)}{g(x)}\text{.}$ So, let's try $g(x) = \frac{1}{e^{x/2}} = e^{-x/2}\text{.}$ Now,

\begin{align*} \lim_{x \to \infty}\frac{f(x)}{g(x)} &= \lim_{x \to \infty}\frac{x}{e^x+\sqrt{x}}\div \frac{1}{e^{x/2}}= \lim_{x \to \infty}\frac{x}{e^{x/2}+\frac{\sqrt{x}}{e^{x/2}}} \end{align*}

Hmm... this looks hard. Instead of dealing with it directly, let's use the squeeze theorem (see CLP-I notes).

\begin{align*} 0 &&\leq&& \frac{x}{e^{x/2}+\frac{\sqrt{x}}{e^{x/2 }}}&&\leq&& \frac{x}{e^{x/2}} \end{align*}

Using l'Hôpital's rule,

\begin{equation*} \lim_{x \to \infty} \underbrace{\frac{x}{e^{x/2}}}_{\atp{\mathrm{num}\to \infty}{\mathrm{den}\to\infty}} = \lim_{x \to \infty}\frac{1}{\frac{1}{2}e^{x/2}}=0 = \lim_{x \to \infty}0 \end{equation*}

So, by the squeeze theorem $\lim\limits_{x \to 0}\frac{\frac{x}{e^x+\sqrt{x}}}{\frac{1}{e^{x/2}}}=0\text{.}$ Since this limit exists, $\frac{1}{e^{x/2}}$ is a reasonable function to use in the limiting comparison test (provided its integral converges). So, we need to show that $\int_0^\infty \frac{1}{e^{x/2}}\,\dee{x}$ converges. This can be done by simply evaluating it:

\begin{equation*} \int_0^\infty \frac{1}{e^{x/2}}\,\dee{x} = \lim_{b \to \infty}\int_0^b {e^{-x/2}}\,\dee{x} = \lim_{b \to \infty} -\frac{1}{2}\big[e^{-x/2}\big]_0^b = \lim_{b \to \infty} -\frac{1}{2}\left[\frac{1}{e^{b/2}}-1\right] = \frac{1}{2} \end{equation*}

So, all together:

• The functions $\frac{x}{e^x+\sqrt{x}}$ and $\frac{1}{e^{x/2}}$ are defined and continuous for all $x \geq 0\text{,}$ and $\frac{1}{e^{x/2}} \ge 0$ for all $x \ge 0\text{.}$
• $\int_0^\infty \frac{1}{e^{x/2}}\,\dee{x}$ converges.
• The limit $\lim\limits_{x \to \infty}\frac{\frac{x}{e^x}+\sqrt{x}}{\frac{1}{e^{x/2}}}$ exists (it's equal to 0).
• So, the limiting comparison test (Theorem 1.12.17) tells us that $\int_0^\infty \frac{x}{e^x+\sqrt{x}}\,\dee{x}$ converges as well.
Exercise25(*)
Hint

There are two things that contribute to your error: using $t$ as the upper bound instead of infinity, and using $n$ intervals for the approximation.

First, find a $t$ so that the error introduced by approximating $\int_0^\infty \frac{e^{-x}}{1+x}\dee{x}$ by $\int_0^t \frac{e^{-x}}{1+x}\dee{x}$ is at most $\frac{1}{2} 10^{-4}\text{.}$ Then, find your $n\text{.}$

$t=10$ and $n= 2042$ will do the job. There are many other correct answers.

Solution

There are two sources of error: the upper bound is $t\text{,}$ rather than infinity, and we're using an approximation with some finite number of intervals, $n\text{.}$ Our plan is to first find a value of $t$ that introduces an error of no more than $\frac{1}{2}10^{-4}\text{.}$ That is, we'll find a value of $t$ such that $\int_t^\infty \frac{e^{-x}}{1+x} \dee{x} \leq \frac{1}{2}10^{-4}\text{.}$ After that, we'll find a value of $n$ that approximates $\int_0^t \frac{e^{-x}}{x+1} \dee{x}$ to within $\frac{1}{2}10^{-4}\text{.}$ Then, all together, our error will be at most $\frac{1}{2}10^{-4} + \frac{1}{2}10^{-4} = 10^{-4}\text{,}$ as desired. (Note we could have broken up the error in another way—it didn't have to be $\frac{1}{2}10^{-4}$ and $\frac{1}{2}10^{-4}\text{.}$ This will give us one of many possible answers.)

