[MUSIC].
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.