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We've seen a bunch of arguments, or perspectives, on why the fundamental

theorem of calculus should be true, but we've also paid a high price in this

course. We've been doing alot of things

rigorously, and we've done the squeeze theorem.

Well with the squeeze theorem in hand, it's possible to give a rigorous formal

proof of the fundamental theorem of calculus.

Well just to remind you this is actually what we're going to prove rigorously

here. I'm going to start with some function

from this closed interval ab to the real numbers.

It's a continuous function, so it's actually integrable.

then I write down the accumulation function big F which takes the integral

from a to x of the function little f. And the big claim here, is that the

derivative of the accumulation function is the function little f.

The statement of the fundamental theorem of calculus is about a derivative, so

we're really trying to do two different things.

I'm trying to show that a limit of a difference quotient exists, that the

derivative exists, and then second, I'm making a claim as to the value of that

derivative. specifically, I want to show that this

derivative, which is defined to be the limit as h approaches 0 of big F of x

plus h minus big F of x over h, I want to show that this is equal to little f of x.

But to make matters a little bit easier, I'm only going to consider the situation

when h is positive. So I'll put a little plus there.

Let's rewrite that difference of two accumulation functions, as a single

definite integral. Well how so?

Well let's just write this down. So, this is the limit as h approaches 0

from the right of big F of x plus h, that's the integral from a to x plus h of

f of t dt minus big F of x, which is the integral from a to x of f of t dt all

over h. But now this difference of integrals, can

be written as a single integral. This is the limit as h approaches 0 from

the right, of, well, if I integrate from a to x plus h and subtract the integral

from a to x, what's left over, is the integral from x to x plus h.

Of f of t dt all over h. Let's think about how large and how small

that integral can be. Well, first some notation.

I'll define little m of h to be the minimum value of the function f on the

interval between x and x plus h. And I'll define big M of h, to be the

maximum value of the function little f on the interval, well the same interval, a

closed interval from x to x plus h. Now you might be wondering, how do I know

that there is a minimum and a maximum value?

Well, little f assumed to continuous, so by the extreme value theorem there is a

minimum and a maximum value on a closed interval, and these are closed intervals.

This gives us a bound on the integral. Specifically let's think about the

integral from x to x plus h of the function little f, and since the function

little f has a maximum value of big M on this interval, this integral is no bigger

than big M of h times h. And conversely, all right, the minimum

value is little m of h, and that means that this integral is at least as big as

little m of h times h. Now let's divide both sides by h, and

that doesn't effect the inequality because I'm only really thinking about

the case when h is positive. Well that gives me little m of h is less

than or equal to the integral from x to x plus h, of f of t dt, over h, is less

than or equal to big M of h. Now, we apply the squeeze theorem.

Well how does that go? Well let's think about the limit as h

approaches 0 from the right of little m of h.

Now, remember what little m is, little m is the minimum value of the function f on

the interval between x and x plus h. And as h approaches 0 from the right,

this minimum value of f It's just the value of f at the point x.

And for the same reasoning, the limit as h approaches 0 from the right of the

maximum value, big M, the maximum value of f on that interval is also little f of

x. So what I've got here is this inequality,

and the left-hand side and the right-hand side both have equal limit as h

approaches 0 from the right, and that means that this middle term also has a

limit that exists, and equals f of x. Let me write that down.

So what I can conclude, is that the limit, as h approaches 0 from the right,

of the integral from x to x plus h, of f of t dt divided by h, well that's equal

to f of x. And if I play the same game but think

about h approaching 0 from the left. I can get that, that limit is also f of

x, and consequently, the two sided limit, as h approaches 0, of this term, is equal

to f of x. And so we've verified that the derivative

exist, and it's equal to f of x.