 Welcome back to part two in our lecture 46 for Math 1210 Calculus One here. In this part of the video, we are gonna talk about the fundamental theorem calculus, part one in particular, there's a second part forthcoming, and we're gonna talk about the proof of this thing. So we wanna try to understand what this thing is. Now, in mathematics, like in a class like calculus one or what have you, there's often a lot of theorems and propositions you can find inside of these textbooks, right? These are statements of mathematical truth, but oftentimes they're just given lame numbers names, like this is theorem seven on page 12, this is theorem 9.6.8, this is theorem 5.3.2. Now these theorems are important, they shouldn't be ignored, but the reason they're just given these numerical names is that you'll probably use them in the homework for that section of the textbook, but probably not beyond that. On the other hand, some theorems are given names, so we can help remember them, like the intermediate value theorem, the squeeze theorem, the extreme value theorem, the Fermat's theorem, the mean value theorem, Roll's theorem, just to name a few examples from calculus one. The reason we give these names is so that we can refer to them in the future because we will use them more than once, not even just in this class, we might continue to use them in future mathematical courses. Now the adjective fundamental is one we use sometimes to describe theorems in mathematics, and that term is used very, very loosely. There's not a lot of fundamental theorems you hear about in your undergraduate mathematics course. The fundamental theorem calculus is one of those things, and so really we should take this to understand that this is a really big deal, this theorem really, really matters, and it's gonna affect the rest of calculus one, but also it's gonna be a driving force for all of calculus two, and any other calculus course that you would take, hence why we call it the fundamental theorem of calculus. So imagine we have a function F, which is continuous on some domain, an interval, closed interval from A to B. Then the function G, we're gonna find as this integral function, G of X is defined as the integral from A to X of the function F of T, D, T, and here A represents just a constant, this is the left bound of this interval right there, and then X on top is our variable, and it's the variable of the function G. So we're looking for the area, so G is the area function that measures the area under F from A to X. And then, I'm sorry about that, there's a typo right here, this should say G. So then F is our, the function G is, we can show that in fact G is then continuous on this interval A to B. This A to B actually serves as the domain of our function G right here. G we can say this function we define using intervals here, enter goals, excuse me. G is continuous on its domain A to B and it'll be differentiable on the open domain A to B. We can't necessarily guarantee differentiability at the end points, because it turns out there could be like some vertical tangents or things going on right there, but the derivative of G is gonna equal the function F of X, the integrand of this function right here. And so what you can see to the left right here is the proof of this statement. We're gonna talk about the proof. And then I'll have some space to the right so that we can actually draw some pictures to better understand it, because the words alone can be sometimes difficult to digest. And the proof of the photomial theorem calculus is actually a fun little guide, because it's sort of like the reunion episode at the end of our series, where like every important result, some of those I mentioned earlier are gonna kind of pop their heads up in this proof. And so it's like the who's who of the calculus world. So the proof is gonna follow in the basic idea that if we want to calculate the derivative of G, we're gonna have to go to the definition of the derivative. So the definition of the derivative, remember, this is gonna be the limit as H approaches zero of the function G of X plus H minus G of X over H. So we have to work with this limit of difference quotients. And so what we're gonna first do is consider the difference quotient of this expression. So let's actually stick with the numerator beginning right here. Let's take G of X plus H minus G of X. Now, G of X by the definition we see up here, we can plug that in immediately for G of X, right? And so we're gonna get that G of X is the integral from A to X of F of TDT. But if we apply that same definition for G of X plus H, we get this integral right here, the integral from A to X plus H of F of TDT. So that, if we just apply the definition of the function G, we're gonna see that's what this difference looks like, this thing right here. Now, for the sake of simplification, we're gonna apply the following property, but actually maybe make a comment about this right here. These numbers X and X plus H, what do we mean by that? It's important to recognize that the domain of our function G isn't itself gonna be an interval. So we have A right here and we have B right here. We're choosing an arbitrary X inside of this domain. And if we go a little bit to the right of that, if we take some sidestep that is itself a distance of H, this will then give us the number X plus H, which we see right here. And so be aware that we're inside the domain of domain of G, we're taking these two values X plus H. And that's part of our difference quotient formula up here, right? And so to get to this next stage, what I wanna do is rewrite this integral right here, take the integral from A, excuse me, integral from A to X plus H of F of TDT. Now we've had a property of integration that says the following, if you integrate from A to B of any function F of TDT, you can break this up into two integrals where you integrate from A to C, F of TDT, plus the integral from C to B, F of TDT. And so we're gonna apply that property right here, where this integral from A to X plus H, we're gonna insert the number X in between it. So this becomes two integrals, which are these two right here. We get the integral from A to X of F of TDT plus integral from X to X plus H of F of TDT. The advantage of using X right here is that you're gonna see that this integral from A to X will cancel with the difference along the difference right here because we have a negative integral from A to F of X, integral from A to X right there. And so this will simplify just to be the following integral right here. G of X plus H minus G of X becomes the integral from X to X plus H of