Subsection 1.3.1 The Fundamental Theorem of Calculus
We have spent quite a few pages (and lectures) talking about definite integrals, what they are (Definition
1.1.9), when they exist (Theorem
1.1.10), how to compute some special cases (Section
1.1.5), some ways to manipulate them (Theorem
1.2.1 and
1.2.3) and how to bound them (Theorem
1.2.13). Conspicuously missing from all of this has been a discussion of how to compute them in general. It is high time we rectified that.
The single most important tool used to evaluate integrals is called “the fundamental theorem of calculus”. Its grand name is justified — it links the two branches of calculus by connecting derivatives to integrals. In so doing it also tells us how to compute integrals. Very roughly speaking the derivative of an integral is the original function. This fact allows us to compute integrals using antiderivatives
1 . Of course “very rough” is not enough — let's be precise.
Theorem 1.3.1. Fundamental Theorem of Calculus.
Let \(a \lt b\) and let \(f(x)\) be a function which is defined and continuous on \([a,b]\text{.}\)
Part 1. Let
\(\ds F(x)=\int_a^x f(t)\dee{t}\) for any
\(x \in[a,b]\text{.}\) Then the function
\(F(x)\) is differentiable and further
\begin{align*}
F'(x) &=f(x)
\end{align*}
Part 2. Let
\(G(x)\) be any function which is defined and continuous on
\([a,b]\text{.}\) Further let
\(G(x)\) be differentiable with
\(G'(x)=f(x)\) for all
\(a \lt x \lt b\text{.}\) Then
\begin{align*}
\int_a^b f(x)\dee{x} &=G(b)-G(a) & \text{or equivalently} &&
\int_a^b G'(x)\dee{x} &=G(b)-G(a)
\end{align*}
Before we prove this theorem and look at a bunch of examples of its application, it is important that we recall one definition from differential calculus — antiderivatives. If \(F'(x) = f(x)\) on some interval, then \(F(x)\) is called an antiderivative of \(f(x)\) on that interval. So Part 2 of the fundamental theorem of calculus tells us how to evaluate the definite integral of \(f(x)\) in terms of any of its antiderivatives — if \(G(x)\) is any antiderivative of \(f(x)\) then
\begin{align*}
\int_a^b f(x) \dee x &= G(b)-G(a)
\end{align*}
The form \(\int_a^b G'(x)\,\dee{x} = G(b) - G(a)\) of the fundamental theorem relates the rate of change of \(G(x)\) over the interval \(a\le x\le b\) to the net change of \(G\) between \(x=a\) and \(x=b\text{.}\) For that reason, it is sometimes called the “net change theorem”.
We'll start with a simple example. Then we'll see why the fundamental theorem is true and then we'll do many more, and more involved, examples.
Example 1.3.2. A first example.
Consider the integral
\(\int_a^b x \dee{x}\) which we have explored previously in Example
1.2.6.
We do not give completely rigorous proofs of the two parts of the theorem — that is not really needed for this course. We just give the main ideas of the proofs so that you can understand why the theorem is true.
Part 1.
We wish to show that if
\begin{align*}
F(x) &= \int_a^x f(t) \dee{t} & \text{then}&&
F'(x) &= f(x)
\end{align*}
Assume that
\(F\) is the above integral and then consider
\(F'(x)\text{.}\) By definition
\begin{align*}
F'(x) &=\lim_{h\rightarrow 0} \frac{F(x+h)-F(x)}{h}
\end{align*}
To understand this limit, we interpret the terms
\(F(x), F(x+h)\) as signed areas. To simplify this further, let's only consider the case that
\(f\) is always nonnegative and that
\(h \gt 0\text{.}\) These restrictions are not hard to remove, but the proof ideas are a bit cleaner if we keep them in place. Then we have
\begin{align*}
F(x+h)&=\text{the area of the region $\big\{\ (t,y)\ \big|\ a\le t\le x+h,\
0\le y\le f(t)\ \big\}$}\\
F(x)&=\text{the area of the region $\big\{\ (t,y)\ \big|\ a\le t\le x,
\phantom{+h\ \,}\ 0\le y\le f(t)\ \big\}$}
\end{align*}
-
Then the numerator
\begin{gather*}
F(x+h)-F(x)=\text{the area of $\big\{\ (t,y)\ \big|\ x\le t\le x+h,\
0\le y\le f(t)\ \big\} $}
\end{gather*}
This is just the more darkly shaded region in the figure
We will be taking the limit
\(h\rightarrow 0\text{.}\) So suppose that
\(h\) is very small. Then, as
\(t\) runs from
\(x\) to
\(x+h\text{,}\) \(f(t)\) runs only over a very narrow range of values
2 , all close to
\(f(x)\text{.}\)
So the darkly shaded region is almost a rectangle of width \(h\) and height \(f(x)\) and so has an area which is very close to \(f(x)h\text{.}\) Thus \(\frac{F(x+h)-F(x)}{h}\) is very close to \(f(x)\text{.}\)
In the limit \(h\rightarrow 0\text{,}\) \(\frac{F(x+h)-F(x)}{h}\) becomes exactly \(f(x)\text{,}\) which is precisely what we want.
