 A useful strategy in life, and in math, is look at a problem from different directions. We've been looking at areas by considering the sum of rectangles oriented vertically, but what if we oriented the rectangles horizontally? Well, if you don't play, you can't win. Let's see what happens. Now, why would we want to do that? Well, let's consider the following integral. We need to find an anti-derivative of log x, but we don't know it. Question in the back. The last time I took this course, we learned this. Well, yes, that's true. We will learn how to do this eventually, but remember, a problem exists whether or not you know how to solve it, so let's try and solve this now. Now, remember, one interpretation of the definite integral is that it's the area between y equals f of x and the x-axis over the interval from a to b. And so this integral corresponds to the area between y equals log x and the x-axis over the interval from 1 to 2. So let's draw a graph of our region, and if it's not written down, it didn't happen. Label the graph. If we try to find the area of the region by summing horizontally-oriented represented rectangles, we get the following. So the length of each rectangle, well, that's end minus beginning. But since our rectangles are running horizontally, these are the x-coordinates of the end and beginning. So regardless of where the rectangle is, the end is always at x equals 2. So that means the end is a constant 2, and we can use the value 2. Meanwhile, the beginning is sum x-coordinate, so we'll just use x. And the height of the rectangle is some small portion of the y-axis dy. So these rectangles have width 2 minus x, height dy, and we'll sum them up from x equals 1 to x equals 2. So remember the differential variable is the only variable allowed since our differential is dy, then everything has to be rewritten in terms of y. Since y equals log x, then x equals e to the y. We also have these limits x equals 1 to x equals 2, and those also have to be rewritten in terms of y. So x equals 1 since x equals e to the y equals means replaceable, then solve. Likewise, x equals 2 becomes, and so we can rewrite our integral as. And at this point, this is an integral we can evaluate.