Let's find a $t$ such that $\int_t^\infty \frac{e^{-x}}{1+x}\dee{x}\le\frac{1}{2} 10^{-4}\text{.}$ For all $x\ge 0\text{,}$ $0 \lt \frac{e^{-x}}{1+x}\le e^{-x}\text{,}$ so

\begin{align*} \int_t^\infty \frac{e^{-x}}{1+x}\dee{x} \le \int_t^\infty e^{-x}\dee{x} =e^{-t}&\stackrel{(*)}{\le}\frac{1}{2} 10^{-4}\\ \hbox{where ($*$) is true if }\quad t&\ge -\log\Big(\frac{1}{2} 10^{-4}\Big)\approx 9.90 \end{align*}

Choose, for example, $t=10\text{.}$

Now it's time to decide how many intervals we're going to use to approximate $\displaystyle\int_0^t \frac{e^{-x}}{x+1} \dee{x}\text{.}$ Again, we want our error to be less than $\frac{1}{2}10^{-4}\text{.}$ To bound our error, we need to know the second derivative of $\frac{e^{-x}}{x+1}\text{.}$

\begin{align*} f(x)=\frac{e^{-x}}{1+x} \implies f'(x)=-\frac{e^{-x}}{1+x}-\frac{e^{-x}}{(1+x)^2} \implies f''(x)&=\frac{e^{-x}}{1+x}+2\frac{e^{-x}}{(1+x)^2} +2\frac{e^{-x}}{(1+x)^3} \end{align*}

Since $f''(x)$ is positive, and decreases as $x$ increases,

\begin{gather*} |f''(x)|\le f''(0) = 5 \implies |E_n|\le\frac{5(10-0)^3}{24n^2}=\frac{5000}{24n^2} =\frac{625}{3n^2} \end{gather*}

and $|E_n|\le \frac{1}{2} 10^{-4}$ if \begin{alignat*}{3} &&\frac{625}{3n^2}&\le \frac{1}{2} 10^{-4}\\ &\iff\qquad & n^2&\ge \frac{1250\times 10^4}{3}\\ &\iff\quad &n&\ge \sqrt{\frac{1.25\times 10^7}{3}} \approx 2041.2 \end{alignat*} So $t=10$ and $n= 2042$ will do the job. There are many other correct answers.

Exercise26
Hint

Look for a place to use Theorem 1.12.20.

Examples 1.2.10 and 1.2.11 have nice results about the area under an even/odd curve.

(a) The integral converges. \qquad (b) The interval converges.

Solution

1. Since $f(x)$ is odd, using the reasoning of Example 1.2.11, \begin{align*} \int_{-\infty}^{-1} f(x) \dee{x}&=\lim_{t \to \infty}\int_{-t}^{-1}f(x) \dee{x}=\lim_{t \to \infty}-\int_{1}^{t} f(x) \dee{x}=-\lim_{t \to \infty} \int_{1}^{t} f(x) \dee{x} \end{align*} Since $\displaystyle\int_1^\infty f(x) \dee{x}$ converges, the last limit above converges. Therefore, $\displaystyle\int_{-\infty\vphantom{\frac12}}^{-1} f(x) \dee{x}$ converges.
2. Since $f(x)$ is even, using the reasoning of Example 1.2.10, \begin{align*} \int_{-\infty}^{-1} f(x) \dee{x}&=\lim_{t \to \infty}\int_{-t}^{-1}f(x) \dee{x}=\lim_{t \to \infty}\int_{1}^{t} f(x) \dee{x}=\lim_{t \to \infty} \int_{1}^{t} f(x) \dee{x}\\ \end{align*}

Since $\displaystyle\int_1^\infty f(x) \dee{x}$ converges, the last limit above converges. Since $f(x)$ is continuous everywhere, by Theorem 1.12.20, $\displaystyle\int_{-1}^\infty f(x) \dee{x}$ converges (note the adjusted lower limit). Then, since

\begin{align*} \int_{-\infty}^\infty f(x) \dee{x}& = \int_{-\infty}^{-1} f(x) \dee{x} + \int_{-1}^\infty f(x) \dee{x} \end{align*} and both summands converge, our original integral converges as well.
Exercise27
Hint

$x$ should be a real number

false

Solution

Define $F(x)= \int_0^x \frac{1}{e^{t}}\dee{t}\text{.}$

\begin{align*} F(x)=\int_0^x \frac{1}{e^{t}}\dee{t}&=\left[-\frac{1}{e^{t}}\right]_0^x = \frac{1}{e^0}-\frac{1}{e^{x}} \lt \frac{1}{e^0}=1 \end{align*}

So, the statement is false: there is no $x$ such that $F(x)=1\text{.}$ For every real $x\text{,}$ $F(x) \lt \frac{1}{e^0}=1\text{.}$

We note here that $\displaystyle\lim_{x \to \infty} \int_0^x\frac{1}{e^{t}}\dee{t}=1\text{.}$ So, as $x$ grows larger, the gap between $F(x)$ and 1 grows infintesimally small. But there is no real value of $x$ where $F(x)$ is exactly equal to 1.

Exercises1.13.1Exercises

Exercise1
Hint

Each option in each column should be used exactly once.

Solution

(A) Note $\int f'(x)f(x)\ \dee{x} = \int u\ \dee{u}$ if we substitute $u=f(x)\text{.}$ This is the kind of integrand described in (I). It's quite possible that a $u$-substitution would work on the others, as well, but (I) is the most reliable kind of integrand for a $u$-substitution.\\ (B) A trigonometric substitution usually allows us to cancel out a square root containing a quadratic function, as in (IV).\\ (C) We can often antidifferentiate the product of a polynomial with an exponential function using integration by parts: see Examples 1.7.1, 1.7.6 in the CLP-II text. If we let $u$ be the polynomial function and $\dee{v}$ be the exponential, as long as we can antidifferentiate $\dee{v}\text{,}$ we can repeatedly apply integration by parts until the polynomial function goes away. So, we go with (II)\\ (D) We apply partial fractions to rational functions, (III).

