How to use the Fundamental Theorem

Our first use of the Fundamental Theorem of Calculus will be to evaluate definite integrals. Let's remember what the theorem says:

If f is a continuous function and $ G(x) $ is an antiderivative of f , then

\[ \int_a^b f(x)~dx = G(b) - G(a) \]

We have already seen some examples when we first learned about the Fundamental Theorem. However, on this page, we would like to show some of the pitfalls which may arise.


Antiderivatives

Before we begin considering examples, let's just record a few simple antiderivatives. There is no real method to finding the antiderivative of a given function---oftentimes, we just use trial and error. In fact, sometimes there is no simple expression which allows us to write down an antiderivative. We will see more of this later. For now, it is easy to verify that the following functions are antiderivatives of the given functions. The quantity $ k $ is a constant.

$ f(x) $ $ G(x) $
$ 1 $ $ x $
$ x^n $ $ \frac{x^{n+1}}{n+1} \mbox{ if } n\neq -1 $
$ \frac 1x $ $ \ln |x| $
$ \sin(kx) $ $ -\frac 1k \cos(kx) $
$ \cos(kx) $ $ \frac 1k \sin(kx) $
$ e^{kx} $ $ \frac 1k e^{kx} $
$ \frac{1}{1+x^2} $ $ \tan^{-1}(x) $

As a shorthand, we will often write $ G(x)|_a^b = G(b) - G(a) $ .


Example 1

We will compute the definite integral $ \int_0^{\pi/2} \sin(x)~dx $ . Notice that since the function is positive on the region of integration, we may interpret this as the area under the graph of the function.

Notice that

\begin{eqnarray*} \int_0^{\pi/2} \sin(x)~dx & = & -\cos(x)|_0^{\pi/2} \\ & = & -\cos(\frac \pi 2) - (-\cos(0)) \\ & = & 1 \end{eqnarray*}

We have now computed that the area under the graph above is 1.


Example 2

Now we will show you a few examples where the Fundamental Theorem cannot be applied in such a straightforward way. In this example, we will compute

\[ \int_0^2 \frac {1}{x^2}~dx \]

What makes this integral unusual is that the function is not continuous on the region of integration since it has a discontinuity at x = 0. We will have to be careful in applying the Fundamental Theorem since it requires that the function we are integrating be continuous.

Graphically, we are trying to compute the area underneath the graph of $ y = \frac{1}{x^2} $ as shown below.

From this picture, it is clear that the area might not be finite and so we might expect that the integral does not exist.

Since the function is not continuous, we cannot compute the integral using the Fundamental Theorem. We can, however, use the Fundamental Theorem to compute the integral over an interval $ [r,2] $ and then consider what happens when we let the left endpoint r approach 0.

The follwing demonstration should give you the idea. The area you see is $ A_r $ , the area under the graph on the interval $ [r,2] $ . As you vary r, you can see that $ A_r $ will fill up the entire area between x = 0 and x = 2 as you let r get close to 0.

Now let's compute:

\begin{eqnarray*} A_r = \int_r^2 \frac{1}{x^2} ~dx & = & -\frac 1x |_r^2 \\ & = & -\frac 12 - (-\frac 1r) \\ & = & \frac 1r - \frac 12 \end{eqnarray*}

Notice that as r approaches 0 , the area $ A_r = \frac 1r - \frac 12 $ becomes arbitrarily large. This means that the area under the curve between x = 0 and x = 2 cannot be finite. We then say that the integral $ \int_0^2 \frac 1{x^2}~dx $ does not exist.

Because of this kind of phenomenon, you will need to be cautious should you see an integral like

\[ \int_{-1}^2 \frac{1}{x^2}~dx \]

You may be tempted to apply the Fundamental Theorem in the most straightforward way---that is, you may say that the integral is equal to $ -\frac 1x|_{-1}^2 = -\frac 12 - 1 = -\frac 32 $ . However, this would be incorrect: the Fundamental Theorem does not apply in this case since the function we are integrating is not continuous on the interval $ [-1,2] $ .


Example 3

We will now consider a similar example: $ \int_0^2 \frac 1{\sqrt{x}}~dx $ . Again, the function has a discontinuity at x = 0 so we can try the same approach that we tried in the last problem: namely, we will first integrate over the interval $ [r,2] $ on which the function is continuous and then we will let r approach 0.

If we compute, we find that

\begin{eqnarray*} A_r = \int_r^2 \frac{1}{\sqrt{x}}~dx & = & \int_r^2 x^{-1/2}~dx \\ & = & 2x^{1/2}|^2_r \\ & = & 2\sqrt{2} - 2\sqrt{r} \end{eqnarray*}

Now as r approaches 0 , we see that $ A_r $ approaches $ 2\sqrt{2} $ . In this case, we can say that the area under the graph between x = 0 and x = 2 is finite and equal to $ 2 \sqrt{2} $ .

You may wonder why these two examples seem to be similar but have such different results. We can understand this by considering their graphs.

Here, you can see that the graph of the reciprocal of the square root is much closer to the y axis. In fact, it is close enough that the area is finite.


Example 4

Finally, we will look at an example in which the function is continuous, but additional care is needed. We will define a function by

\[ f(x) = \left\{\begin{array}{ll} 1 & \mbox{ if } x \leq 1 \\ x & \mbox{ if } x \geq 1 \end{array}\right. \]

We will compute the integral $ \int_0^2 f(x)~dx $ which may be interpreted as the area shown below.

This integral may be computed as

\begin{eqnarray*} \int_0^2 f(x)~dx & = & \int_0^1 f(x)~dx + \int_1^2 f(x)~dx \\ & = & \int_0^1 1 ~dx + \int_1^2 x~dx \\ & = & x|_0^1 + \frac 12 x^2|_1^2 \\ & = & \frac 52 \end{eqnarray*}