Welcome to calculus.
I'm professor Ghrist. We're about to begin lecture 26, bonus
material. In our main lesson, we covered the
fundamental theorem of integral calculus. We saw how it was used, and what it
meant. But we did not say why it is true.
Let's sketch a proof, but how do we proceed?
Our goal is to show that the definite integral of f from a to b is the
indefinite integral evaluated at the limits.
We won't prove that directly rather, we will prove a lemma, a preparatory step.
The lemma is going to look very different, but is really closely related.
It states the following, the integral of f of t dt, as t goes from a to x is what?
Well, let's see, this is going to be a function of x.
If we differentiate that with respect to x, what will we get?
We will get f of x, the integrand evaluated at x.
Now this seems a little unusual since you're differentiating a limit of a
definite integral, but lets explore, see if we can make sense of it.
Lets make sure we're not completely crazy and check that this works in a simple
case. Lets differentiate the integral as t goes
from a to x of a constant dt. Well, we can do the definite integral of
a constant, c. And that's just going to give us c times
the upper limit x minus c times the lower limit, a.
If we differentiate that function of x, what do we get?
We get c, the constant which is the integrand we began with.
So, this is not a clearly crazy thing to do.
Well, let's see if we can prove this lemma.
Consider what the integrand f of t looks like, we need to compute the definite
integral from a to x. And the claim is that the derivative of
the definite integral is f evaluated at x.
Well, lets see, let's denote by capital F, the definite integral as a function of
x. Then, if we want to differentiate that
with respect to x, what should we do? Let's look at what happens when we
increase x by a small amount. Let's call that h.
Well, according to the definition, that is the integral of f of t dt, as t goes
from a to x plus h. Now of course, we could write that as the
integral as t goes from a to x, plus the integral as t goes from x to x plus h.
That comes from additivity of the integral.
And since by definition, the integral from a to x is simply capital F at x, we
see something that should start looking like the definition of a derivative.
Namely capital F at x plus h equals capital F at x plus something.
What is that something? That something is going to have the
derivative of capital F built into it. Now, what we need to focus on is this
interval from x to x plus h. And here, I'm going to I have to make a
little bit of a more restrictive assumption about the integrand f namely,
that not only is it continuous but it also is reasonable.
lets say has a a Taylor series associated to it.
Now, for this definite integral, lets choose a partition with width h.
What is the height? Well, the height if we choose the left
hand endpoint would be f of x. The width is h now, this is not exactly
what the definite integral is and it's just an approximation, there's some
leftover stuff that we haven't accounted for.
How can we estimate that? Well, if the intagrand f has a a
reasonable form to it, let's say it has a well defined derivative in the sense of a
Taylor series. This, change in the height, this change
in little f, is going to be big O of h. That is, it shrinks to zero linearly as h
goes to zero. So, the worst thing that could happen,
the estimate for what we left out here is something that is big O of h times the
width h. Now, if we take big O of h times h, of
course that's big O of h squared. And now stepping back we see that we've
computed the derivative of capital F of x because capital F of x plus h is capital
F of x plus something times h, plus something in big O of h squared.
What is that coefficient in front of the h term?
That first-order variation is, little f evaluated at x.
That is what we're trying to prove. And now let's see what that does for us.
The indefinite integral of f is what? Well, from this lemma, we see that the
definite integral of f of t dt as t goes from a to x is an indefinite integral
because it's derivative is little f of x. Therefore, the general indefinite
integral of f is, this definite integral of f of t as t goes from a to x plus an
arbitrary constant. We now have an explicit form, the
antiderivative in terms of a definite integral.
Now, what do we have to do? Well, if we were to evaluate that
indefinite integral as x goes from a to b, what would we get?
We would get the integral from a to x of f of t dt plus some constant, c evaluated
at b, minus the same thing evaluated at a.
Lets fill that in. When we valuate at b, we get the integral
of f of t dt. As t goes from a to b, plus a constant.
We subtract off the integral as t goes from a to a of f of t dt plus a constant.
I'm going to ignore the plus c, because we first added and then subtracted.
Now what do we notice? Well, the integral from a to a is always
zero, and so we're left with the integral from a to b.
But, notice that we're using t whereas as we stated our theorem in terms of x.
But of course, that doesn't matter a bit. You can use any symbol you want for the
definite integral. And that is exactly what we were trying
to show, the the indefinite integral evaluated from a to b is the definite
integral from a to b. Well, that's nice as a proof, but one of
the wonderful things about this proof is that it gives us a new tool for computing
some interesting derivatives. Derivatives of definite integrals with
functions in the limits. Let's do a more complicated example.
Consider the derivative with respect to x evaluated at x equals 1 of the following
function. The integral of e to the t squared dt, as
t goes from log of x to square root of x. And this ostensibly is difficult.
You can't find the antiderivative of e to the t squared, easily if at all.
But, we can still compute this derivative.
How do we do that? Well, we've gotta be a bit careful.
Our lemma only applies in the case where the limits are from a to x.
And here we have functions of x. So, let's think for a moment.
What we know, is that the derivative, the integral from a to x of f of t dt is f of
x. What is going to help us, is to have the
function of x at the upper limit. So, lets rewrite this integral using
additivity as the integral from 1 to square root of x plus the integral from
log of x to 1. And now, using the orientation property,
we can reverse the limits of the second integral and subtract it.
So, we're subtracting the integral from 1 to log of x.
Now, 1 plays the role of a, the constant at the lower limit.
At the upper limits, we don't have x, we have a function of x and so, what we're
going to do is rethink our lemma in terms of some function u of x, at the upper
limit. But we must be careful to use the chain
rule to get the derivative with respect to x of the integral from a to u of f of
t dt is f of u du dx. And now, we can proceed if we
differentiate the integral from 1 to square root of x of e to the t squared
dt. We get e to the square root of x squared,
times the derivative of square root of x. When we subtract off the second integral,
that is e to the log squared of x, times the derivative of log of x.
This is the derivative, that complicated-looking integral, we were
asked to evaluate it at x equals 1. And the rest is simply algebra.
We take e to the x over 2 root x minus e to the log squared x over x.
Evaluate at x equals 1 and you can check that we get one half e minus 1.
You will find this method useful in several problems.
This is just one approach to the proof of the fundamental theorem of integral
calculus. We made some [UNKNOWN] restrictive
assumptions in using our Taylor series understanding of what a derivative is.
There are other proofs using more sophisticated means, limits perhaps that
will give you a more general result.