 Once we recognize the definite integral as an expression for finding the area of a region, we can evaluate many of these definite integrals directly using simple geometry. So remember our notation for the definite integral corresponds to the area of the region under the graph of y equals f of x, above the graph of y equals g of x, and over the interval from a to b. Now there's one useful variation on this. If I want to find the area of the region under the graph of y equals f of x, above the x-axis, from x equals a to x equals b, then my second curve is y equals zero, and so this f of x minus g of x just becomes f of x, and my definite integral looks like this. And we usually omit the x equals part and simply write this expression. However, the important thing to remember is that when we see this expression, this requires that the limits of integration are in terms of the differentiable variable dx. In other words, this a and this b should be read as x equals a and x equals b. So let's see if we can evaluate some definite integrals. How about this one? So we know the definite integral represents the area of some region, so let's figure out what that region is. So we need to know the under, above, from, and to. So to begin with, we're under y equals 2x and above the x-axis. And we're going to consider the region from x equals zero to x equals four. And since there's no nearby band of rogue mathematicians we have to impress, let's go ahead and sketch this region. And we see that this region is a triangle and so we can calculate the area of this region if we can find the dimensions of this triangle. And we find this triangle as width four and height eight. So its area is going to be one-half base times height, or 16, which will also be the value of the definite integral. How about this definite integral? So again, we're describing a region that's under something, above something, from someplace to someplace. And in this form, the under and the above are easy to pick off and the from and the to are just as easy. The only thing that's difficult is trying to figure out what our region is going to describe. For that we can do a little bit of algebra. We have y equals square root of 25 minus x squared and square roots are generally hard to deal with, so let's square both sides to get rid of them. And if we rearrange our terms a little bit, we'll get the following equation. So now we need to figure out what region is being described by this equation. And the equation itself describes a circle with a radius of five centered at the origin. But because we're looking at the region that's under y equals square root 25 minus x squared and above the x axis, the curve is the upper half circle with radius five and the region itself is going to be one quadrant of the circle. And that means the area of this region will be 25 pi over four.