 So, we're going to now discuss the fundamental theorem of calculus, which is going to help us find the area of any region that we can describe. Now, before we actually introduce the fundamental theorem of calculus, let's actually frame the question. Being able to write about something, being able to write down the question is actually a fairly important part of the process. So let's take a look at that. So suppose we're looking to find the area of some region, and we have to describe where that region is. So maybe it's between some curve y equals f of x, y equals g of x, and well, there's my top bottom. I need my left and right, so I will also have my left boundary x equals a, and my right boundary x equals b, and that forms a region that I can talk about. Now, with the Riemann sums, we had approximated the area of the region by adding up a whole bunch of rectangles. So let's take a look at what those rectangles might look like. So I'll draw a representative rectangle, as if I were going to find a Riemann sum for this region. So I want to find the area of that rectangle, which means I need to know the height, and I need to know the width. Well, the height of the rectangle, here's the top on y equals f of x, here's the bottom on y equals g of x, and the height is just going to be top minus bottom. So that's just going to be f of x, top minus g of x, bottom, so there's the height of the rectangle. Meanwhile, the width of that rectangle, well, I can describe that width as being a tiny part of the x-axis, this little wedge here, and I'll represent that using this differential dx, and that tells us the little tiny portion, there's the d part, of the x-axis, and that's where the x part comes in. So I have the height of the rectangle, I have the width of the rectangle, and so now I can find the area of the rectangle, and that's just going to be the product of the height, f of x minus g of x, and the width, dx, and that's going to be the area of this one little rectangle in here. Well, if I want the area of the entire region, I can find that by summing all of those rectangles, and I can express that sum as follows, and this gives me something that we call the definite integral. Now there's four key parts of this definite integral. The first part, what's called the integrand, this portion, is actually just going to be the height of that representative rectangle, it's top minus bottom. The second part, the width of the rectangle, little portion of the x-axis, that's going to be our differential, and then we also indicate where the interval starts, x equals a, and where the interval ends, x equals b. And so those are the four parts of the definite integral that correspond to the height of a rectangle, the width of the rectangle, where our region starts, where our region ends. And here's an important check. If the differential, dx, and the limits of the definite integral, x equal to a, x equal to b, they have to be in terms of the same variable. If they are not, you cannot evaluate the integral. Well, not this semester anyway. That's a topic for second semester calculus. Let's take a look at an example. So maybe we want to talk about the area between y equals x squared and y equals four, and let's see if we can express this as a definite integral. Now if you want to do this problem the hard way and quite probably get the wrong answer, you can skip the first step. However, if you want to do this the easy way and most likely set yourself up to find the correct answer, the first step is kind of important. Let's go ahead and graph that. And so here we have our graph of the two curves, y equals x squared, y equals four. And note that our two curves intersect at two points. Now we actually have to find those intersection points there where the curves y equals x squared, y equals four intersect each other. So that's an algebraic process that we're not going to go into here, but you should be able to do that. And let's see. So I have, have I described a region? Well, I have to have a top curve, y equals four. I have to have a bottom curve, y equals x squared. I have to have a left point over here someplace. I have to have a right point over here someplace. So now I have a complete description of the region. Again, if you want to make your life easier and set up the answer correctly, then you probably want to sketch a representative rectangle. And again, these correspond to the rectangles whose areas you'd be summing to find the Riemann sum. So let's go ahead and draw that rectangle. And the importance of this rectangle is it helps to further identify top and bottom. So I want to know the area of that rectangle. So I need to know the height, which is going to be top minus bottom. Top is on y equals four. Bottom is on this curve, y equals x squared. So the height is going to be four minus x squared. The width of that rectangle is going to be some tiny bit of the x axis. Again, we'll represent that as dx. And the area of that we can represent as four minus x squared. That's height times dx, that's width. And so our area of that one little rectangle four minus x squared dx. Well, for the Riemann sum, I want to sum up all of those rectangles. Well, how'd that get in there? Actually, this is how an S would have been written in the latter part of the 16th century, early part of the 17th century. Oddly enough, that's when calculus was being invented. So this symbol, which we usually read as an integral symbol, or maybe as an anti-derivative symbol, it's an S. It is a 17th century S and really refers back to the idea that what we're looking at is a sum of a whole bunch of rectangles from differential and limits have to match, so our differential is x, we have to talk about our x values. We're going to sum those rectangles from x equals negative two up to x equals positive two. So there's x equals negative two to x equals two and our notation is going to look like that. And here's 90% of the problem of finding the area between y equals x squared and y equals four. Once you've set up the definite integral, actually evaluating the definite integral is fairly straightforward. And we'll discuss that in the next lecture.