Part 2.
We want to show that \(\int_a^b f(t)\dee{t}=G(b)-G(a)\text{.}\) To do this we exploit the fact that the derivative of a constant is zero.
Let
\begin{align*}
H(x) &= \int_a^x f(t)\dee{t} -G(x)+G(a)
\end{align*}
Then the result we wish to prove is that
\(H(b)=0\text{.}\) We will do this by showing that
\(H(x)=0\) for all
\(x\) between
\(a\) and
\(b\text{.}\)
We first show that
\(H(x)\) is constant by computing its derivative:
\begin{align*}
H'(x) &= \diff{}{x}\int_a^x f(t)\dee{t} - \diff{}{x}\left( G(x) \right)+ \diff{}{x}\left( G(a) \right)\\
\end{align*}
Since \(G(a)\) is a constant, its derivative is \(0\) and by assumption the derivative of \(G(x)\) is just \(f(x)\text{,}\) so
\begin{align*}
&= \diff{}{x}\int_a^x f(t)\dee{t} - f(x)\\
\end{align*}
Now Part 1 of the theorem tells us that this derivative is just \(f(x)\text{,}\) so
\begin{align*}
&= f(x) - f(x) = 0
\end{align*}
Hence
\(H\) is constant.
To determine which constant we just compute
\(H(a)\text{:}\)
\begin{align*}
H(a) &= \int_a^a f(t)\dee{t} - G(a)+G(a)\\
&= \int_a^a f(t)\dee{t} & \text{by Theorem }\knowl{./knowl/thm_Intdomain.html}{\text{1.2.3}}\text{(a)}\\
&=0
\end{align*}
as required.
The simple example we did above (Example
1.3.2), demonstrates the application of part 2 of the fundamental theorem of calculus. Before we do more examples (and there will be many more over the coming sections) we should do some examples illustrating the use of part 1 of the fundamental theorem of calculus. Then we'll move on to part 2.
Example 1.3.3. \(\diff{}{x}\int_0^x t \dee{t}\).
Consider the integral
\(\int_0^x t\,\dee{t}\text{.}\) We know how to evaluate this — it is just Example
1.3.2 with
\(a = 0\text{,}\) \(b = x\text{.}\) So we have two ways to compute the derivative. We can evaluate the integral and then take the derivative, or we can apply Part 1 of the fundamental theorem. We'll do both, and check that the two answers are the same.
First, Example
1.3.2 gives
\begin{gather*}
F(x) = \int_0^x t\,\dee{t} =\frac{x^2}{2}
\end{gather*}
So of course \(F'(x) = x\text{.}\) Second, Part 1 of the fundamental theorem of calculus tells us that the derivative of \(F(x)\) is just the integrand. That is, Part 1 of the fundamental theorem of calculus also gives \(F'(x) = x\text{.}\)
In the previous example we were able to evaluate the integral explicitly, so we did not need the fundamental theorem to determine its derivative. Here is an example that really does require the use of the fundamental theorem.
Example 1.3.4. \(\diff{}{x}\int_0^x e^{-t^2}\dee{t}\).
We would like to find
\(\diff{}{x}\int_0^x e^{-t^2}\dee{t}\text{.}\) In the previous example, we were able to compute the corresponding derivative in two ways — we could explicitly compute the integral and then differentiate the result, or we could apply part 1 of the fundamental theorem of calculus. In this example we do not know the integral explicitly. Indeed it is not possible to express
3 the integral
\(\int_0^x e^{-t^2}\dee{t}\) as a finite combination of standard functions such as polynomials, exponentials, trigonometric functions and so on.