Note: without knowing more about the functions, there's no guarantee that the methods we chose will be the best methods, or even that they will work (with the exception of (I)). With practice, you gain intuition about likely methods for different integrals. Luckily for you, there's lots of practice below.

Exercise2
Hint

The integrand is the product of sines and cosines. See how this was handled with a substitution in Section 1.8.1 of the CLP-II text.

After your substitution, you should have a polynomial expression in $u$—but it might take some simplification to get it into a form you can easily integrate.

$\dfrac{1}{5}-\dfrac{2}{7}+\dfrac{1}{9} =\dfrac{8}{315}$

Solution

The integrand is a product of powers of sine and cosine. Since cosine has an odd power, we want to substitute $u=\sin x\text{,}$ $\dee{u}=\cos x \dee{x}\text{.}$ Therefore, we should:

• reserve one cosine for the derivative of sine in our substitution, and
• change the rest of the cosines to sines using the identity $\sin^2x+\cos^2x=1\text{.}$
\begin{align*} \int_0^{\pi/2}\sin^4x\cos^5x\dee{x}&= \int_{0}^{\pi/2}\sin^4x(\cos^2x)^2\cos x\dee{x}\\ &=\int_{0}^{\pi/2}\sin^4x(1-\sin^2x)^2\underbrace{\cos x\dee{x}}_{du}\\ &= \int_{\sin(0)}^{\sin(\pi/2)}u^4(1-u^2)^2du\\ &= \int_{0}^{1}u^4(1-2u^2+u^4)du\\ &= \int_{0}^{1}(u^4-2u^6+u^8)du\\ &=\left[ \frac{1}{5}u^5-\frac{2}{7}u^7+\frac{1}{9}u^9 \right]_{u=0}^{u=1}\\ &= \left(\frac{1}{5}-\frac{2}{7}+\frac{1}{9}\right) - 0\\ &=\dfrac{8}{315} \end{align*}
Exercise3
Hint

We notice that the integrand has a quadratic polynomial under the square root. If that polynomial were a perfect square, we could get rid of the square root: try a trig substitution, as in Section 1.9 of the CLP-II text.

The identity $\sin(2\theta)=2\sin\theta\cos\theta$ might come in handy.

$\dfrac{3}{2\sqrt{5}} \arcsin\left(x\sqrt{\dfrac{5}{3}}\right) + \dfrac{x}{2}\sqrt{3-5x^2} +C$

Solution

We notice that there is a quadratic equation under the square root. If that equation were a perfect square, we could get rid of the square root: so we'll mould it into a perfect square using a trig substitution.

Our candidates will use one of the following identities:

\begin{equation*} 1-\sin^2\theta=\cos^2\theta \hspace{1cm} \tan^2\theta+1=\sec^2\theta \hspace{1cm} \sec^2\theta-1=\tan^2\theta \end{equation*}

We'll be substituting $x=$(something), so we notice that $3-5x^2$ has the general form of (constant)$-$(function), as does $1-\sin^2\theta\text{.}$ In order to get the constant right, we multiply through by three:

\begin{equation*} 3-3\sin^2\theta = 3\cos^2\theta \end{equation*}

Our goal is to get $3-5x^2=3-3\sin^2\theta\text{;}$ so we solve this equation for $x$ and decide on the substitution

\begin{equation*} x=\sqrt{\frac{3}{5}}\sin\theta,\hspace{1cm} \dee{x}=\sqrt{\frac{3}{5}}\cos\theta \dee{\theta} \end{equation*}

Now we evaluate our integral.

\begin{align*} \int\sqrt{3-5x^2}\dee{x}&=\int\sqrt{3-5\left( \sqrt{\frac{3}{5}}\sin\theta\right)^2}\sqrt{\frac{3}{5}}\cos\theta d \theta\\ &=\int \sqrt{3-3\sin^2\theta}\sqrt{3/5}\cos\theta \dee{\theta}\\ &=\int \sqrt{3\cos^2\theta}\sqrt{3/5}\cos\theta \dee{\theta}\\ &=\int \sqrt{3}\cos\theta\sqrt{3/5}\cos\theta \dee{\theta}\\ &=\frac{3}{\sqrt{5}}\int \cos^2\theta \dee{\theta}\\ &=\frac{3}{\sqrt{5}}\int \frac{1+\cos 2\theta}{2} \dee{\theta}\\ &=\frac{3}{2\sqrt{5}}\int (1+\cos2\theta) \dee{\theta}\\ &= \frac{3}{2\sqrt{5}}\left[ \theta+\frac{1}{2}\sin(2\theta) \right]+C\\ &=\frac{3}{2\sqrt{5}}\left[ \theta + \sin\theta\cos\theta \right]+C \end{align*}

