 Okay, so now we have a definite integral. What can we do with it? Well, our basic application of the definite integral involves finding the area of a region. So let's say I want to find the area of the region between, oh, I don't know, how about y equals 12 minus x squared and y equals 4x. And again, the way to do this problem wrong is to skip the first step. The way to do the problem correctly is to start off by graphing the region. So there's our region y equals 12 minus x squared and y equals 4x. And note that these two curves, y equals 12 minus x squared, y equals 4x, intersect in two places. Here we find the intersection point is 2,8. Here we find the intersection point is negative 6, negative 24. So we need to graph the region if we're going to try and get the area. We'll sketch a representative rectangle, and we'll find the height of that rectangle. So the height is always top minus bottom. So here the top is the curve 12 minus x squared. The bottom is this curve y equals 4x. So top minus bottom, 12 minus x squared minus 4x. The width, well, that width is going to be a tiny portion of the x axis. That's going to be called dx. And so that says that if I want to find the area, I'm going to take the areas of all these representative rectangles, height of 12 minus x squared minus 4x times width dx, small portion of the x axis. That's the area of one representative rectangle, and sum up all of those rectangles from x equal to negative 6 to x equal to 2. And so there's our limits of integration. And we have a definite integral, and so at this point it's just a question of evaluating it. So I find any antiderivative, how about that one, and I'm going to evaluate the antiderivative at x equals negative 6, x equals 2, and I'm going to find the difference between the two of them, and after all the dust settles I end up with 85 and 1 third as my answer.