 So here's a way of approaching certain integration problems that will make your life much, much, much easier. And in this particular instance, it follows the general strategy of we can go up to go down. In many cases, it's actually worth changing the direction of integration. For example, suppose I have some curve uv, I don't know what that's going to look like, so here's my generic drawing of a nice curve, and I want to find the integral of uv, the geometrically the area of the region bounded above by the curve, by the axis, and from some point to some place else, it may look something like that. Now, I could find, one of the ways that I might do this is I could find the remaining region, the integral vdu, and then do a little bit of geometry. Again, the key to geometry is scotch tape and scissors. If you want a region, you can figure out how you can cut it from a larger region, or you can decide how you can paste it together from two other regions. So let's see how that might work in practice. So here's a nice problem, find the area between y equals log x and the x axis over some interval, and so I want to do this problem in the hardest way possible and run the risk of getting the wrong answer. So what I'm going to do is I'm going to write down the integral 1 to 4 log x dx, and then I'll evaluate it. Well, maybe I don't want to do this the hardest way possible. Maybe I actually want to not run the risk of getting the wrong answer, so maybe I'll try something a little bit easier. And so our standard approach will begin by graphing our region y equals log x, and again it's a region, so there's a top bottom right and left. There's our top, we go from x equals 1 to x equals 4, and there's the region that we're looking at. And I see that in fact, if I do want to find the area of this region, I do actually want to evaluate the integral for 1 to 4 log x dx. Great, except I might not know what the antiderivative of log x is. So I'm kind of stuck, but let's try a little bit of geometry. The area that I'm interested in, this area here, I can actually find that by starting with a larger area, something like this, and getting rid of the portion that I don't want, getting rid of this bit right here. Now you say, well, that makes the problem harder, doesn't it? I mean, I have to go through all this trouble of not only computing the area of a region here, but oh my god, there's this complicated problem I have to find the area of this region here. Oh, wait, that's a rectangle. I can figure that out pretty easily. So while we've added one extra problem here, it's actually something that's very easy to solve. Now the challenge, of course, is finding the area of the region that we don't want. So I need to express this in terms of something whose bounds I can find. Now let's see, this is the graph of y equals log of x, but it's also the graph of x equals e to the y. And there's a reason why I want to make that change that'll become apparent in a second. Now, again, regions are always bounded by top, bottom, right, and left. So now I have to find that top y equals log of 4. This is x equals 4. The graph is x equals e to the y, y equals log of x. So I know the y value. The bottom is going to be my x-axis, y equals 0, and there's my region. Now, again, if you try to do definite integrals without drawing the representative rectangles, you will probably get them wrong at some point. And they will certainly make your life easier by drawing them. So let's go ahead and do that. I know it's a tedious task to draw a rectangle, but it does make our lives significantly easier. And let's figure out what the area of that rectangle is going to be. So in order to find the area of that rectangle, I need length. And let's see, I have my right-hand side here, which is at some... Well, it's an x value, and my left-hand side is also an x value. Now, I don't want to write x minus x, because that doesn't really tell me anything useful, but I do know that x is equal to e to the y. So this length over here, I'm at e to the y. Over here, I'm at an x value, well, x value 1. So my length of that rectangle, e to the y minus 1, the width of that rectangle is this tiny bit here that's a tiny portion of our y-axis. And if I want to find the area, I'm going to sum up all of those rectangles from my starting point y equals 0 to my ending point y equals log 4. Now, before we go any further, remember that the differential controls the integral. So here, my differential, dy, everything has to be expressed in terms of y. So I have y equals 0, y equals log 4, e to the y. We're good, and this is an integral that will give us the area of the region that we don't want to include. Now, realize that this doesn't really solve our problem. It rather gives us a different problem to solve. So previously, when I wanted to calculate the area directly, I had to find the integral of log of x dx. And again, maybe I didn't know how to find the antiderivative of that function. Now, I have to find the integral of e to the y minus 1, and then I have to do a few other steps. But the question is, can I find the integral of e to the y minus 1? Well, that's actually the world's easiest antiderivative. So yeah, I think I could do that. So the area that I want, remember, this only gives us this area here. So remove this area from the rectangular region here. So the area that I actually want is going to be the whole thing. That is a rectangle, the width of that rectangle. Here I am at x equals 4. Here I am at x equals 1. The width of that rectangle, 4 minus 1. The height of that rectangle, well again, here's x equals 4. So this height must be log 4. So there's the whole thing, and I'll pull out my scotch tape and scissors. I don't actually need the scotch tape. I'll just use the scissors and clip off the portion that I don't want. And that's going to give me the area. And at this point, it is a matter of evaluating a particular algebraic and calculus expression. 4 minus 1 log 4 minus a certain definite integral. And I'll go through the steps on that. Nothing too exciting here. 4 minus 1 is 3. Log 4, this is the world's easiest antiderivative. Well, almost the world's easiest. That minus 1 introduces a complication. But we find our antiderivative. We evaluate at the end points. Log 4 and at 0, we subtract. And after all the dust settles, we end up with 4 log 4 minus 3 as the area of the region.