 So let's find the areas of some regions. So here's a bounded region, and suppose we want to partition it into four equal-width subregions, and then use these subregions to approximate the area by finding the upper sum, the lower sum, the left sum, and the right sum. Now, since area is a geometric concept, the hardest way possible to solve this problem is to not draw a picture. And if you want to get street cred among wandering bands of rogue mathematicians, you can try to do it that way. But you might instead want to do this problem the easier way and start by sketching a graph of the region. Since we want to partition this region into four equal-width subregions, we might want to figure out how wide those individual subregions are going to be. Now we know the right boundary is at x equals 3 and the left boundary is at x equals 1, so the entire region is 3 minus negative 1, four units wide. And so to partition it into four equal-width subregion, each subregion needs to be one unit wide. And this means that our partition points are going to be located at x equals negative 1, that's the start of the region, 0, 1, 2, and 3. So let's go ahead and draw some vertical lines through those points. Now it'll also be useful to know where these vertical lines intersect the top and bottom curves. So we'll locate those points. Now for this problem, the bottom is always going to be along the x-axis and the top is always going to be along our curve. So we'll locate these points of intersection. And now we're ready to draw some rectangles. To form the upper sum, we want to find the smallest rectangle in each subregion that still contains the subregion. So in this first subregion, we might want to start with a big rectangle and then shrink it down so that it just barely contains the region. And if we do that for each of our subregions, then we get... And now we want to find the area of each of these rectangles. So for the upper sum, there are going to be four rectangles, and to find the area of any rectangle, we need to know its width and height. So the first rectangle has width 1. And since we went to the trouble of making all of these regions equal in width, then all of these rectangles are going to have width 1. What about the height? For the first rectangle, the bottom is on the x-axis and the top runs through this point negative 1, 3. So the height is going to be 3, and so the area of that first rectangle is going to be 3. The second rectangle also has width 1, and the top of the rectangle runs through this point 1, 3, and so the height of the rectangle is also going to be 3. And its area will also be 3. And we can make a similar observation for the height of the third and fourth rectangle, and we can find their areas. And so we can now compute the upper sum as the sum of these areas, which will be 23. Next, to find the lower sum, we'll form the largest rectangle in each region that's still inside the sub-region. So we might imagine starting in the region and expanding until we run out of region. And we find the areas of each of these rectangles. And once we have those areas, we can add them together to form the lower sum. For the left sum, we'll use the largest rectangle whose left side is in the sub-region. And so one way to do this is to go to the left side of the sub-region and then expand from the left side to get our rectangle. So we might do that like this, the other left side. And now we have a bunch of rectangles whose areas we can calculate and add together to form the left sum. And finally, the right sum will be the largest rectangle whose right side is in the sub-region. And so we might sketch out what those rectangles look like and then compute the areas of the individual rectangles. Then, adding these together will give us our right sum. Now all of these give us an approximation to the area of the region. The upper sum is guaranteed to be greater than the area. The lower sum is guaranteed to be less than the area. And the left and right sums are guaranteed to be somewhere around the area. Though we don't know whether they are too big or too small. But these are only approximations. So let's see if we can try to improve our approximations. Suppose we take the lower sums. And we'll make two important simplifications. First, we'll assume that all of our partitions have the same width. And why not? That makes things a lot easier. The other simplifying assumption is that our graph is monotonic. And that's a fancy way of saying it's only going in one direction. Either it's always increasing or it's always decreasing. And while most graphs do increase or decrease at different points, all this really means is that we'll break our graph apart into portions that are monotonically increasing and portions that are monotonically decreasing. So let's partition our region and then draw the lower rectangles. The error of our approximation is the difference between the areas of the rectangles and the actual area of the region. While our lower sum will leave out the portion between the tops of the rectangle and the curve. So we'll represent that gap here. And so a question we might want to ask ourselves is, self, can we find a bound for this area? And rather surprisingly, we can. And this is most obvious through the magic of CGI. And so what we're going to do is we're going to slide all of these error pieces over to one side. And we note that all of these error pieces fit into a rectangle. And so this gives us an upper bound for the error. It's the height of the rectangle times the partition width. Now if you look at this rectangle, the height of the rectangle is based on where the lowest and highest points of the curve are located. And so the height of the rectangle is the same no matter how wide our rectangles are. And so this means we can conclude the following, which is sometimes referred to as Newton's theorem. Suppose f is increasing. Then the error in using the lower sum to approximate the area decreases as the width of the subregions decrease. And with relatively minor modifications, we can see that this also applies if f is decreasing or even piecewise monotonic, or if we're using the upper left or right sum. So it really doesn't matter whether we use the upper sum, the left sum, the lower sum, or the right sum, pick one and work with it. And this suggests the following general approach. We'll partition the interval so that no piece is wider than some amount we'll call delta. And for convenience, we'll make all of our subintervals have this width delta. And then we can find the upper, lower, left, or right sum, your choice. And as long as delta is small, the upper, lower, left, or right sum will approximate the area. Or will it? Well, that's actually a topic for a later class. But for right now and for this class, it turns out that as long as delta is small, we will get a good approximation to the area.