 Alright, so let's take a look at finding the areas of regions using what's called a Riemann sum. Important pronunciation note, this word, this is actually the name of a 19th century German mathematician whose, because he was German, pronounced his name Riemann. This is not pronounced Riemann. If you pronounce it Riemann, everybody will know that you didn't learn it from somebody who knew what they were talking about. So the basic problem here is we want to find the area of a region, and so here's a general approach to finding that area. So here I have some regions. So there's my top, bottom, left, right. So it's a bounded region. I can talk about what the area is. And so a typical approach is something like the following. I'm going to divide it up into a whole bunch of little pieces. I'm going to, in each of these pieces, I'm going to draw the largest rectangle that fits inside the sub-region. But here I'm drawing the rectangles that fit inside these sub-regions that I've formed by my partition. Why draw rectangles? Well, the reason is that I know how to find the area of a rectangle, and so I'll find the areas of all those rectangles. And when I add up those rectangles, when I sum those rectangles, the sum of the areas of the rectangles will be less than the actual area of the region. And so in other words, this sum of areas gives me an approximation to the area. And because it's less than the actual area, it's what I'm going to call the lower sum or the lower Riemann sum. And likewise, I can find the upper sum. And it's not going to be too much different. I can find the upper Riemann sum in almost exactly the same way. I'm going to divide the region up into a bunch of pieces. This time I'm going to draw the smallest rectangle that contains the sub-region. So imagine that you're shipping these pieces. You want to find a box that fits. So I'm going to find the smallest box that I can use to ship the pieces, and those boxes might look something like that. And again, these are rectangles, so I know how to find the area of a rectangle. And again, the significance here is the sum of the areas of the rectangles is going to be greater than the actual area of the region. And so I get a second approximation, the Riemann upper sum. Well, what do I know? The area of the region is less than the upper sum, but it's greater than the lower sum. So the actual area of the region will be somewhere between them. Now there's an important qualifier here, which is that we have to have a region that actually has an area. And that's a topic that you will discuss at a later point in mathematics. For right now we'll assume that we have areas, we have regions that actually have something that we could call an area. And in that case, we know that the actual area is going to be someplace between the upper sum, which is a little bit larger than the region, and the lower sum, which will be a little bit smaller than the region. Well, let's find it. So let's find a Riemann sum. Let's approximate the area that's bounded by y equals x squared in the x-axis over the interval 1, 2, 7. And for variety, we'll use the upper sum using n equals 3 subregions. And we'll also find the lower sum over the same subregions. So let's go ahead and sketch what that region looks like. So this is y equals x squared under y equals x squared above the x-axis from x equals 1 to x equals 7. There's top, bottom, left, right. We have a bounded region, so I can at least ask the question, what is the area? Now I want to use the upper sum with n equals 3 subregions. So I want to divide this region into three pieces. So I'm going to divide it out of how about here and here. So I need what are called partition points. So maybe my first partition, I'll occur it, how about x equals 2? That's a nice number. And then I need a second partition point someplace between 2 and 7. How about x equals pi over the square root of 11 times, how about x equals 5, just to make our lives easier? So we'll partition the region at x equals 2, x equals 5, and to find the upper sum, I'm going to find the smallest rectangle that contains each of the subregions. And again, the idea to think about is I want to ship this piece here, this piece in the first subregion. What's the smallest box that I can use? Likewise, to ship this piece, what's the smallest box I can use? And likewise for that piece. So what do those boxes look like? They look a little bit like that. So now I need to know what the areas of these rectangles are. Well, how do you find the area of a rectangle? Well, I need to know the height of the rectangle and I need to know the width of the rectangle. And if I knew both of those, I could find the area. So the first thing to notice is that these rectangles will generally have one vertex that's on the curve. So here, this very first rectangle has a vertex that's right here at this point on the curve of y equals x squared. So if only I knew the coordinates of that point. If only I knew what the x value was at that point. Oh, here it is, x equals 2. I know that x equals 2, y equals x squared. So that tells me that this point right here has coordinates 2 comma 4. And so there's the vertex of that first rectangle. There's one of the vertices of that rectangle. And so while the y value is 4, so that tells me the height distance above the x axis is going to be 4. And the width, this triangle, this rectangle goes from x equals 1 up to x equals 2. So since it extends from x equals 1 to x equals 2, the width is going to be end minus beginning. That's going to have width 1. And what's the area of a rectangle with height 4, width 1, that area is going to be 4. And I'll go ahead and mark that down on my first rectangle. Well, that worked out pretty well. Let's take a look at our second rectangle. So that's going to be a rectangle that covers this region here. And the top point of that rectangle is going to be right here at this point, which is on the graph of y equals x squared at the point where x is equal to 5. So that vertex is going to be located at 525. And my second rectangle will look something like this. Height is 25. Width is, I'm going from x equals 2 to x equals 3. The width is going to be 5 minus 2, 3. And so I have a rectangle with height 25, width 3, in the area 75. And finally, that last rectangle. So that last rectangle is going to have the vertex right up here on y equals x squared at the point where x is equal to 7. So that vertex is going to be the point 749. And the area of that rectangle, height is 49. Width is 7 minus 5. And so the area length times width 98. That's the area of the last rectangle. And if I sum up those areas, I will get an approximation to the area of the region. So 4 plus 75 plus 98 is 177. And that will be my upper sum, which again, is going to be a little bit larger than the actual area of the region. Well, how about the lower sum? And for the lower sum, I'm going to do almost exactly the same thing. I do want to use the same subregions, but this time I'm going to find the largest rectangles that can fit into each of those subregions. So those rectangles look something like that, and I go through the same process. So that very first rectangle, the vertex occurs at a point on y equals x squared, where x is equal to 1. So I know the coordinates of that first point, I know the height of that rectangle. And I know the width of that rectangle, so I know the area of the rectangle. Likewise, second rectangle, the vertex is going to be at a point on the graph of y equals x squared, where x is equal to 2. And I know the coordinates, so that gives me the height. I know the width, and so I know the area. And then that third rectangle, the vertex, is going to be at this point on the graph y equals x squared, where x equals 5. So that tells me what the coordinates are. I know the height, I know the width, I know the area. And the sum of these areas is going to be my lower sum. And a nice picture to keep in mind. I am interested in the area of the region. So here's the area that I want. Well, it's going to be greater than the lower sum. Again, these rectangles that form the lower sum are all completely contained by the region. So whatever this area is, it'll be less than the area of the region I want. And equivalently, also, I'm also going to have this area to be less than the upper sum. Because these rectangles that form the upper sum are larger than the region. They contain the region, and then a little bit more. And because I know those two numbers, the area of the region itself is going to be some place between 63 and 177. Now that leaves a fairly large region interval of possible values for the area. But we'll see if we can make that approximation more precise in the next video.