Area = 2x(1-x) dx – 3(x-1) x dx Part (a) Remember, the AP Test graders don’t expect you to show the work for these integrations, although they do require you to show them your integrals. After that, you can sit back, relax, and let your graphing calculator do the dirty work! Area = f(x) dx – g(x) dx 1 Area = 2x(1-x) dx – 3(x-1) x dx 1 Area = 1.133 units2

We’ll need to use the WASHER method. . . Part (b) y = 2 R = 2 – g(x) r = 2 – f(x) We’ll need to use the WASHER method. Volume = p dx 1 [2 – g(x)] 2 [2 – f(x)] – Volume = p dx 1 [2 – 3(x-1) x ] 2 [2 – 2x(1-x)] – Volume = 16.179 units3

. . Part (c) h(x) = kx (1-x) The key to this part is to not be intimidated by the extreme wordiness!

We don’t know much about h(x), but we do know this… . . Part (c) h(x) = kx (1-x) We don’t know much about h(x), but we do know this… 1) The entire function is above the x-axis between 0 and 1. Since k>0, k must be positive. x is also positive when x is between 0 and 1. Thus, kx must be positive. The quantity (1-x) is positive as well while x is between 0 and 1.

We don’t know much about h(x), but we do know this… . . Part (c) h(x) = kx (1-x) We don’t know much about h(x), but we do know this… 1) The entire function is above the x-axis between 0 and 1. 2) h(x) is a downward-opening parabola. You can tell by distributing the “kx”, which gives us h(x) = kx – kx2. 3) h(x) is incredibly similar to f(x). In fact, if you wanted a visual aide here, you could replace the f(x) graph with an h(x) and still not lose any necessary accuracy.

. . Part (c) y = h(x) h(x) = kx (1-x) h(x) – g(x) h(x) – g(x) The area of this square is [h(x)-g(x)]2. The solid we’re dealing with is made up of many other squares of different sizes, but this area formula works for all of them.

Volume = [h(x) – g(x)]2 dx . . Part (c) y = h(x) h(x) = kx (1-x) h(x) – g(x) h(x) – g(x) You may have noticed that there’s no p here. That’s because we’re not using the disk, washer, or shell method. We’re simply adding up the areas of all the squares. We can do this addition by integrating the area formula. The volume of this region can be calculated by integrating the area formula from 0 to 1. Volume = [h(x) – g(x)]2 dx 1

Volume = [h(x) – g(x)]2 dx Part (c) . . Volume = [h(x) – g(x)]2 dx 1 Part (c) Since we already know that the volume is 15, here’s our final answer. Remember, we don’t need to solve the equation. [ kx(1-x) – 3(x-1) x ] 2 dx = 15 1 Volume = [h(x) – g(x)]2 dx 1