 So what happens if one of our boundary curves is not the x-axis? So for example, as opposed to we want to find the area between the graph of y equals 2x plus 3 and y equals x squared minus 2x minus 2. So again, we'll begin by graphing these two and we see we do have a top, the line, the bottom, the parabola, and our left and right are going to be determined by the intersection point. And so we locate the intersection points which will have the same y values. So y has to equal y, but equals means replaceable. So y is x squared minus 2x minus 2 and y is also 2x plus 3. So we solve. And remember the differential variable is controlling. Everything has to be expressed in terms of whatever the differential variable is. So we'll jump about half a minute into the future and see what our differential variable will be. Some, the areas of those rectangles. Oh wait, it looks like my past self has arrived. You'll say, sorry to interrupt, but I just need to check on the differential variable. Wait a minute. I could say something different, you know. Well, you could, but I'd know about it because I would have said it. Oh right, time travel. Sorry to interrupt, but I just need to check on the differential variable. Yep, it's dx. So back to my own time. So thanks for our handy time machine. We know that we only need to worry about the x values. So we don't actually need the y-coordinates, so we won't compute them. But again, if it's not written down, it didn't happen. Let's go ahead and label the beginning and ending with those x values. And we'll draw our representative rectangle. And here we see the top is on the line y equals 2x plus 3. The bottom is on the parabola y equals x squared minus 2x minus 2. And so the height, top, minus bottom. The width is dx. And so the rectangles will have area negative x squared plus 4x plus 5 dx. And we'll sum the areas of those rectangles. Oh wait, it looks like my past self has arrived. You'll say, sorry to interrupt, but I just need to check on the differential variable. Wait a minute, I could say something different, you know. Well, you could, but I'd know about it because I would have said it. Oh right, time travel. Sorry to interrupt, but I just need to check on the differential variable. Yep, it's dx. So back to my own time. You know, I wish I could remember to ask about stock tips. Anyway, where were we? Oh, right. So we can sum the areas of the rectangles from where we start at x equals negative 1 up to where we end at x equals 5. We'll anti-differentiate and then evaluate. Or maybe let's find the area between the graphs of y equals x cubed minus x squared plus 3x and y equals 5x. Now, since every other problem of this type began with graphing, we'll do something completely different. Wait, no, that's not a good idea. Let's go ahead and graph and find the intersection points. So again, where the graphs intersect, the y values are equal, but equals means replaceable. We know that y is equal to x cubed minus x squared plus 3x and y is equal to 5x. So we replace and solve and we get solutions 0, 2, and negative 1. So notice that the area of interest actually consists of two regions. Now in the first region, we draw our representative rectangle and we see that it has a top on y equals x cubed minus x squared plus 3x, a bottom on y equals 5x and a height top minus bottom. The width is a tiny portion of the x-axis, dx. So the rectangles will have area x cubed plus x squared minus 2x times dx and we'll sum the areas of these rectangles from x equals negative 1 to x equals 0. We also have this second region. So again, if we draw our representative rectangle, we see the top is on the line y equals 5x. The bottom is on the line y equals x cubed minus x squared plus 3x and the height top minus bottom negative x cubed plus x squared plus 2x. The width is a tiny portion of the x-axis, dx, and so the rectangles will have area negative x cubed plus x squared plus 2x times dx. And again, we'll sum the areas of these rectangles from x equals 0 up to x equals 2. We'll anti-differentiate, we'll evaluate, and we'll sum to find the area.