Part b, the area of r is going to be equal to the integral from 0, that's this point of intersection right over here, to 1 half. So let me make it clear, this is 0 to 1 half. And then the function on the top, so we could just take the area of that, but then we're going to have to subtract from that the area underneath the function below. Or one way to think about it is, the integral from 0 to 1 half of the top function is g of x, which is sine of pi x. So sine of pi x. But if we just evaluated this integral, let me just put a dx over here. If we just evaluated this, we would get the area of this entire region.