Despite this, we can find its derivative by just applying the first part of the fundamental theorem of calculus with \(f(t)=e^{-t^2}\) and \(a=0\text{.}\) That gives
\begin{align*}
\diff{}{x}\int_0^x e^{-t^2}\dee{t} &=\diff{}{x}\int_0^x f(t)\dee{t}\\
&=f(x) = e^{-x^2}
\end{align*}
Let us ratchet up the complexity of the previous example — we can make the limits of the integral more complicated functions. So consider the previous example with the upper limit \(x\) replaced by \(x^2\text{:}\)
Example 1.3.5. \(\diff{}{x}\int_0^{x^2} e^{-t^2}\dee{t}\).
Consider the integral \(\int_0^{x^2} e^{-t^2}\dee{t}\text{.}\) We would like to compute its derivative with respect to \(x\) using part 1 of the fundamental theorem of calculus.
The fundamental theorem tells us how to compute the derivative of functions of the form \(\int_a^x f(t)\dee{t}\) but the integral at hand is not of the specified form because the upper limit we have is \(x^2\text{,}\) rather than \(x\text{,}\) — so more care is required. Thankfully we can deal with this obstacle with only a little extra work. The trick is to define an auxiliary function by simply changing the upper limit to \(x\text{.}\) That is, define
\begin{align*}
E(x) &= \int_0^x e^{-t^2}\dee{t}\\
\end{align*}
Then the integral we want to work with is
\begin{align*}
E(x^2) &= \int_0^{x^2} e^{-t^2}\dee{t}
\end{align*}
The derivative
\(E'(x)\) can be found via part 1 of the fundamental theorem of calculus (as we did in Example
1.3.4) and is
\(E'(x)= e^{-x^2}\text{.}\) We can then use this fact with the chain rule to compute the derivative we need:
\begin{align*}
\diff{}{x} \int_0^{x^2} e^{-t^2}\dee{t} &= \diff{}{x}E(x^2) &\text{use the chain rule}\\
&= 2x E'(x^2)\\
&= 2x e^{-x^4}
\end{align*}
What if both limits of integration are functions of
\(x\text{?}\) We can still make this work, but we have to split the integral using Theorem
1.2.3.
Example 1.3.6. \(\diff{}{x}\int_x^{x^2} e^{-t^2}\dee{t}\).
Consider the integral
\begin{gather*}
\int_x^{x^2} e^{-t^2}\dee{t}
\end{gather*}
As was the case in the previous example, we have to do a little pre-processing before we can apply the fundamental theorem.
This time (by design), not only is the upper limit of integration \(x^2\) rather than \(x\text{,}\) but the lower limit of integration also depends on \(x\) — this is different from the integral \(\int_a^x f(t)\dee{t}\) in the fundamental theorem where the lower limit of integration is a constant.
Fortunately we can use the basic properties of integrals (Theorem
1.2.3(b) and (c)) to split
\(\int_x^{x^2} e^{-t^2}\dee{t}\) into pieces whose derivatives we already know.
\begin{align*}
\int_x^{x^2} e^{-t^2}\dee{t}
&=\int_x^0 e^{-t^2}\dee{t}+\int_0^{x^2} e^{-t^2}\dee{t}
&\text{by Theorem }\knowl{./knowl/thm_Intdomain.html}{\text{1.2.3}}\text{(c)}\\
&=-\int^x_0 e^{-t^2}\dee{t}+\int_0^{x^2} e^{-t^2}\dee{t}
&\text{by Theorem }\knowl{./knowl/thm_Intdomain.html}{\text{1.2.3}}\text{(b)}
\end{align*}
With this pre-processing, both integrals are of the right form. Using what we have learned in the previous two examples,
\begin{align*}
\diff{}{x}\int_x^{x^2} e^{-t^2}\dee{t}
&=\diff{}{x}\left( -\int_0^{x} e^{-t^2}\dee{t} + \int_0^{x^2} e^{-t^2}\dee{t} \right)\\
&=-\diff{}{x}\int^x_0 e^{-t^2}\dee{t} +\diff{}{x}\int_0^{x^2} e^{-t^2}\dee{t}\\
&=- e^{-x^2} +2x e^{-x^4}
\end{align*}
Before we start to work with part 2 of the fundamental theorem, we need a little terminology and notation. First some terminology — you may have seen this definition in your differential calculus course.
