The Fundamental Theorem of Calculus

Through the examples we looked at for the area under graphs of functions, we were led to an interesting observation: there seems to be a relationship between the process of integration, which is just a fancy way of performing sums, and the process of differentiation. In fact, this observation is the one basic fact which underlies almost all of our work in this course. Consequently, we will give it a name which indicates its importance: The Fundamental Theorem of Calculus.


What does the Fundamental Theorem mean?

Before we jump in and tell you about the theorem, we will try and give you an intuitive feel for it through a demonstration. We have seen already that the definite integral of a positive function can be interpreted as the area under the graph of the function.

But what about functions which are negative? There's a pretty simple explanation in that case as well. Remember that the definite integral is given by a sum

\[ \int_a^b f(x) ~dx = \lim_{n\to\infty} \sum_{k=0}^{n-1} f(x_k)~\Delta x \]

where the points $ x_k $ are formed by breaking the interval $ [a,b] $ into n pieces of width $ \Delta x $ . When the function f was positive, we could interpret each term $ f(x_k)~\Delta x $ as the area of a very thin rectangle. However, remember that area is always considered to be positive and so if the function f is negative, the term $ f(x_k)~\Delta x $ represents the additive inverse of the area of a small rectangle.

This means that when we perform the sum $ \sum_{k=0}^{n-1} f(x_k)~\Delta x $ , we are actually computing the additive inverse of the area of all the rectangles and so the area A is given by

\[ A = - \int_a^b f(x) ~dx ~~~~\mbox{ or } \int_a^b f(x)~dx = -A \]

In other words, the definite integral measures the additive inverse of the area between the graph and the x axis.

Now when the function f has some regions where it is positive and others where it is negative, we may compute the definite integral by integrating over the regions where the function is positive and add that to the integral over the region where the function is negative. This leads to the observation that

\[ \int_a^b f(x)~dx = A_1 - A_2 \]

where $ A_1 $ is the area bounded by the region where the function is positive and $ A_2 $ is the area bounded by the region where the function is negative.

So in this most general case, the definite integral can still be thought of as measuring area, but it does so by measuring some areas as negative and others as positive.

With this interpretation, we can try to convey graphically the main idea behind the Fundamental Theorem of Calculus. Below, you will see the graph of a function f on the left. The graph on the right is the function

\[ F(x) = \int_0^x f(t)~dt \]

Now when you drag the ball on the graph to the left, you will see a region of area which contributes to the definite integral $ F(x) $ . Yellow area contributes positively, while orange area contributes negatively.

Notice that as you first drag the ball to the right away from the origin, you are in a region where $ f(x) $ is positive. Consequently, as you increase x , the new area which is exposed is weighted positively in the definite integral. This means that the function $ F(x) $ is increasing in that region. When you reach the point at which $ f(x) = 0 $ , you have uncovered all of the positive area. This produces a maximum value for $ F(x) $ . Increasing the value of x even further causes the value of the function $ f(x) $ to become negative. Then the new area which is exposed is weighted negatively and this causes the function $ F(x) $ to decrease.

To summarize, you can see that

  1. $ f(x) > 0 $ implies that $ F(x) $ is increasing at x .
  2. $ f(x) = 0 $ implies that x is a critical point for $ F(x) $ .
  3. $ f(x) < 0 $ implies that $ F(x) $ is decreasig at x .

These three relationships between $ F(x) $ and $ f(x) $ are precisely those enjoyed by a function and its derivative. It leads us to speculate that

\[ F^\prime(x) = f(x) \]

We will now verify that this is in fact the case.


The Fundamental Theorem of Calculus, Part I

The first part of the Fundamental Theorem of Calculus says that our earlier observation is indeed correct. Namely, if $ f(x) $ is a continuous function and $ F(x) = \int_0^x f(t) ~dt $ , then

\[ F^\prime(x) = f(x) \]

This is really quite a remarkable statment and is often not fully appreciated by students when they first study Calculus. Let's think about what it says. Remember that a definite integral is really just some kind of fancy sum---that is, the definite integral is obtained, in some sense, by adding together many very small numbers. What the statement above says is that this process of summing is the inverse process to differentiation. You may find it surprising that there is such a simple, elegant relationship between these two processes which at first appear unrelated.

