In the last lecture we considered the problem of finding the area under the graph of a function f(x) over an interval [a,b]. The approach we took was to divide the interval into smaller subintervals and place a rectangle over each subinterval whose height was given by the value of the function at some point in the subinterval. This lead to an approximation formula for the area.
Here cn is a sample point located somewhere in the sub-interval [xn , xn+1] and 
is the width of the nth subinterval.
In the limit as the number of subdivisions becomes very large and each subdivision becomes very small, the approximate area becomes an exact area, called the definite integral of f(x) over the interval [a,b].
Our goal in this lecture is to develop an alternative way to compute areas. The first step along the way is to turn the area problem into a problem concerning some function and its derivative. To this end we will introduce the so-called area function:
The most important fact about this area function involves its derivative. If 
is continuous at x, it turns out that
The only way to compute this derivative is via the definition:
Since the numerator in the fraction is the difference of two areas, we can rewrite the limit
I am going to give a more detailed proof below, but for now we can give a quick and dirty proof that this limit should be 
. The key observation is that as h gets very small, 
does not vary much in value between x and x+h. Thus, as a first approximation we can say that the area in the numerator is approximately 
. Thus
and we are done.
In order to give a more rigorous proof of the last result, we will first need to prove the following.
Theorem Let f(x) be a continuous function on the interval [a,b]. Then there exists a c in [a,b] for which
To make more sense of this equation, it is better to write it as
In this form it says that if you take the area under f(x) over the interval [a,b] and ‘flatten it out’, you get a rectangle whose height is given by the value of the function at some point c in the interval.
Proof Because f(x) is continuous on [a,b], it must have a maximum and a minimum somewhere in that interval.
This leads to the inequality for areas
or
or
Since f(x) is continuous, by the Intermediate Value Theorem it takes every possible value between m and M. In particular, there is at least one place c at which the function f(x) has a value equal to
This proves the result.
We are now in a position to give a better proof of the theorem about the derivative of the area function. What happens when we try to compute the derivative of the area function?
The only way to compute this derivative is via the definition:
Since the numerator in the fraction is the difference of two areas, we can rewrite the limit
| (1) |
Recall the theorem we just proved.
If we apply that theorem to the fraction in (1) with
a = x
b = x+h
b - a = h
we get
where c is somewhere in the interval [x,x+h]. In the limit as h goes to 0, c gets squeezed down to x. Because f(x) is continuous we have that
If you do not see this right away, remember the definition of continuity:
exists
The bottom line is that
We have just established that the derivative of the area function
is f(x). The next step is to figure out what F(x) itself is. To help us find F(x) we will need to make use of the following theorem that we proved in chapter 4.
Theorem If 
and 
have the same derivative for all x, f(x) and g(x) differ by a constant.
Here is how we will use this result to find F(x).
.G(a) - F(a) = G(a) - 0 = C
G(a) = C