From our substitution $x=\sqrt{3/5}\sin\theta\text{,}$ we glean $\sin\theta = x\sqrt{5/3}\text{,}$ and $\theta = \arcsin \left(x\sqrt{5/3} \right)\text{.}$ To figure out $\cos\theta\text{,}$ we draw a right triangle. Let $\theta$ be one angle, and since $\sin\theta = \dfrac{x\sqrt{5}}{\sqrt{3}}\text{,}$ we let the hypotenuse be $\sqrt{3}$ and the side opposite $\theta$ be $x\sqrt{5}\text{.}$ By Pythagorus, the missing side (adjacent to $\theta$) has length $\sqrt{3-5x^2}\text{.}$

Therefore, $\cos\theta = \frac{\mbox{adj}}{\mbox{hyp}} = \dfrac{\sqrt{3-5x^2}}{\sqrt{3}}\text{.}$ So our integral evaluates to:

\begin{align*} \frac{3}{2\sqrt{5}}\left[ \theta + \sin\theta\cos\theta \right]+C&= \frac{3}{2\sqrt{5}}\left[ \arcsin(x\sqrt{5/3}) + x\sqrt{5/3}\cdot \frac{\sqrt{3-5x^2}}{\sqrt{3}} \right]+C\\ &= \frac{3}{2\sqrt{5}} \arcsin(x\sqrt{5/3}) + \frac{x}{2}\cdot {\sqrt{3-5x^2}} +C \end{align*}
Exercise4
Hint

Notice the integral is improper. When you compute the limit, l'H\^{o}pital's rule might help.

If you're struggling to think of how to antidifferentiate, try writing $\dfrac{x-1}{e^x} = (x-1)e^{-x}\text{.}$

0

Solution

First, we note the integral is improper. So, we'll need to replace the top bound with a variable, and take a limit. Second, we're going to have to antidifferentiate. The integrand is the product of an exponential function, $e^{-x}\text{,}$ with a polynomial function, $x-1\text{,}$ so we use integration by parts with $u=x-1\text{,}$ $\dee{v}=e^{-x}\dee{u}\text{,}$ $\dee{u}=\dee{x}\text{,}$ and $v = -e^{-x}\text{.}$

\begin{align*} \int \dfrac{x-1}{e^x} \dee{x}&=-(x-1)e^{-x} + \int e^{-x} \dee{x}\\ &=-(x-1)e^{-x} -e^{-x}+C =-xe^{-x}+C\\ \text{So,}\qquad\int_0^\infty \dfrac{x-1}{e^x} \dee{x} & = \lim_{b \to \infty}\int_0^b \dfrac{x-1}{e^x} \dee{x}\\ & = \lim_{b \to \infty}\left[-\frac{x}{e^x}\right]_0^b = \lim_{b \to \infty}\bigg[-\underbrace{\frac{b}{e^b}}_{\atp{\mathrm{num}\to\infty}{\mathrm{den}\to\infty}}\bigg]\\ & \stackrel{(*)}{=} \lim_{b \to \infty}\frac{1}{e^b}=0 \end{align*}

(In the equality marked ($*$), we used l'H\^{o}pital's rule.)

So, $\displaystyle\int_0^\infty \dfrac{x-1}{e^x}\dee{x}=0\text{.}$

Remark: this shows that, interestingly, $\displaystyle\int_0^\infty \dfrac{x}{e^x}\dee{x}=\displaystyle\int_0^\infty \dfrac{1}{e^x}\dee{x}\text{.}$

Exercise5
Hint

Which method usually works for rational functions (the quotient of two polynomials)?

$\log\left|\dfrac{x+1}{3x+1}\right|+C$

Solution

• Solution 1: Notice the denominator factors as $(x+1)(3x+1)\text{.}$ Since the integrand is a rational function (the quotient of two polynomials), we can use partial fraction decomposition.

\begin{align*} \frac{-2}{3x^2+4x+1}&=\frac{-2}{(x+1)(3x+1)}\\ &=\frac{A}{x+1}+\frac{B}{3x+1}\\ &=\frac{A(3x+1)+B(x+1)}{(x+1)(3x+1)}\\ &=\frac{(3A+B)x+(A+B)}{(x+1)(3x+1)}\\ \end{align*}

So:

\begin{align*} -2&=(3A+B)x+(A+B)\\ 0&=3A+B \mbox{ and }-2=A+B\\ B&=-3A \mbox{ and hence } -2=A+(-3A)\\ {\color{red}A}&{\textcolor{red}{=1}} \mbox{ so then } {\color{red}B=-3}\\ \end{align*}

So now:

\begin{align*} \frac{-2}{3x^2+4x+1}&=\frac{1}{x+1}-\frac{3}{3x+1}\\ \int \frac{-2}{3x^2+4x+1}\dee{x}&=\int\left(\frac{1}{x+1}-\frac{3}{3x+1}\right)\dee{x}\\ &=\log|x+1|-\log|3x+1|+C\\ &=\log\left|\dfrac{x+1}{3x+1}\right|+C \end{align*}
• Solution 2: The previous solution is probably the nicest. However, for the foolhardy or the brave, this integral can also be evaluated using trigonometric substitution.

