[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.