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.