We start by completing the square on the denominator.

\begin{align*} 3x^2+4x+1 &= 3\left(x^2+\frac{4}{3}x+\frac{1}{3}\right)\\ &=3\left(x^2+2\cdot\frac{2}{3}x+\frac{4}{9}-\frac{4}{9}+\frac{1}{3}\right)\\ &=3\left(\left(x+\frac{2}{3}\right)^2-\frac{4}{9}+\frac{3}{9}\right)\\ &=3\left(\left(x+\frac{2}{3}\right)^2-\frac{1}{9}\right)\\ &=3\left(x+\frac{2}{3}\right)^2-\frac{1}{3}\\ \end{align*}

This has the form of a function minus a constant, which matches the trigonometric identity $\sec^2\theta - 1 = \tan^2\theta\text{.}$ Multiplying through by $\frac{1}{3}\text{,}$ we see we can use the identity $\frac{1}{3}\sec^2\theta - \frac{1}{3} = \frac{1}{3}\tan^2\theta\text{.}$ So, to get the substitution right, we want to choose a substitution that makes the following true:

\begin{align*} 3\left(x+\frac{2}{3}\right)^2-\frac{1}{3}&=\frac{1}{3}\sec^2\theta-\frac{1}{3}\\ 3\left(x+\frac{2}{3}\right)^2&=\frac{1}{3}\sec^2\theta\\ 9\left(x+\frac{2}{3}\right)^2&=\sec^2\theta\\ \color{red}3x+2&=\color{blue}\sec\theta\\ \end{align*}

And, accordingly:

\begin{align*} \color{red}3\dee{x}&=\color{blue}\sec\theta\tan\theta \dee{\theta}\\ \end{align*}

Now, let's simplify a little and use this substitution on our integral:

\begin{align*} \int\dfrac{-2}{3x^2+4x+1}\dee{x}&=\int\dfrac{-2}{3\left(x+\frac{2}{3}\right)^2-\frac{1}{3}}\dee{x}\\ &=\int\dfrac{-2}{9\left(x+\frac{2}{3}\right)^2-1}3\dee{x}\\ &=\int\dfrac{-2}{\left(\color{red}{3x+2}\right)^2-1}\textcolor{red}{3\dee{x}}\\ &=\int\dfrac{-2}{\left(\color{blue}{\sec\theta}\right)^2-1}\color{blue}{\sec\theta\tan\theta \dee{\theta}}\\ &=\int\dfrac{-2}{\tan^2\theta}\sec\theta\tan\theta \dee{\theta}\\ &=\int -2\dfrac{\sec\theta}{\tan\theta}\dee{\theta}\\ &=\int -2\dfrac{1}{\cos\theta}\cdot\dfrac{\cos\theta}{\sin\theta} \dee{\theta}\\ &=\int -2\dfrac{1}{\sin\theta} \dee{\theta}\\ &=\int -2\csc\theta \dee{\theta}\\ \end{align*}

Using the result of Example 1.8.20 in the CLP-II text, or a table of integrals:

\begin{align*} &=2\log\left|\csc\theta+\cot\theta \right|+C \end{align*}

Our final task is to translate this back from $\theta$ to $x\text{.}$ Recall we used the substitution $3x+2=\sec\theta\text{.}$ Using this information, and $\sec\theta = \dfrac{\mathrm{hypotenuse}}{\mathrm{adjacent}}\text{,}$ we can fill in two sides of a right triangle with angle $\theta\text{.}$ The Pythagorean theorem tells us the third side (opposite to $\theta$) has measure $\sqrt{(3x+2)^2-1}=\sqrt{9x^2-12x+3}\text{.}$

\begin{align*} 2\log\left|\csc\theta+\cot\theta \right|+C&= 2\log\left|\dfrac{3x+2}{\sqrt{9x^2+12x+3}}+\dfrac{1}{\sqrt{9x^2+12x+3}} \right|+C\\ &= 2\log\left|\dfrac{3x+3}{\sqrt{9x^2+12x+3}}\right|+C\\ &= \log\left|\dfrac{(3x+3)^2}{\sqrt{9x^2+12x+3}^2}\right|+C\\ &= \log\left|\dfrac{(3x+3)^2}{9x^2+12x+3}\right|+C\\ &= \log\left|\dfrac{9(x+1)^2}{3(3x+1)(x+1)}\right|+C\\ &= \log\left|\dfrac{3(x+1)^2}{(3x+1)(x+1)}\right|+C\\ &= \log\left|\dfrac{3(x+1)}{3x+1}\right|+C\\ &= \log\left|\dfrac{x+1}{3x+1}\right|+\log3+C\\ \end{align*}

Since $C$ is an arbitrary constant, we can write our final answer as

\begin{align*} & \log\left|\dfrac{x+1}{3x+1}\right|+C \end{align*}
Exercise6
Hint

It would be nice to replace logarithm with its derivative, $\dfrac{1}{x}\text{.}$

$\dfrac{8}{3}\log2-\dfrac{7}{9}$

Solution

We see that we have two functions multiplied, but they don't simplify nicely with each other. However, if we differentiate logarithm, and integrate $x^2\text{,}$ we'll get a polynomial. So, let's use integration by parts.

\begin{equation*} u=\log x \hspace{1cm} \dee{v}=x^2\dee{x} \end{equation*}

\begin{equation*} \dee{u}=(1/x)\dee{x} \hspace{1cm} v=x^3/3 \end{equation*}

First, let's antidifferentiate. We'll deal with the limits of integration later.

