The fundamental theorem of calculus

As we have seen, finding the (signed) area is hard work, and with the exception of a few examples almost impossible to do. To illustrate this, take the area under the graph of f(x)=xf(x)=\sqrt{x}, for example, that is,

∫01x dx\int_0^1 \sqrt{x}\, dx

Forming the sum of bars nn bars to get an approximation, we obtain

∫01x dx≈∑k=1nf(xk)Δx=f(x1)Δx+f(x2)Δx+...+f(xn)Δx=1n1n+2n1n+...+nn1n=1n1.5+1n1.5+...nn1.5=1n1.5(1+2+...+n)⏟??\begin{array}{lll}\int_0^1 \sqrt{x}\, dx &\approx& \sum_{k=1}^n f(x_k)\Delta x\\ &=& f(x_1)\Delta x+f(x_2)\Delta x+...+f(x_n)\Delta x\\ &=& \sqrt{\frac{1}{n}}\frac{1}{n}+\sqrt{\frac{2}{n}}\frac{1}{n}+...+\sqrt{\frac{n}{n}}\frac{1}{n}\\ &=& \frac{\sqrt{1}}{n^{1.5}}+\frac{\sqrt{1}}{n^{1.5}}+...\frac{\sqrt{n}}{n^{1.5}}\\ &=& \frac{1}{n^{1.5}}\underbrace{\left(\sqrt{1}+\sqrt{2}+...+\sqrt{n}\right)}_{??}\\ \end{array}

where x1=1n,x2=2n,...,xn=nn=1x_1=\frac{1}{n}, x_2=\frac{2}{n}, ..., x_n=\frac{n}{n}=1 are the locations on the xx-axis of the nn bars and Δx=1n\Delta x=\frac{1}{n} is the width of each of these bases.

But now what? Remember, what we ultimately want to know is the exact value towards which the sum of the bar areas converge for n→∞n\rightarrow \infty. But without a computer we cannot even find the sum for, say 10 00010\,000 bars:

∫01x dx≈110 0001.5(1+2+...+10 000⏟??)\begin{array}{lll}\int_0^1 \sqrt{x}\, dx &\approx& \frac{1}{10\,000^{1.5}}(\underbrace{\sqrt{1}+\sqrt{2}+...+\sqrt{10\,000}}_{??})\\ \end{array}

Using the computer, and writing some lines of code, we find that this sum is

0.6667164...0.6667164...

This is certainly a good approximation, as the number of used bar is quiet high, but it is nowhere near a precise value.

It is therefore astonishing to learn that there is actually a really simple solution to this problem. So simple, that in this case I could find the exact area with a small mental calculation ... it is

∫01x dx=23⋅13/2−23⋅03/2=23\int_0^1 \sqrt{x}\, dx = \frac{2}{3}\cdot 1^{3/2}-\frac{2}{3}\cdot 0^{3/2}=\frac{2}{3}

Comparing this to the approximation we calculated with the computer, we see that the two values are indeed close together. So what is the secret? Well, it took several hundred years of people dedicated to maths and physics to develop the right framework and concepts, to find the right questions to ask, and to put everything together. Once that was done, the emerging rule was simple. It is called the fundamental theorem of calculus and works like this:

Theorem 1: Fundamental theorem of calculus

The integral ∫abf(x) dx\int_a^b f(x)\, dx can be found as follows:

  1. Find a new function FF such that its derivative is ff: F′=fF'=f.

  2. The integral is then given by subtracting F(a)F(a) from F(b)F(b):

    ∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b)-F(a)

So, let's apply the fundamental theorem to our example.

Example 1

Find the integral ∫01xdx\int_0^1 \sqrt{x} dx.

Solution

  1. find a function FF with F′(x)=xF'(x)=\sqrt{x}. With a little trial and error we get

    F(x)=23x3/2F(x)=\frac{2}{3} x^{3/2}

    In fact, for the derivative of FF it is true that

    F′(x)=23⋅32x3/2−1=x1/2=xF'(x)=\frac{2}{3}\cdot\frac{3}{2} x^{3/2-1}=x^{1/2}=\sqrt{x}

  2. we thus have

    F(1)=23⋅13/2=23F(1)=\frac{2}{3}\cdot 1^{3/2}=\frac{2}{3}

    and

    F(0)=23⋅03/2F(0)=\frac{2}{3}\cdot 0^{3/2}

    Thus, according to the main theorem

    ∫01x dx=F(1)−F(0)=23−0=23\int_0^1 \sqrt{x}\, dx = F(1)-F(0)=\frac{2}{3}-0=\frac{2}{3}

So, finding the integral of ff is suddenly dead easy ... given that we can find a function FF whose derivative is ff. Any such function FF is called an antiderivative of ff.

Definition 1

Consider a function ff. Any function FF with

F′=fF'=f

is called an antiderivative of ff.

In many cases it is simple to find an antiderivative, but sometimes it is not trivial at all.

In a later section we will discuss some rules for finding the antiderivative of functions. But let us continue with some examples of the fundamental theorem of calculus in the form of exercises, and then its proof in the next section.

Exercise 1
Q1

Show that F(x)=x4F(x)=x^4 is an antiderivative of f(x)=4x3f(x)=4 x^3. Use this fact to determine the integral

∫134x3 dx\int_1^3 4x^3\, dx

with the help of the fundamental theorem of calculus.

Q2

Determine the integral

∫0π2cos⁡(x) dx\int_0^\frac{\pi}{2} \cos(x)\, dx

using the fundamental theorem of calculus.

Solution
A1

FF is an antiderivative of ff because

F′(x)=4x3F'(x)=4x^3

It follows from the fundamental theorem of calculus that

∫134x3 dx=F(3)−F(1)=34−14=81−1=80‾\begin{array}{lll} \int_1^3 4x^3\, dx &=& F(3)-F(1)\\ &=& 3^4-1^4\\ &=& 81-1\\ &=& \underline{80} \end{array}
A2

We know that the derivative of the sine is the cosine:

sin⁡(x)↓′cos⁡(x)\begin{array}{ccc} \sin(x)\\ \downarrow '\\ \cos(x) \end{array}

so sin⁡(x)\sin(x) is an antiderivative of cos⁡(x)\cos(x), where xx has to be in radians. Using the fundamental theorem of calculus we get

∫0π2cos⁡(x) dx=sin⁡(π/2)−sin⁡(0)=1−0=1‾\begin{array}{lll} \int_0^\frac{\pi}{2} \cos(x)\, dx &=& \sin(\pi/2)-\sin(0)\\ &=& 1-0\\ &=& \underline{1} \end{array}