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Welcome to calculus. I'm professor Ghrist.

Â We're about to begin lecture 26, bonus material.

Â In our main lesson, we covered the fundamental theorem of integral calculus.

Â We saw how it was used, and what it meant.

Â But we did not say why it is true. Let's sketch a proof, but how do we

Â proceed? Our goal is to show that the definite

Â integral of f from a to b is the indefinite integral evaluated at the

Â limits. We won't prove that directly rather, we

Â will prove a lemma, a preparatory step. The lemma is going to look very

Â different, but is really closely related. It states the following, the integral of

Â f of t dt, as t goes from a to x is what? Well, let's see, this is going to be a

Â function of x. If we differentiate that with respect to

Â x, what will we get? We will get f of x, the integrand

Â evaluated at x. Now this seems a little unusual since

Â you're differentiating a limit of a definite integral, but lets explore, see

Â if we can make sense of it. Lets make sure we're not completely crazy

Â and check that this works in a simple case.

Â Lets differentiate the integral as t goes from a to x of a constant dt.

Â Well, we can do the definite integral of a constant, c.

Â And that's just going to give us c times the upper limit x minus c times the lower

Â limit, a. If we differentiate that function of x,

Â what do we get? We get c, the constant which is the

Â integrand we began with. So, this is not a clearly crazy thing to

Â do. Well, let's see if we can prove this

Â lemma. Consider what the integrand f of t looks

Â like, we need to compute the definite integral from a to x.

Â And the claim is that the derivative of the definite integral is f evaluated at

Â x. Well, lets see, let's denote by capital

Â F, the definite integral as a function of x.

Â Then, if we want to differentiate that with respect to x, what should we do?

Â Let's look at what happens when we increase x by a small amount.

Â Let's call that h. Well, according to the definition, that

Â is the integral of f of t dt, as t goes from a to x plus h.

Â Now of course, we could write that as the integral as t goes from a to x, plus the

Â integral as t goes from x to x plus h. That comes from additivity of the

Â integral. And since by definition, the integral

Â from a to x is simply capital F at x, we see something that should start looking

Â like the definition of a derivative. Namely capital F at x plus h equals

Â capital F at x plus something. What is that something?

Â That something is going to have the derivative of capital F built into it.

Â Now, what we need to focus on is this interval from x to x plus h.

Â And here, I'm going to I have to make a little bit of a more restrictive

Â assumption about the integrand f namely, that not only is it continuous but it

Â also is reasonable. lets say has a a Taylor series associated

Â to it. Now, for this definite integral, lets

Â choose a partition with width h. What is the height?

Â Well, the height if we choose the left hand endpoint would be f of x.

Â The width is h now, this is not exactly what the definite integral is and it's

Â just an approximation, there's some leftover stuff that we haven't accounted

Â for. How can we estimate that?

Â Well, if the intagrand f has a a reasonable form to it, let's say it has a

Â well defined derivative in the sense of a Taylor series.

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This, change in the height, this change in little f, is going to be big O of h.

Â That is, it shrinks to zero linearly as h goes to zero.

Â So, the worst thing that could happen, the estimate for what we left out here is

Â something that is big O of h times the width h.

Â Now, if we take big O of h times h, of course that's big O of h squared.

Â And now stepping back we see that we've computed the derivative of capital F of x

Â because capital F of x plus h is capital F of x plus something times h, plus

Â something in big O of h squared. What is that coefficient in front of the

Â h term? That first-order variation is, little f

Â evaluated at x. That is what we're trying to prove.

Â And now let's see what that does for us. The indefinite integral of f is what?

Â Well, from this lemma, we see that the definite integral of f of t dt as t goes

Â from a to x is an indefinite integral because it's derivative is little f of x.

Â Therefore, the general indefinite integral of f is, this definite integral

Â of f of t as t goes from a to x plus an arbitrary constant.

Â We now have an explicit form, the antiderivative in terms of a definite

Â integral. Now, what do we have to do?

Â Well, if we were to evaluate that indefinite integral as x goes from a to

Â b, what would we get? We would get the integral from a to x of

Â f of t dt plus some constant, c evaluated at b, minus the same thing evaluated at

Â a. Lets fill that in.

Â When we valuate at b, we get the integral of f of t dt.

Â As t goes from a to b, plus a constant. We subtract off the integral as t goes

Â from a to a of f of t dt plus a constant. I'm going to ignore the plus c, because

Â we first added and then subtracted. Now what do we notice?

Â Well, the integral from a to a is always zero, and so we're left with the integral

Â from a to b. But, notice that we're using t whereas as

Â we stated our theorem in terms of x. But of course, that doesn't matter a bit.

Â You can use any symbol you want for the definite integral.

Â And that is exactly what we were trying to show, the the indefinite integral

Â evaluated from a to b is the definite integral from a to b.

Â Well, that's nice as a proof, but one of the wonderful things about this proof is

Â that it gives us a new tool for computing some interesting derivatives.

Â Derivatives of definite integrals with functions in the limits.

Â Let's do a more complicated example. Consider the derivative with respect to x

Â evaluated at x equals 1 of the following function.

Â The integral of e to the t squared dt, as t goes from log of x to square root of x.

Â And this ostensibly is difficult. You can't find the antiderivative of e to

Â the t squared, easily if at all. But, we can still compute this

Â derivative. How do we do that?

Â Well, we've gotta be a bit careful. Our lemma only applies in the case where

Â the limits are from a to x. And here we have functions of x.

Â So, let's think for a moment. What we know, is that the derivative, the

Â integral from a to x of f of t dt is f of x.

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What is going to help us, is to have the function of x at the upper limit.

Â So, lets rewrite this integral using additivity as the integral from 1 to

Â square root of x plus the integral from log of x to 1.

Â And now, using the orientation property, we can reverse the limits of the second

Â integral and subtract it. So, we're subtracting the integral from 1

Â to log of x. Now, 1 plays the role of a, the constant

Â at the lower limit. At the upper limits, we don't have x, we

Â have a function of x and so, what we're going to do is rethink our lemma in terms

Â of some function u of x, at the upper limit.

Â But we must be careful to use the chain rule to get the derivative with respect

Â to x of the integral from a to u of f of t dt is f of u du dx.

Â And now, we can proceed if we differentiate the integral from 1 to

Â square root of x of e to the t squared dt.

Â We get e to the square root of x squared, times the derivative of square root of x.

Â When we subtract off the second integral, that is e to the log squared of x, times

Â the derivative of log of x. This is the derivative, that

Â complicated-looking integral, we were asked to evaluate it at x equals 1.

Â And the rest is simply algebra. We take e to the x over 2 root x minus e

Â to the log squared x over x. Evaluate at x equals 1 and you can check

Â that we get one half e minus 1. You will find this method useful in

Â several problems. This is just one approach to the proof of

Â the fundamental theorem of integral calculus.

Â We made some [UNKNOWN] restrictive assumptions in using our Taylor series

Â understanding of what a derivative is. There are other proofs using more

Â sophisticated means, limits perhaps that will give you a more general result.

Â