\begin{align*} \int x^2\log x \dee{x} &=\underbrace{(\log x)}_{u}\underbrace{(x^3/3)}_{v})-\int\underbrace{(x^3/3)}_{v}\underbrace{(1/x)\dee{x}}_{\dee{u}}\\ &=\frac{1}{3}x^3\log x - \frac{1}{3}\int x^2\dee{x}\\ &=\frac{1}{3}x^3\log x -\frac{1}{3}\cdot\frac{1}{3}x^3+C \\ &=\frac{1}{3}x^3\log x - \frac{1}{9}x^3+C\\ \end{align*}

We use the Fundamental Theorem of Calculus Part 2 to evaluate the definite integral.

\begin{align*} \int_1^2 x^3\log x \dee{x}&= \left[\frac{1}{3}x^3\log x - \frac{1}{9}x^3\right]_1^2\\ &=\left[\frac{1}{3}2^3\log 2 - \frac{1}{9}2^3 \right] - \left[\frac{1}{3}1^3\log 1 - \frac{1}{9}1^3 \right]\\ &=\frac{8\log 2}{3}-\frac{8}{9}-0+\frac{1}{9}\\ &=\frac{8}{3}\log2-\frac{7}{9} \end{align*}
Exercise7(*)
Hint

The integrand is a rational function, so it is possible to use partial fractions. But there is a much easier way!

$\dfrac{1}{2}\log\big|x^2-3\big| + C$

Solution

The derivative of the denominator shows up in the numerator, only differing by a constant, so we perform a substitution. Specifically, substitute $u=x^2-3\text{,}$ $\dee{u}=2x\,\dee{x}\text{.}$ This gives

\begin{gather*} \int\frac{x}{x^2-3}\dee{x} =\int\frac{\dee{u}/2}{u} =\frac{1}{2}\log|u| + C =\frac{1}{2}\log\big|x^2-3\big| + C \end{gather*}
Exercise8(*)
Hint

You should prepare your own personal internal list of integration techniques ordered from easiest to hardest. You should have associated to each technique your own personal list of signals that you use to decide when the technique is likely to be useful.

(a) $2$ \qquad (b) $\dfrac{2}{15}$ \qquad (c) $\dfrac{3e^4}{16}+\dfrac{1}{16}$

Solution

(a) Although a quadratic under a square root often suggests trigonometric substitution, in this case we have an easier substitution. Specifically, let $y=9+x^2\text{.}$ Then $\dee{y}=2x\dee{x},\ x\dee{x}=\frac{\dee{y}}{2},\ y(0)=9,$ and $y(4)=25\text{.}$

\begin{align*} \int_0^4\frac{x}{\sqrt{9+x^2}}\,\dee{x} &=\int_9^{25}\frac{1}{\sqrt{y}}\,\frac{\dee{y}}{2} =\frac{1}{2}\cdot\frac{\sqrt{y}}{1/2}\Big|_9^{25} =5-3= 2 \end{align*}

(b) The power of cosine is odd, so we can reserve one cosine for the differential and change the rest to sines. Substituting $y=\sin x,$ $\dee{y}=\cos x\text{,}$ $\dee{x},$ $y(0)=0,$ $y(\pi/2)=1,$ $\cos^2x=1-y^2\text{:}$

\begin{align*} \int_0^{\pi/2}\hskip-.2in\cos^3x\ \sin^2x\,\dee{x} &=\int_0^{\pi/2}\hskip-.2in\cos^2x\ \sin^2x\ \cos x\,\dee{x} =\int_0^1\!(1-y^2)y^2\,\dee{y} =\int_0^1(y^2-y^4)\,\dee{y} \\ &=\bigg[\frac{y^3}{3}-\frac{y^5}{5}\bigg]_0^1 =\frac{1}{3}-\frac{1}{5} \\ &= \frac{2}{15} \end{align*}

(c) The integrand is the product of two different kinds of functions, with no obvious substitution or simplification. If we differentiate $\log x\text{,}$ it will match better with the polynomial nature of the rest of the integrand. So, integrate by parts with $u(x)=\log x$ and $\dee{v}=x^3\,\dee{x}\text{,}$ then $\dee{u}=\frac{1}{x}\,\dee{x}$ and $v=x^4/4\text{.}$

\begin{align*} \int_1^{e}x^3\log x\,\dee{x} &=\frac{x^4}{4}\log x\bigg|_1^e -\int_1^{e}\frac{x^4}{4}\cdot\frac{1}{x}\,\dee{x} =\frac{e^4}{4}-\int_1^{e}\frac{x^3}{4}\,\dee{x} =\frac{e^4}{4}-\frac{x^4}{16}\bigg|_1^e \\ &=\frac{3e^4}{16}+\frac{1}{16} \end{align*}
Exercise9(*)
Hint

Despite both containing a trig function, the two integrals are easiest to evaluate using different methods.