Let's now see why this relationship holds. To study the derivative $ F^\prime(x) $ , let's first study

\begin{eqnarray*} F(x+h) - F(x) & = & \int_0^{x+h} f(t) ~dt - \int_0^x f(t)~dt \\ & = & \int_0^x f(t) ~dt + \int_x^{x+h} f(t)~dt - \int_0^x f(t)~dt \\ & = & \int_x^{x+h} f(t)~dt \end{eqnarray*}

So the difference can itself be described by a definite integral. Since we are eventually interested in what happens as h becomes very small, we want to study the integral $ \int_x^{x+h} f(t)~dt $ for small h .

In the following demonstration, you can move the ball to change the value of h .

Notice that as h becomes small, the definite integral can be thought of as the area of a very thin strip. In particular, the area of this strip may be approximated by the area of a rectangle. This means that

\[ F(x+h) - F(x) = \int_x^{x+h} f(t)~dt \approx f(x) h \]

or in other words,

\[ \frac{F(x+h) - F(x)}{h} \approx f(x) \]

Now as h becomes very small, the approximation becomes even better and so we have

\[ F^\prime(x) = \lim_{h\to 0} \frac{F(x+h) - F(x)}{h} = f(x) \]

thus verifying the relationship we have been led to expect. This completes the argument which justifies the relationship

\[ F^\prime(x) = f(x) \]


The Fundamental Theorem of Calculus, Part II

So far we have found a very interesting relationship between the definite integral and differentiation. Now we will see how to use this relationship to actually compute definite integrals.

Notice that

\[ \int_a^b f(x)~dx = \int_0^b f(x)~dx - \int_0^a f(x)~dx = F(b) - F(a) \]

In fact, this relationship holds for any function $ G(x) $ which satsifies $ G^\prime(x) = f(x) $ (such a function is called an antiderivative of $ f(x) $ ). To see why this is so, let's compute

\[ (F-G)^\prime(x) = F^\prime(x) - G^\prime(x) = f(x) - f(x) = 0 \]

In other words, the derivative of the function $ F - G $ is constantly zero which means that $ F - G $ is itself constant (since its tangent lines are always horizontal). We can then conclude that there is a constant C such that $ F(x) - G(x) = C $ for all x . In other words, $ F(x) = G(x) + C $ .

Now we know how to use $ F(x) $ to evaluate definite integrals. We saw that

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

This gives us the second part of the Fundamental Theorem of Calculus: if f is a continuous function and G is an antiderivative of f , then

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


A few examples

In the page on area , we computed a few examples through the summing process. Now we can show how to easily perform those computations using the Fundamental Theorem of Calculus.

Example 1: We will compute the area under the straight line $ y = f(x) = x $ between x = 0 and x = 1 . In other words, we want to evaluate $ \int_0^1 x~dx $ .

An antiderivative of the function $ f(x) = x $ is given by the function $ G(x) = \frac{x^2}{2} $ (you can check that $ G^\prime(x)=f(x) $ ). Now applying the Fundamental Theorem of Calculus, we have

\[ \int_0^1 x~dx = G(1) - G(0) = \frac 12 \]

This is in fact the same result as we found those the laborious process of summing.

Example 2: Let's compute $ \int_0^1 x^2 ~dx $ . An antiderivative for the function $ f(x) = x^2 $ is the function $ G(x) = \frac{x^3}{3} $ . This means that

\[ \int_0^1 x^2 ~dx = G(1) - G(0) = \frac 13 \]

Example 3: Finally, we compute $ \int_0^1 e^x~dx $ . An antiderivative for the function $ f(x) = e^x $ is the function $ G(x)=e^x $ . Consequently, it follows that

\[ \int_0^1 e^x ~dx = G(1) - G(0) = e^1 - e^0 = e - 1 \]

This is clearly a much easier way to evaluate definite integrals! In fact, we will spend quite a bit of time learning how to find antiderivatives so that we may compute lots of definite integrals in this way.


Summary

The Fundamental Theorem of Calculus is remarkable for it expresses a relationship between integration, which is a sophisticated kind of addition, and differentiation. To be more specific:

If f is a continuous function

  1. and $ F(x) = \int_0^x f(t)~dt $ , then

    \[ F^\prime(x) = f(x) \]
  2. and G is any antiderivative of f , then

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