Definition 1.3.7. Antiderivatives.
Let \(f(x)\) and \(F(x)\) be functions. If \(F'(x)=f(x)\) on an interval, then we say that \(F(x)\) is an antiderivative of \(f(x)\) on that interval.
As we saw above, an antiderivative of \(f(x)=x\) is \(F(x) = x^2/2\) — we can easily verify this by differentiation. Notice that \(x^2/2 + 3\) is also an antiderivative of \(x\text{,}\) as is \(x^2/2 + C\) for any constant \(C\text{.}\) This observation gives us the following simple lemma.
Lemma 1.3.8.
Let \(f(x)\) be a function and let \(F(x)\) be an antiderivative of \(f(x)\text{.}\) Then \(F(x)+C\) is also an antiderivative for any constant \(C\text{.}\) Further, every antiderivative of \(f(x)\) must be of this form.
Proof.
There are two parts to the lemma and we prove each in turn.
Let
\(F(x)\) be an antiderivative of
\(f(x)\) and let
\(C\) be some constant. Then
\begin{align*}
\diff{}{x}\left( F(x) + C \right)
&= \diff{}{x}\left( F(x) \right) + \diff{}{x}\left( C \right)\\
&= f(x) + 0
\end{align*}
since the derivative of a constant is zero, and by definition the derivative of
\(F(x)\) is just
\(f(x)\text{.}\) Thus
\(F(x)+C\) is also an antiderivative of
\(f(x)\text{.}\)
Now let
\(F(x)\) and
\(G(x)\) both be antiderivatives of
\(f(x)\) — we will show that
\(G(x) = F(x)+C\) for some constant
\(C\text{.}\) To do this let
\(H(x) = G(x)-F(x)\text{.}\) Then
\begin{align*}
\diff{}{x}H(x)
&= \diff{}{x}\left( G(x)-F(x) \right)
= \diff{}{x} G(x) - \diff{}{x}F(x)\\
&= f(x) - f(x) = 0
\end{align*}
Since the derivative of
\(H(x)\) is zero,
\(H(x)\) must be a constant function
4 . Thus
\(H(x)=G(x)-F(x)=C\) for some constant
\(C\) and the result follows.
Based on the above lemma we have the following definition.
Definition 1.3.9.
The “indefinite integral of \(f(x)\)” is denoted by \(\int f(x)\dee{x}\) and should be regarded as the general antiderivative of \(f(x)\text{.}\) In particular, if \(F(x)\) is an antiderivative of \(f(x)\) then
\begin{align*}
\int f(x)\dee{x} &= F(x) + C
\end{align*}
where the \(C\) is an arbitrary constant. In this context, the constant \(C\) is also often called a “constant of integration”.
Now we just need a tiny bit more notation.
Definition 1.3.10.
The symbol
\begin{gather*}
\left.\int f(x)\dee{x} \right|_{a}^{b}
\end{gather*}
denotes the change in an antiderivative of \(f(x)\) from \(x=a\) to \(x=b\text{.}\) More precisely, let \(F(x)\) be any antiderivative of \(f(x)\text{.}\) Then
\begin{align*}
\left.\int f(x)\dee{x} \right|_{a}^{b}
&= \left.F(x)\right|_a^b = F(b) - F(a)
\end{align*}
Notice that this notation allows us to write part 2 of the fundamental theorem as
\begin{align*}
\int_a^b f(x)\dee{x} &= \left.\int f(x)\dee{x} \right|_{a}^{b}\\
&= \left.F(x)\right|_a^b = F(b) - F(a)
\end{align*}
Some texts also use an equivalent notation using square brackets:
\begin{align*}
\int_a^b f(x)\dee{x} &= \Big[F(x)\Big]_a^b = F(b) - F(a).
\end{align*}
You should be familiar with both notations.
We'll soon develop some strategies for computing more complicated integrals. But for now, we'll try a few integrals that are simple enough that we can just guess the answer. Of course, any antiderivative that we can guess we can also check — simply differentiate the guess and verify you get back to the original function:
\begin{align*}
\diff{}{x} \int f(x)\dee{x} &= f(x).