(a) $1$ \qquad (b) $\dfrac{8}{15}$

Solution

(a) Integrate by parts with $u=x$ and $\dee{v}=\sin x\,\dee{x}$ so that $\dee{u}=\dee{x}$ and $v=-\cos x\text{.}$

\begin{align*} \int x\sin x\,\dee{x}&= -x\cos x-\int (-\cos x)\,\dee{x} =-x\cos x+\sin x+C\\ \mbox{So, }\quad\int_0^{\pi/2} x\sin x\,\dee{x}&=\Big[-x\cos x+\sin x\Big]_0^{\pi/2} =1 \end{align*}

(b) The power of cosine is odd, so we can reserve one cosine for $\dee{u}$ and change the rest into sines. Make the substitution $u=\sin x\text{,}$ $\dee{u}=\cos x\,\dee{x}\text{.}$

\begin{align*} \int_0^{\pi/2} \cos^5 x\,\dee{x} &=\int_0^{\pi/2} \big(1-\sin^2x\big)^2\cos x\,\dee{x} =\int_0^1\big(1-u^2\big)^2\,\dee{u} =\int_0^1\big(1-2u^2+u^4\big)\,\dee{u}\cr &=\left[u-\frac{2}{3}u^3+\frac{1}{5}u^5\right]_0^1 =1-\frac{2}{3}+\frac{1}{5} =\frac{8}{15} \end{align*}
Exercise10(*)
Hint

For the integral of secant, see See Section 1.8.3 or Example 1.10.5 in the

CLP-II text.

In (c), notice the denominator is not yet entirely factored.

(a) $e^2+1$ \qquad (b) $\log(\sqrt{2}+1)$ \qquad (c) $\log\frac{15}{13}\approx 0.1431$

Solution

(a) This is a classic integration-by-parts example. If we integrate $e^x\text{,}$ it doesn't change, and if we differentiate $x$ it becomes a constant. So, let $u=x$ and $\dee{v}=e^x\,\dee{x}\text{,}$ so that $\dee{u}=\dee{x}$ and $v=e^x\text{.}$

\begin{gather*} \int_0^2 xe^x\,\dee{x}=\Big[xe^x\Big]_0^2-\int_0^2 e^x\,\dee{x} =2e^2-\Big[e^x\Big]_0^2 =e^2+1 \end{gather*}

(b) We have a quadratic function underneath a square root. In the absence of an easier substitution, we can get rid of the square root with a trigonometric substitution. Substitute $x=\tan y\text{,}$ $\dee{x}=\sec^2 y\,\dee{y}\text{.}$ When $x=0\text{,}$ $\tan y=0$ so $y=0\text{.}$ When $x=1\text{,}$ $\tan y=1$ so $y=\frac{\pi}{4}\text{.}$ Also $\sqrt{1+x^2} =\sqrt{1+\tan^2y}=\sqrt{\sec^2 y}=\sec y\text{,}$ since $\sec y\ge 0$ for all $0\le y\le\frac{\pi}{4}\text{.}$

\begin{align*} \int_0^1\frac{1}{\sqrt{1+x^2}}\,\dee{x} &=\int_0^{\pi/4} \frac{\sec^2y\,\dee{y}}{\sec y} =\int_0^{\pi/4} \sec y\,\dee{y} =\Big[\log|\sec y+\tan y|\Big]_0^{\pi/4}\\ &=\log\left| \sec \frac{\pi}{4} + \tan\frac{\pi}{4}\right|-\log\left| \sec 0 + \tan 0\right|\\ &=\log\left| \sqrt{2} + 1\right|-\log\left| 1+ 0\right|=\log(\sqrt{2}+1)\\ \text{So,}\qquad\int_0^1\frac{1}{\sqrt{1+x^2}}\,\dee{x}&=\log(\sqrt{2}+1) \end{align*}

(c) The integral is a rational function. In the absence of an obvious substitution, we use partial fractions.

\begin{align*} \frac{4x}{(x^2-1)(x^2+1)} &=\frac{4x}{(x-1)(x+1)(x^2+1)} =\frac{a}{x-1}+\frac{b}{x+1}+\frac{cx+d}{x^2+1} \\ \end{align*}

Multiplying by the denominator,

\begin{align*} 4x&=a(x+1)(x^2+1)+b(x-1)(x^2+1)+(cx+d)(x-1)(x+1)\tag{*} \end{align*}

Setting $x=1$ gives $4a=4\text{,}$ so \textcolor{red}{$a=1$}. Setting $x=-1$ gives $-4b=-4\text{,}$ so \textcolor{red}{$b=1$}. Substituting in $a=b=1$ in ($*$) gives:

\begin{align*} 4x&=(x+1)(x^2+1)+(x-1)(x^2+1)+(cx+d)(x-1)(x+1)\cr 4x&= 2x(x^2+1)+(cx+d)(x-1)(x+1)\cr 4x-2x(x^2+1)&=(cx+d)(x-1)(x+1)\cr -2x(x^2-1)&=(cx+d)(x^2-1)\cr -2x&=cx+d\cr \color{red} c&\color{red}=-2,\ d=0\cr \end{align*}

So,

\begin{align*} \int_3^5\frac{4x}{(x^2-1)(x^2+1)}\,\dee{x} &=\int_3^5\Big(\frac{1}{x-1}+\frac{1}{x+1}-\frac{2x}{x^2+1}\Big)\,\dee{x} \\ &=\Big[\log|x-1|+\log|x+1|-\log(x^2+1)\Big]_3^5\\ &=\log 4+\log 6-\log 26-\log 2-\log 4+\log 10 \\ &=\log\frac{6\times 10}{26\times 2} =\log\frac{15}{13}\approx 0.1431 \end{align*}
Exercise11(*)
Hint

Part (a) can be done by inspection — use a little highschool geometry! Part (b) is reminiscent of the antiderivative of logarithm—how did we find that one out? Part (c) is an improper integral.

