All right, welcome back. Now we're going to continue our conversation with the first fundamental theorem calculus, but we're going to talk about part two. Before we do that though, let's warm up a little bit with an example with the question here. So here's the question. So state an antiderivative of the functionf(x) equals cosine of x squared. So here's the jeopardy game. I'm thinking of cosine x squared. This is the answer. Find for me a function whose derivative is cosine of x squared. Now, this is challenging a little bit. This is really hard. So positive for a second and think it through before we move on to the answer. And when hidden, of course, it's going to use the fundamental theorem of calculus part one. All right, ready for the answer? Pause video quick, quick, quick. All right, here we go. The fundamental theorem of calculus part one says that to find an antiderivative, you can take the net accumulation function. So let's take the net accumulation function from any value in the domain. Here's beauty of it. It doesn't matter, so we can fix zero just because we'll put t as our upper bound and we will put cosine of x squared dx. This function I claim is an antiderivative of cosine x squared. So let's see that. So the derivative of the accumulator function, which is ddt of the integral definite interal t of cosine of x squares dx. Here it comes, get ready. It is cosine. This is a fundamental calculus, cosine of t squared. So I found a function whose derivative, who's anti, sorry, final function whose derivative is the cosine question. Other variables is a dummy variable. It doesn't matter, it is still the same function and it works. So any smooth function you give me, cosine x square and x squared function, you can find an antiderivative by using the net accumulator function. One more example just as a warm up as similar, but with a different flavor. So state an antiderivative of the function f(x)=e to the x squared with A and will use the capital letter A for accumulator functions as we know that we're going to use, but I want a particular one. I want A(2) to equal 0. I want A(2)=0. So again, pause the video. If you can, try to work this out before moving on, ready? Here we go. So I know it's an accumulator function, right? Antiderivatives related to the fundamental calculus with net accumulated functions. So I want A(2) to be 0. So I want some base such that when I plug into to be 0. So the trick, of course is if I plug in 2, the accumulation from 2 to 2 is going to be 0. So that condition, A(2)=0 tells you that you don't get to pick an arbitrary base like we did in the first example. You have to pick 2 and then you have the f(x) function here which in this case is e to the x squared dx. And you can check if you take a derivative, you do get e to the t squared. And if you plug in 0, A(2) also get 0. So everything works out accordingly. So there's a little warm up for you before we actually start talking about part two of this fundamental theorem of calculus. Okay, so part of the fund theorem calculus. It will finish our sort of understanding of the relationship between differentiation and integration. So let's put it down. So same thing if I take a function f(x) which is smooth on the closed interval from A to B. The closer from A to B and (x), capital of X is any antiderivative of f(x), then the given function there then the integral from A to B the definite integral from A to B. The net accumulation from A to b of f(x) dx is equal to the antiderivative plugged into b minus the anti derivative plugged in at a. Now, this is a deep theorem. So we're going to push the proof off for a bit, but remember this. What's the point of all this? It highlights the very delicate connection, the very sort of deep connection between what we originally seem to be just different things between finding slopes of tangent lines and net accumulation and just to give credit where credit's due. This result here is usually given to Newton. Sir Isaac Newton knighted and gottfried Wilhelm von Leibniz, and this was back in the day when there was no Internet and stuff so they were building these things up independently. This was their idea. We're going to work through some examples and use this first before sort of talking about why it's true. Let's use this, and this tells you, by the way, what this gives you in terms of just calculations, it gives a way, it gives a process, let's say, a process for evaluating definite integrals. It tells you it's a two-step process, definite integrals, and the process is as followed. So this is, you want to solve, we're going to hand you lots of these, a and b of f(x)dx. And so step one is you find an antiderivative, find an antiderivative, any one you want, F(x), capital F(x), of little f(x), and we'll get better at that in our course. That's not, that's easier said than done sometimes. And then you calculate F(b)-F(a). And once you find that, now you subtract the values, that is the number, that is the answer for the definite integral. And we have some notation that sort of helps us out here, I don't know if you've seen this before. I think I've used it in prior videos. If I have the definite integral of F(x)dx, sometimes we write it as the antiderivative of x but then evaluated from a to b, so this long evaluation bar means plug in the top, and then subtract, and plug in the bottom. Watch out for this bar, this evaluation bar. You plug in the top and then you plug in the bottom. Let's do some examples to do this. We'll start with some easy stuff and again this will get hard and getting better at finding antiderivatives is a big goal for the rest of this course. Let's do an example. Let's take the definite integral from -1 to three of of the line 2x+5. Now again, if you break this off, you can show this geometrically, and we've done this before, but let's actually do this using antiderivatives. The function 2x+5 is our little f(x) and I would love to find a big F(x), a big antiderivative. We can do this because it's an easy function and you can probably do it in your head. What function has derivative 2x? Well that's x^2, derivative of x^2 is 2x. And then what function has the derivative of five? That's 5x. So a nice antiderivative is x^2+5x, and so we'll evaluate that from -1 to three, and all that means is we're just going to plug in. And then we do some arithmetic, so we get (3^2+15), use parentheses here, minus ((-1)^2-5). And you work that out, you get 9+15, so 24, minus one, minus five. Clean it all up and you get 28. So our final answer here is 28. We'll do more of these, but this is just kind of what they look like, the hard part of course is finding antiderivatives, and we will get better at that and come up with rules for that in the next video, but I just want to talk about this use of the constant, where did that C go, and the idea is the freedom that we have to pick any antiderivative, why didn't I pick a plus C? Well, when you do these things you tend to pick the antiderivative, you pick the anti derivative with C=0. And the reason for that, actually it doesn't matter in all honesty, because if I had another arbitrary constant when I workout from a to b, the integral from a to b of f(x)dx, if I had some arbitrary constant, let's just say whatever it is, C is not zero, you can, it'll be F(b), and then you'll have your +C. But then remember you subtract, so let's put that in parantheses. You subtract (F(a)+C). And when you do that, you can rearrange some things. You get F(b)-F(a)+C-C. So the C's you pick, in all honesty, it doesn't matter, they just cancel. They just cancel, so that's why you tend not to see [LAUGH] get it, you tend not to see the constant C, it doesn't impact the value of the definite integral, so we just pick C to be zero to simplify the process because they're going to cancel anyway, so that's why there was no C in the prior one. That's usually a common question that people ask, like where is the C? Let's just put it all together, and then we'll go off and do lots and lots more problems. What is the whole point, what's the power of understanding this thing? The beauty of the fundamental theorem. If I have the definite integral Of f(x)dx, we saw that this is equal to the integral of the derivative of the anti derivative, so it's weird to say, but let me put it in symbols and actually might make more sense. The derivative of the anti derivative dx, and this is, by the fundamental theorem part 2, the anti derivative F(b)- F(a). So from this perspective, you have completed the sort of the beauty of the relationship between a derivatives and integrals. So you see that like definite integral operation, it undoes the operation of differentiation. It completes the argument that the derivative and the definite integral, are inverse operations, and if you remember just to compare it, so this was part 2, just to compare it to see it all on one screen, one of the most beautiful slides in all of calculus. Let's put it altogether, if I have sum, the derivative of the net accumulator function f(x)dx I should say then this is f(t). So these two slides, so remember that this is part 1, the one I just have here and this is part 2, these two relationships, they put it all together. This is sort of this is beautiful, so add this to the things that most beautiful things you've ever seen. You put it right up there with like the most gorgeous sunset, or I don't know some Vista, but this is it. And again like the relationship between tangent lines, and an end approximation, it's not immediately obvious, right, why this is there? So this bridge, is why it's called the fundamental theorem of calculus.