\end{align*}
We do these examples in some detail to help us become comfortable finding indefinite integrals.
Example 1.3.11. Compute the definite integral \(\int_1^2 x\dee{x}\).
Compute the definite integral \(\int_1^2 x\dee{x}\text{.}\)
Solution: We have already seen, in Example
1.2.6, that
\(\int_1^2 x\dee{x}=\frac{2^2-1^2}{2}=\frac{3}{2}\text{.}\) We shall now rederive that result using the fundamental theorem of calculus.
The main difficulty in this approach is finding the indefinite integral (an antiderivative) of
\(x\text{.}\) That is, we need to find a function
\(F(x)\) whose derivative is
\(x\text{.}\) So think back to all the derivatives you computed last term
5 and try to remember a function whose derivative was something like
\(x\text{.}\)
This shouldn't be too hard — we recall that the derivatives of polynomials are polynomials. More precisely, we know that
\begin{align*}
\diff{}{x}x^n &= n x^{n-1}
\end{align*}
So if we want to end up with just
\(x = x^1\text{,}\) we need to take
\(n=2\text{.}\) However this gives us
\begin{align*}
\diff{}{x}x^2 &= 2x
\end{align*}
This is pretty close to what we want except for the factor of
\(2\text{.}\) Since this is a constant we can just divide both sides by
\(2\) to obtain:
\begin{align*}
\frac{1}{2}\cdot \diff{}{x}x^2 &= \frac{1}{2}\cdot 2x &\text{which becomes}\\
\cdot \diff{}{x}\frac{x^2}{2}&= x
\end{align*}
which is exactly what we need. It tells us that
\(x^2/2\) is an antiderivative of
\(x\text{.}\)
Once one has an antiderivative, it is easy to compute the indefinite integral
\begin{align*}
\int x\dee{x} &= \frac{1}{2}x^2+C
\end{align*}
as well as the definite integral:
\begin{align*}
\int_1^2 x\dee{x}
&= \left.\frac{1}{2}x^2 \right|_1^2 &\text{since $x^2/2$ is an antiderivative of $x$}\\
&=\frac{1}{2} 2^2- \frac {1}{2}1^2 =\frac{3}{2}
\end{align*}
While the previous example could be computed using signed areas, the following example would be very difficult to compute without using the fundamental theorem of calculus.
Example 1.3.12. Compute \(\int_0^{\frac{\pi}{2}} \sin x\dee{x}\).
Compute \(\int_0^{\frac{\pi}{2}} \sin x\dee{x}\text{.}\)
Example 1.3.13. Compute \(\int_1^2 \frac{1}{x}\dee{x}\).
Find \(\int_1^2 \frac{1}{x}\dee{x}\text{.}\)
Solution:
Once again, the crux of the solution is guessing a function whose derivative is
\(\frac{1}{x}\text{.}\) Our standard way to differentiate powers of
\(x\text{,}\) namely
\begin{gather*}
\diff{}{x} x^n= n x^{n-1},
\end{gather*}
doesn't work in this case — since it would require us to pick
\(n=0\) and this would give
\begin{align*}
\diff{}{x} x^0 &= \diff{}{x} 1 = 0.
\end{align*}
Fortunately, we also know
6 that
\begin{gather*}
\diff{}{x}\log x = \frac{1}{x}
\end{gather*}
which is exactly the derivative we want.
We're now ready to compute the prescribed integral.
\begin{align*}
\int_1^2 \frac{1}{x}\dee{x}
&= \left. \log x \right|_1^2 & \text{since $\log x$ is an antiderivative of $1/x$}\\
&= \log 2 - \log 1 & \text{since $\log 1 = 0$}\\
&= \log 2
\end{align*}
Example 1.3.14. \(\int_{-2}^{-1} \frac{1}{x}\dee{x}\).
Find \(\int_{-2}^{-1} \frac{1}{x}\dee{x}\text{.}\)
Solution:
As we saw in the last example,
\begin{gather*}
\diff{}{x}\log x = \frac{1}{x}
\end{gather*}
and if we naively use this here, then we will obtain
\begin{align*}
\int_{-2}^{-1} \frac{1}{x}\dee{x} &= \log(-1)-\log(-2)
\end{align*}
which makes no sense since the logarithm is only defined for positive numbers
7 .
