Subsection1.3.1The 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
You learned these near the end of your differential calculus course. Now is a good time to revise — but we’ll go over them here since they are so important in what follows.
. Of course “very rough” is not enough — let’s be precise.
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
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 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.
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*}
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
Notice that if \(f\) were discontinuous, then this might be false.
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 \(f(x)\text{.}\)
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.
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.
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.
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 \(\int_0^x e^{-t^2} \dee{t}\) is closely related to the “error function” which is an extremely important function in mathematics. While we cannot express this integral (or the error function) as a finite combination of polynomials, exponentials etc, we can express it as an infinite series \(\int_0^x e^{-t^2}\dee{t}
= x - \frac{x^3}{3\cdot 1} + \frac{x^5}{5\cdot 2} - \frac{x^7}{7\cdot 3!} +
\frac{x^9}{9\cdot 4!} +\cdots + (-1)^k \frac{x^{2k+1}}{(2k+1)\cdot k!} + \cdots\text{.}\) But more on this in Chapter 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
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 \(x^2\text{:}\)
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
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:
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.
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.
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.
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.
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
Since the derivative of \(H(x)\) is zero, \(H(x)\) must be a constant function 4
This follows from the Mean Value Theorem. Indeed, fix any number \(x_0\text{.}\) Then, for each \(x\ne x_0\text{,}\) the MVT gives us a number \(c\) between \(x_0\) and \(x\) with \(H(x)-H(x_0)= H'(c)\,(x-x_0) = 0 \text{,}\) since the derivative of \(H\) is zero everywhere. Thus \(H(x)=H(x_0)\) for all \(x\) and \(H(x)\) is a constant function.
. Thus \(H(x)=G(x)-F(x)=C\) for some constant \(C\) and the result follows.
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”.
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
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:
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
Of course, this assumes that you did your differential calculus course last term. If you did that course at a different time then please think back to that point in time. If it is long enough ago that you don’t quite remember when it was, then you should probably do some revision of derivatives of simple functions before proceeding further.
and try to remember a function whose derivative was something like \(x\text{.}\)
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.
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
Recall that in most mathematics courses (especially this one) we use \(\log x\) without any indicated base to denote the natural logarithm — the logarithm base \(e\text{.}\) Many widely used computer languages, like Java, C, Python, MATLAB, \(\cdots\text{,}\) use \(\log(x)\) to denote the logarithm base \(e\) too. But many texts also use \(\ln x\) to denote the natural logarithm \(\log x = \log_e x = \ln x.\) The reader should be comfortable with all three notations for this function. They should also be aware that in different contexts — such as in chemistry or physics — it is common to use \(\log x\) to denote the logarithm base 10, while in computer science often \(\log x\) denotes the logarithm base 2. Context is key.
that
\begin{gather*}
\diff{}{x}\log x = \frac{1}{x}
\end{gather*}
which makes no sense since the logarithm is only defined for positive numbers 7
This is not entirely true — one can extend the definition of the logarithm to negative numbers, but to do so one needs to understand complex numbers which is a topic beyond the scope of this course.
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?
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
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.
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.
For completeness we’ll show how to evaluate this integral by sneaking up on the point of discontinuity in the interval of integration. As noted above, we will give a fuller explanation of such integrals later in the text.
Rather than evaluating the integral directly, we will approximate the integral using definite integrals on intervals that avoid the discontinuity. In the current example, the original domain of integration is \(-1\le x\le 1\text{.}\) The domains of integration of the approximating integrals exclude from \([-1,1]\) small intervals around \(x=0\text{.}\)
The shaded area in the figure below illustrates a typical approximating integral, whose domain of integration consists of the original domain of integration, \([-1,1]\text{,}\) but with the interval \([-t,T]\) excluded.
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.
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.
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
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:
So far, we have been able to guess many antiderivatives. Often, however, antiderivatives are very difficult to guess. In Questions 16 through 19, we will find some antiderivatives that might appear in a table of integrals. Coming up with the antiderivative might be quite difficult (strategies to do just that will form a large part of this semester), but verifying that your antiderivative is correct is as simple as differentiating.
Suppose that \(f(x)\) is a function and \(F(x) = e^{(x^2-3)} + 1\) is an antiderivative of \(f(x)\text{.}\) Evaluate the definite integral \(\displaystyle\int_1^{\sqrt5} f(x)\,\dee{x}\text{.}\)
If \(f(x)\) is continuous on \([1, \pi]\) and differentiable on \((1,\pi)\text{,}\) then \(\displaystyle\int_1^\pi f'(x)\,\dee{x} = f(\pi)-f(1)\text{.}\)
As in the proof of the Fundamental Theorem of Calculus, let \(F(x) = \int_{a}^x f(t)\dee{t}\text{.}\) In the diagram below, shade the area corresponding to \(F(x+h)-F(x)\text{.}\)
Evaluate and simplify \(\diff{}{x}\{x\log(ax)-x\}\text{,}\) where \(a\) is some constant and \(\log(x)\) is the logarithm base \(e\text{.}\) What antiderivative does this tell you?
Evaluate and simplify \(\diff{}{x}\left\{\log\left|x+\sqrt{x^2+a^2}\right|\right\}\text{,}\) where \(a\) is some constant. What antiderivative does this tell you?
Evaluate and simplify \(\displaystyle\diff{}{x}\left\{\sqrt{x(a+x)}-a\log\left(\sqrt{x}+\sqrt{a+x}\right)\right\}\text{,}\) where \(a\) is some constant. What antiderivative does this tell you?
Evaluate \(\displaystyle \int_1^5 f(x)\,\dee{x}\text{,}\) where \(\displaystyle f(x)= \begin{cases} 3 &\text{ if $x\le 3$} \\
x &\text{ if $x\ge 3$}
\end{cases}\text{.}\)
A car traveling at \(30\,\textrm{m}/\textrm{s}\) applies its brakes at time \(t=0\text{,}\) its velocity (in \(\textrm{m}/\textrm{s}\)) decreasing according to the formula \(v(t) = 30 - 10t\text{.}\) How far does the car go before it stops?
Define the function \(F(x)=\displaystyle\int_0^{x^2}\sin(\sqrt{t})\,\dee{t}\) on the interval \(0 \lt x \lt 4\text{.}\) On this interval, where does \(F(x)\) have a maximum?
Evaluate \(\lim\limits_{n\rightarrow\infty}\dfrac{\pi}{n}\displaystyle\sum\limits_{j=1}^n
\sin\left(\frac{j\pi}{n}\right)\) by interpreting it as a limit of Riemann sums.
Below is the graph of \(y=f(t)\text{,}\)\(-5 \leq t \leq 5\text{.}\) Define \(F(x) = \displaystyle\int_{0}^x f(t)\dee{t}\) for any \(x\) in \([-5,5]\text{.}\) Sketch \(F(x)\text{.}\)
Let \(f(x)\) be an even function, defined everywhere, and let \(F(x)\) be an antiderivative of \(f(x)\text{.}\) Is \(F(x)\) even, odd, or not necessarily either one? (You may use your answer from Section 1.2, Question 1.2.3.20. )