F of TDT. And this becomes the numerator over the difference quotient. So if we come over to the difference quotient, that's part of the derivative definition, we get G of X plus H minus G of X over H. Well, as the numerator is just this integral, we can pop it in down here and our difference quotient will look like one over X times the integral the integral from X to X plus H of F of TDT. So this is a pretty important observation as we're working with this definition of the derivative. We're now ready to work with our integral, our difference quotient, which is a difference quotient of integral function. Now we have to start considering what's the limit of this thing as H goes to zero. In the present form, I want you to be aware that this is an indeterminate form because if we plug in X equals zero, we're gonna look like one over zero times the integral from X to X of F right, TDT. But whenever you take an integral from a number to itself, that always ends up with zero. So this thing is gonna look like zero zero, which is a typical thing that happens when we're working with difference quotients. So we're gonna have to figure out some way of resolving that, hopefully it's a removal discontinuity, but we have to deal with that indeterminate form. So let's consider the situation where H is positive. Negative is gonna be handled very similar here. So if H is positive, I want you to consider this interval we had before. If H is positive, that's exactly the situation we see in this picture up here. X plus H will be a little bit bigger than X by amount of H. And because the function is continuous, because F is continuous from A to B, it'll also be continuous from X to X plus H. And because it's continuous from X plus H on a closed interval, we can apply the extreme value theorem. The extreme value theorem will guarantee there is an absolute minimum and an absolute maximum. We're gonna call those values M and little M for the minimum and capital M for the maximum right there. And so I want you to think of the following type of picture, slide this up a little bit. We have our domain, right? We're going from A to B and our function F, it's continuous, so maybe it's doing something like this. Here's our F. And so we're picking some specific values, X and X plus H inside of that. So if we look for the maximum, that appears to be happening right here at X. This is our maximum M right there. And then we have some minimum right here as well, little M. Now it doesn't necessarily have to happen at the X and X plus H. That's just because I happened to pick an interval for which it was decreasing. On the other hand, we had something like the following, like maybe this is our function F right here. The maximum value looks like it's right here. This is our capital M. And we're gonna say that the X coordinate associated to that value we're gonna call that V. And then let's say the minimum it seems to be about right here, this is little M, we'll say that the X coordinate associated to that is U. So we get that little F of U equals little M, the minimum. And little F of V equals capital M, the maximum. So this is our max and this is our min. The extreme value theorem guarantees that these numbers exist. So on the interval X to X plus H, we know that F of T will sit between little M and capital M. And this is an integral right here. So if we use the comparison tests for integrals, this tells us that the integral that goes from X to X plus H of F of T, it'll sit below the integral from X to X plus H of little M and it'll sit above the corresponding integral for capital M. Cause these give us upper bounds and be aware these are rectangular regions, right? The little M says, oh, the area under the curve is gonna be less than this rectangle and that's for little M. And for capital M it's saying, oh, the area under the curve will be less than this rectangle right here. That's all these statements are saying and be aware that the thickness of these rectangles in both cases, that's gonna be this H value. And so you're gonna get that the lower, that this integral is bigger than little M H and it's gonna be less than big M H, right? So, but the thing is little M is just F of U times H and then capital M here is just F of V H right here. So what we have right now is we have, we're using the comparison test, we have now sandwiched our integral which represents the numerator of our difference quotient between F of U H and F of V H. If you divide everything by H because H is a positive number, you'll get the very important inequality which is highlighted right here. F of U is less than or equal to one over H integral from X to X plus H of F of T DT which is less than F of V. This is a pretty fun situation to be in because we've taken an indeterminate form zero over zero and we've sandwiched it between two functions. And so, and by similar reason I wanna mention that we get something like this when H is negative, the inequalities are turned around but it's basically the same type of thing. And so when you don't know how to calculate a limit but you can sandwich it between things which we do know how to compute, we can actually use the squeeze theorem to calculate our limit. Didn't I say all our fundamental theorems are coming out? The extreme value theorem, the comparison test, now we're using the squeeze theorem. And so to calculate this limit we're gonna take the limit as H goes to zero. Now as H goes to zero, what's gonna happen to F of V? Well, V is some number that sits between X plus H and X. So as H goes to zero, we're gonna see that X plus H is gonna go towards X which is gonna squeeze V into becoming X itself. Now because F is continuous, we get that F of V is gonna converge towards F of X. And so the right hand side of this limit right here is gonna go to F of X. Now by similar reason, F of U is also gonna go towards F of X right here. And so applying the squeeze theorem, if we take the limit as H goes to zero of this difference quotient, we're gonna see that that is equal to F of X. And that then proves our fundamental theorem of calculus part one, that the derivative of this integral function is equal to the original function F. And it seems kind of crazy, complicated, but at the same time it all kind of makes sense that the squeeze theorem, the screen value theorem, the comparison test, all of these fun things, continuity, differentiability are all coming to play and there's one really amazing proof right here. And so we're gonna, in the next video, I'm gonna show you how to compute some derivatives using this fundamental theorem of calculus. Please stay tuned for those.