The fundamental theorem of calculus

As we have seen, finding integral is hard work, and with the exception of a few examples almost impossible to do. It is therefore surprising to learn that a really simple method exists to find the integal, but it relies on differential calculus. Before showing the method, we need a definition:

Definition 1

Consider a function ff. Any function FF whose derivative is ff is called an antiderivative of ff:

Equation 1
F′=fF'=f

The antiderivative of a function

In a sense, the antiderivative is just the opposite of the derivative, and in german the antiderivative is also called "aufleiten" (as opposed to "ableiten", that is, taking the deriviative).

Example 1

The antiderivate of f(x)=x2f(x)=x^2 is F(x)=13x3F(x)=\frac{1}{3}x^3, because F′(x)=x2F'(x)=x^2

Using the antiderivative, it suddently becomes much simpler to calculate the integral:

Theorem 1

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

  1. Find an antiderivative FF of ff.
  2. The integral is then given by subtracting F(a)F(a) from F(b)F(b):
Equation 2
∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b)-F(a)

The fundamental theorem of calculus (FTC).

This is called the Fundamental Theorem of Calculus (FTC).

The proof is surprisingly simple, and is shown in the next chapter. The problem of determining the value of an integral is now shifted to the problem of finding the antiderivative of a function. Once we have this antiderivative, we are basically done! But finding the antiderivative can be difficult. In a later section we will discuss some rules for finding them. But let's finish with an example and some exercises.

Example 2

Find the integral ∫01x2 dx\int_0^1 x^2\, dx.

Solution

The antiderivative of f(x)=x2f(x)=x^2 is F(x)=13x3F(x)=\frac{1}{3}x^3. Thus, per fundamental theorem of calculus, we get

∫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}
Exercise 1

  1. 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.

  2. Determine the integral

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

    using the fundamental theorem of calculus.

Solution
  • 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}
  • 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}

    • How many antiderivatives can a function ff have? It turns out infinitely many of them! Can you explain why?

    Exercise 2

    Consider the function f(x)=x2f(x)=x^2.

    1. Find an antiderivative of ff.
    2. Find another antiderivative of ff.
    3. How many antiderivatives are there, and how do they differ from each other?
    4. Does it matter, which antiderivative we take for finding the integral
    ∫abx2 dx ?\int_a^b x^2\, dx \text{ ?}
    Solution

    1. To find an antiderivative of ff, we have to find a function FF whose derivative is ff. That is, find FF with

      F′(x)=x2F'(x)=x^2

      Trying out a bit, we see that the one such possible function is

      F(x)=13x3F(x)=\frac{1}{3}x^3

      Indeed,

      F′(x)=13⋅3x2=x2F'(x)=\frac{1}{3}\cdot 3x^2 = x^2

    2. Clearly, the function

      G(x)=13x3+1G(x)=\frac{1}{3}x^3+1

      is also an antiderivative of ff, because the derivative of a constant, the 11, is 00:

      G′(x)=13⋅3x2+0=x2G'(x)=\frac{1}{3}\cdot 3x^2 + 0 =x^2

    3. Indeed, we can add any constant we like to FF, and get another antiderivative of ff. Thus there are infinitely many antiderivatives of x2x^2, and they all differ by a constant from each other:

      13x3,13x3+1,13x3−1.34,13x3+100000.01,...\frac{1}{3}x^3, \frac{1}{3}x^3+1, \frac{1}{3}x^3-1.34, \frac{1}{3}x^3+100000.01, ...

      Indeed, it can be shown that all the antiderivatives of ff must differ by a constant. But we do not show that here.

    4. No, it does not matter, because the constant is subtracted away. Let us denote the constant by cc, where cc can be any number, thus our antiderivative is

      F(x)=13x3+cF(x)=\frac{1}{3} x^3+c

      We then have

      ∫abx2 dx=F(b)−F(a)=(13b3+c)−(13a3+c)=13b3−13a3+c−c=13b3−13a3\begin{array}{lll} \int_a^b x^2\, dx &=& F(b)-F(a)\\ &=& \left(\frac{1}{3} b^3 +c \right) - \left(\frac{1}{3} a^3 +c \right)\\ &=& \frac{1}{3} b^3 - \frac{1}{3} a^3 +c -c\\ &=& \frac{1}{3} b^3 - \frac{1}{3} a^3 \end{array}

      So you see, the constant cc has not effect.