(a) $\dfrac{9}{4}\pi$ \qquad (b) $\log 2-2+\dfrac{\pi}{2}\approx 0.264$ \qquad (b) $2\log 2-\half\approx0.886$

Solution

(a) $\int_0^3\sqrt{9-x^2}\,\dee{x}$ is the area of the portion of the disk $x^2+y^2\le 9$ that lies in the first quadrant. It is $\frac{1}{4}\pi 3^3={\frac{9}{4}\pi}\,\text{.}$ Alternatively, you could also evaluate this integral using the substitution $x=3\sin y\text{,}$ $\dee{x}=3\cos y\,\dee{y}\text{.}$

\begin{align*} \int_0^3\sqrt{9-x^2}\,\dee{x} &=\int_0^{\pi/2}\sqrt{9-9\sin^2y}\ (3\cos y)\,\dee{y} =9\int_0^{\pi/2}\cos^2 y\,\dee{y} \\ &=\frac{9}{2}\int_0^{\pi/2}[1+\cos(2y)]\,\dee{y} =\frac{9}{2}{\Big[y+\frac{\sin(2y)}{2}\Big]}_0^{\pi/2} \\ &=\frac{9}{4}\pi \end{align*}

(b) It's not immediately obvious what to do with this one, but remember we found $\int \log x \dee{x}$ using integration by parts with $u=\log x$ and $\dee{v}=\dee{x}\text{.}$ Let's hope a similar trick works here. Integrate by parts, using $u=\log(1+x^2)$ and $\dee{v}=\dee{x}\text{,}$ so that $\dee{u}=\frac{2x}{1+x^2}\,\dee{x}\text{,}$ $v=x\text{.}$

\begin{align*} \int_0^1\log(1+x^2)\,\dee{x} &=\Big[x\log(1+x^2)\Big]_0^1-\int_0^1 x\frac{2x}{1+x^2}\,\dee{x} =\log 2-2\int_0^1 \frac{x^2}{1+x^2}\,\dee{x} \\ &=\log 2-2\int_0^1\Big(1- \frac{1}{1+x^2}\Big)\,\dee{x} =\log 2-2\big[x-\arctan x\big]_0^1 \\ &=\log 2-2+\frac{\pi}{2}\approx 0.264 \end{align*}

(c) The integrand is a rational function with no obvious substitution, so we use partial fractions.

\begin{align*} \frac{x}{(x-1)^2(x-2)} &=\frac{a}{(x-1)^2}+\frac{b}{x-1}+\frac{c}{x-2} =\frac{a(x-2)+b(x-1)(x-2)+c(x-1)^2}{(x-1)^2(x-2)} \\ \\ \end{align*}

Multiply by the denominator.

\begin{align*} x&=a(x-2)+b(x-1)(x-2)+c(x-1)^2 \end{align*}

Setting $x=1$ gives \textcolor{red}{$a=-1$}. Setting $x=2$ gives \textcolor{red}{$c=2$}. Substituting in $a=-1$ and $c=2$ gives

\begin{align*} b(x-1)(x-2)&=x+(x-2)-2(x-1)^2=-2x^2+6x-4=-2(x-1)(x-2) \\ \implies \color{red}b&\color{red}=-2 \end{align*}

Hence

\begin{align*} \int_3^\infty\frac{x}{(x-1)^2(x-2)}\,\dee{x} &=\lim_{M\rightarrow\infty} \int_3^M\left(-\frac{1}{(x-1)^2}-\frac{2}{x-1}+\frac{2}{x-2}\right)\,\dee{x} \\ &=\lim_{M\rightarrow\infty} \left[\frac{1}{x-1}-2\log|x-1|+2\log|x-2|\right]_3^M\\ &=\lim_{M\rightarrow\infty} \bigg[\frac{1}{x-1}+2\log\bigg|\frac{x-2}{x-1}\bigg|\bigg]_3^M\\ &=\lim_{M\rightarrow\infty} \bigg[\frac{1}{M-1}+2\log\bigg|\frac{M-2}{M-1}\bigg|\bigg] -\bigg[\frac{1}{3-1}+2\log\bigg|\frac{3-2}{3-1}\bigg|\bigg] \\ &=2\log 2-\frac{1}{2}\approx0.886 \end{align*}

since

\begin{align*} \lim_{M\rightarrow\infty}\log\frac{M-2}{M-1} &=\lim_{M\rightarrow\infty}\log\frac{1-2/M}{1-1/M} =\log 1=0\\ \text{and}\qquad \log\frac{1}{2}&=-\log2 \end{align*}
Exercise12
Hint

Use the substitution $u=\sin\theta\text{.}$