Find the area of the region closed by the curves Y equal cosine of the quantity Pi X / 2 and Y equal 1 -, X ^2. When I set those two equations equal, we don't have any good rules as of yet as to how to solve a trig function when it's equaling a polynomial. So a different strategy would be to look at the graph of them. When I look at the graph, I get the Y equal 1 -, X ^2 being a parabola going down, intersecting the X axis at -1 and one and intersecting the Y axis at 0. When we graph Y equal cosine Pi X / 2, we can see that that graph is actually going to go through -1 and one also for the X axis and go through one on the Y axis. Remember that from our trig days, the period of cosine is 2π and then we divide by whatever the coefficient is on the X. So 2π / π halves would give a new period of four. So we would be crossing at -1 and one low point at 2, cross again at 3, high point at 4:00. If we put in more of our graph, that was a three. So that's a one. So seeing some symmetry, if we look at from zero to 1 and multiply by two. Remember if I stick in a negative X for both of them, I'm going to realize that they come out with the exact same equation as the original. Cosine is an even function. So if I stick in a negative X, it just comes out as cosine of the positive of that. And then for the other equation, if I stuck in negative XA, negative X ^2 is really just a positive X ^2. So those are both even functions. We're going to have symmetry. So if I go from zero to 1 and multiply by two, that'll give me the same answers if I went from -1 to one. We need to figure out the top equation. By our graph, we can see the top ones, the parabola 1 -, X ^2 minus, the bottom minus, cosine Pi halves X, all in terms of DX. Taking the integral, we'd get X -, 1/3 X cubed minus sine. Derivative of sine gives me out cosine. But then remember the chain rule. We have to figure out that derivative of the inside piece that Pi halves X which would be π halves and then we divide by it. So we get 2 / π. When we stick in our upper bound minus our lower bound, we get 2 * 1 - 1/3 - 2π sine of π halves minus the quantity 0 - 0 - 2π times sine 0, where we end up with Four Thirds -4 Pi.