 Now we introduce the Riemann sum as the area of a region. This means we can use the fundamental theorem of calculus to find the area of a bounded region. So for example, let's say we want to find the area of the region between y equals 4x plus 2 and the x-axis over the interval between 0 and 4, and let's use calculus. Now remember a bounded region has a top, bottom, left, and right. So let's graph the region. So let's draw our representative rectangle. And we see that our representative rectangle has height y with dx, and so its area, height times width, y dx. The area of the region will be the sum of these areas from where we start, x equals 0, up to where we end, x equals 4. And so we can write our integral, and the differential variable is controlling, so we have to write everything in terms of x. So our upper and lower limits are already in terms of x, x equals 0 and x equals 4. y, on the other hand, well, equals means replaceable, and here we know that y is the same as 4x plus 2. And now we can apply the fundamental theorem of calculus. We'll find our antiderivative of 4x plus 2, and since we're working with a definite integral, we can omit the constant of integration. Let's simplify that just a little bit, and the fundamental theorem of calculus says we can evaluate at 4 and at 0, and then find the difference to get the value of the integral, which gives us the area of the region. Or let's take another one, let's find the area between the graphs of y equals negative x squared plus 2x plus 3, and the x-axis. So we'll graph the region. Now the top is the parabola, at the bottom is the x-axis, and there is a left and right side, which are determined by the intersection points, so we need to find them. So these are going to be the x-intercepts, and they'll occur when y is equal to 0. So we'll solve and we find, and if it's not written down, it didn't happen. Let's go ahead and label these points on our graph. Now let's draw our representative rectangle. No, no. Oh, come on, you're not even trying. Can we get some new interns here who know what a rectangle looks like? There, that wasn't so difficult. So we see that our representative rectangle has height y, with dx, and so its area is y dx, and the area of the region will be the sum of these areas from x equals negative 1 up to x equals 3. And again, the differential variable is controlling, so everything has to be written in terms of x, which means that this y has to be replaced with its equivalent in terms of x. And now we find our antiderivative, and then we evaluate at the endpoints, f of 3 and f of negative 1. And finally, we subtract the two different values.