We can work around this problem using a slight variation of the logarithm — \(\log|x|\text{.}\)
When
\(x \gt 0\text{,}\) we know that
\(|x|=x\) and so we have
\begin{align*}
\log |x| &= \log x & \text{differentiating gives us}\\
\diff{}{x}\log|x| &= \diff{}{x} \log x = \frac{1}{x}.
\end{align*}
When
\(x \lt 0\) we have that
\(|x|=-x\) and so
\begin{align*}
\log |x| &= \log(-x) \qquad \text{differentiating with the chain rule gives}\\
\diff{}{x}\log|x| &= \diff{}{x} \log(-x)\\
&= \frac{1}{(-x)} \cdot (-1) = \frac{1}{x}
\end{align*}
Indeed, more generally we should write the indefinite integral of
\(1/x\) as
\begin{align*}
\int \frac{1}{x} \dee{x} &= \log |x| + C
\end{align*}
which is valid for all positive and negative
\(x\text{.}\) It is, however, undefined at
\(x=0\text{.}\)
We're now ready to compute the prescribed integral.
\begin{align*}
\int_{-2}^{-1} \frac{1}{x}\dee{x}
&= \log|x| \bigg|_{-2}^{-1}
\qquad \text{since $\log|x|$ is an antiderivative of $1/x$}\\
&= \log|-1| - \log|-2| = \log 1-\log 2\\
&= -\log 2 = \log\frac12.
\end{align*}
This next example raises a nasty issue that requires a little care. We know that the function \(1/x\) is not defined at \(x=0\) — so can we integrate over an interval that contains \(x=0\) and still obtain an answer that makes sense? More generally can we integrate a function over an interval on which that function has discontinuities?
Example 1.3.15. \(\int_{-1}^1\frac{1}{x^2}\dee{x}\).
Find \(\int_{-1}^1\frac{1}{x^2}\dee{x}\text{.}\)
Solution: Beware that this is a particularly nasty example, which illustrates a booby trap hidden in the fundamental theorem of calculus. The booby trap explodes when the theorem is applied sloppily.
The sloppy solution starts, as our previous examples have, by finding an antiderivative of the integrand. In this case we know that
\begin{gather*}
\diff{}{x}\frac{1}{x} = -\frac{1}{x^2}
\end{gather*}
which means that
\(-x^{-1}\) is an antiderivative of
\(x^{-2}\text{.}\)
This suggests (if we proceed naively) that
\begin{align*}
\int_{-1}^1 x^{-2}\dee{x}
&= \left.-\frac{1}{x}\right|_{-1}^1
& \text{since $-1/x$ is an antiderivative of $1/x^2$}\\
&= -\frac{1}{1}-\Big(-\frac{1}{-1}\Big)\\
&=-2
\end{align*}
Unfortunately,
At this point we should really start to be concerned. This answer cannot be correct. Our integrand, being a square, is positive everywhere. So our integral represents the area of a region above the \(x\)-axis and must be positive.
So what has gone wrong? The flaw in the computation is that the fundamental theorem of calculus, which says that
\begin{gather*}
\text{if } F'(x)=f(x) \text{ then } \int_a^b f(x)\dee{x}=F(b)-F(a),
\end{gather*}
is
only applicable when
\(F'(x)\) exists and equals
\(f(x)\) for
all \(x\) between
\(a\) and
\(b\text{.}\)
In this case \(F'(x)=\frac{1}{x^2}\) does not exist for \(x=0\text{.}\) So we cannot apply the fundamental theorem of calculus as we tried to above.
An integral, like \(\int_{-1}^1\frac{1}{x^2}\dee{x}\text{,}\) whose integrand is undefined somewhere in the domain of integration is called improper. We'll give a more thorough treatment of improper integrals later in the text. For now, we'll just say that the correct way to define (and evaluate) improper integrals is as a limit of well-defined approximating integrals. We shall later see that, not only is \(\int_{-1}^1\frac{1}{x^2}\dee{x}\) not negative, it is infinite.
