# Difference between revisions of "Fundamental Theorem of Calculus"

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## Revision as of 23:12, 24 March 2009

The **Fundamental Theorem of Calculus** establishes a link between the two central operations of calculus: differentiation and integration.

## Contents

## Introductory Problems

This section is for people who know what integrals are but don't know the Fundamental Theorem of Calculus yet, and would like to try to figure it out. (Actually there are two different but related Fundamental Theorems of Calculus. Questions 0 through 5 correspond to the "first" Fundamental Theorem of Calculus. The last question corresponds to the "second" Fundamental Theorem of Calculus.)

- Evaluate and . (The next few questions are meant as hints for how to do this.)

- An object is moving along a straight line, and its velocity at time is meters/second. (Yes, probably no object really moves this way, but just pretend.)
*Approximately*how far does the object move between times second and seconds? (I picked because I wanted a function that doesn't have a nice anti-derivative.) Interpret the distance that the object travels between times and geometrically, as an area under a curve. - An object is moving along a straight line, and its velocity at time is m/s. Exactly how far does the object go between times sec and sec? Interpret this distance geometrically, as an area under a curve.
- Same as question 3, but this time the object's velocity at time is , and you want to find out exactly how far the object moved between times and . Interpret the distance that the object moved geometrically, as an area under a curve.

Can you do the main problem now?

- Here's a slightly different way to think about the main problem, that doesn't use physics. How much does the function change over the interval from to ? Obviously, the answer is . But, there's another way to look at it:
*the total change is the sum of all the little changes*. Break the interval from to up into little subintervals. Using the derivative, tell me: approximately how much does change over each of these little subintervals? Adding up all of these little changes gives you a Riemann sum that approximates the total change of over the whole interval. This Riemann sum also approximates a certain integral. What is that integral?

The remaining question deals with the "second" Fundamental Theorem of Calculus.

- Let be continuous. Define for all . Draw a picture to explain this definition of . . Assuming is a small number, represent geometrically in your picture. Using your picture, approximately how large is ? Now the final question: What is ?

## Statement

First Fundamental Theorem of Calculus:

Let , , . Suppose is differentiable on the whole interval (using limits from the right and left for the derivatives at and , respectively), and suppose that is Riemann integrable on . Then .

In other words, "the total change (on the right) is the sum of all the little changes (on the left)."

Second Fundamental Theorem of Calculus:

Let , , . Suppose is continuous on the whole interval . Let for all . Then is differentiable on the whole interval (using limits from the right and from the left for the derivatives at a and b, respectively), and for all .

## Intuitive explanation

The first Fundamental Theorem of Calculus basically says that "the total change is the sum of all the little changes."

How much does a function change over an interval ? Obviously, the answer is . But there's another way to look at it. Break the interval up into a whole bunch of tiny subintervals, each having a tiny width . Let's say has been broken up into subintervals, so . Let .

How much does change over the tiny interval ? Of course, the answer is . But this is approximately . If is the amount that changes on the interval , then .

The total change of over the interval , , is exactly equal to . Thus,

.

What we have here on the right is a Riemann sum. It approximates a certain integral. What integral does it approximate? Well, it approximates .

If we repeat this process, using more and more subintervals, then our approximations will get better and better, and the Riemann sums will approximate that integral better and better, and in the limit we will find that .

This can be made into a rigorous proof, if you use the Lagrange's mean value theorem. The standard proof of the first Fundamental Theorem of Calculus, using the Mean Value Theorem, can be thought of in this way.

In order to get an intuitive understanding of the second Fundamental Theorem of Calculus, I recommend just thinking about problem 6. The idea presented there can also be turned into a rigorous proof. (The standard proof can be thought of in this way.)

## Proof

Here's a proof of the first Fundamental Theorem of Calculus.

Let , , . Suppose is differentiable on the whole interval (using limits from the right and left for the derivatives at and , respectively), and suppose that is Riemann integrable on .

Let . There exists such that if is a partition of , and , and for all , , and is in for all , , THEN the Riemann sum

is within of .

(In fact, this is sometimes taken as the definition of the statement " is Riemann integrable on the interval .")

So, let be a partition of such that for all , .

For each , , let be the amount that changes over the interval . Then

.

According to the Mean Value Theorem, for each , , there exists such that .

(This is similar to the part of the intuitive argument where we said that . However, this is better. The Mean Value Theorem, fortunately, gives us exact equality, rather than just an approximation.)

Thus,

.

This last expression, on the right, is a Riemann sum, and it is within of .

THEREFORE, we have found that is within of .

But was arbitrary. The only way that could be within of for any is if is actually equal to .

So that means .

That concludes the proof of the first Fundamental Theorem of Calculus.

For a proof of the second Fundamental Theorem of Calculus, I recommend looking in the book *Calculus* by Spivak. (Hopefully I or someone else will post a proof here eventually.)