The above examples have illustrated how we can use the fundamental theorem of calculus to convert knowledge of derivatives into knowledge of integrals. We are now in a position to easily build a table of integrals. Here is a short table of the most important derivatives that we know.
| \(F(x)\) |
\(1\) |
\(x^n\) |
\(\sin x\) |
\(\cos x\) |
\(\tan x\) |
\(e^x\) |
\(\log_e|x|\) |
\(\arcsin x\) |
\(\arctan x\) |
| \(f(x)=F'(x)\) |
\(0\) |
\(nx^{n-1}\) |
\(\cos x\) |
\(-\sin x\) |
\(\sec^2 x\) |
\(e^x\) |
\(\frac{1}{x}\) |
\(\frac{1}{\sqrt{1-x^2}}\) |
\(\frac{1}{1+x^2}\) |
Of course we know other derivatives, such as those of \(\sec x\) and \(\cot x\text{,}\) however the ones listed above are arguably the most important ones. From this table (with a very little massaging) we can write down a short table of indefinite integrals.
Theorem 1.3.17. Important indefinite integrals.
| \(f(x)\) |
\(F(x)=\int f(x)\dee{x}\) |
| \(1\) |
\(x+C\) |
| \(x^n\) |
\(\frac{1}{n+1}x^{n+1}+C\text{ provided that }n \ne-1\) |
| \(\dfrac{1}{x}\) |
\(\log_e|x|+C\) |
| \(e^x\) |
\(e^x+C\) |
| \(\sin x\) |
\(-\cos x+C\) |
| \(\cos x\) |
\(\sin x+C\) |
| \(\sec^2 x\) |
\(\tan x+C\) |
| \(\dfrac{1}{\sqrt{1-x^2}}\) |
\(\arcsin x+C\) |
| \(\dfrac{1}{1+x^2}\) |
\(\arctan x+C\) |
Example 1.3.18. Using Theorem 1.3.17 to compute some integrals.
Find the following integrals
\(\displaystyle \int_2^7 e^x \dee{x}\)
\(\displaystyle \int_{-2}^2 \frac{1}{1+x^2} \dee{x}\)
\(\displaystyle \int_0^3 (2x^3+7x-2)\dee{x}\)
Solution: We can proceed with each of these as before — find the antiderivative and then apply the fundamental theorem. The third integral is a little more complicated, but we can split it up into monomials using Theorem
1.2.1 and do each separately.
An antiderivative of
\(e^x\) is just
\(e^x\text{,}\) so
\begin{align*}
\int_2^7 e^x \dee{x} &= e^x\bigg|_2^7\\
&= e^7-e^2 = e^2(e^5-1).
\end{align*}
An antiderivative of
\(\frac{1}{1+x^2}\) is
\(\arctan(x)\text{,}\) so
\begin{align*}
\int_{-2}^2 \frac{1}{1+x^2} \dee{x} &= \arctan(x) \bigg|_{-2}^2\\
&= \arctan(2) - \arctan(-2)\\
\end{align*}
We can simplify this a little further by noting that \(\arctan(x)\) is an odd function, so \(\arctan(-2)= -\arctan(2)\) and thus our integral is
\begin{align*}
&= 2\arctan(2)
\end{align*}
We can proceed by splitting the integral using Theorem
1.2.1(d)
\begin{align*}
\int_0^3 (2x^3+7x-2)\dee{x} &= \int_0^3 2x^3\dee{x} + \int_0^3 7x\dee{x} - \int_0^3 2\dee{x}\\
&= 2\int_0^3 x^3\dee{x} + 7\int_0^3 x\dee{x} - 2\int_0^3 \dee{x}\\
\end{align*}
and because we know that \(x^4/4, x^2/2, x\) are antiderivatives of \(x^3, x, 1\) respectively, this becomes
\begin{align*}
&= \left[\frac{x^4}{2}\right]_0^3 + \left[\frac{7x^2}{2}\right]_0^3 - \left[2x\right]_0^3\\
&= \frac{81}{2} + \frac{7\cdot 9}{2} -6\\
&= \frac{81 + 63 - 12}{2} = \frac{132}{2} = 66.
\end{align*}
We can also just find the antiderivative of the whole polynomial by finding the antiderivatives of each term of the polynomial and then recombining them. This is equivalent to what we have done above, but perhaps a little neater:
\begin{align*}
\int_0^3 (2x^3+7x-2)\dee{x} &= \left[ \frac{x^4}{2} + \frac{7x^2}{2} - 2x \right]_0^3\\
&= \frac{81}{2} + \frac{7\cdot 9}{2} -6 = 66.
\end{align*}
