 In this video we provide the solution to question number four for practice exam number one for math 1210 in which case we have to find the area bounded by the curves y equals sine of x, y equals cosine of x, x equals zero and x equals pi halves. So it might actually be beneficial to try to sketch a picture of what's happening real quick. So if we have the x-axis here and the y-axis here, we'll also of course be interested in what happens at x equals two halves. So we get something like that. So this first line is when x equals zero, the y-axis. This line here is when x equals pi halves. And so let's see what happens for sine and cosine. If we do sine of x, which comes first, it'll start at the origin, zero, zero, and then as it moves over towards pi halves, it'll reach the level of one. So you get a graph that looks something like the following. Cosine on the hand does the exact opposite. It starts at one. When you head over towards pi halves, it's actually going to decrease towards that x-intercept. And so you get a picture that looks something like this. These two functions will intersect each other. And sure enough, they're going to intersect at x equals pi fourths. That's when sine and cosine are both equal to each other. And there actually turns out to be a symmetry argument in play here. Because the region between zero to pi fourths is actually the same as the region from pi fourths to pi halves. Because the graphs cross each other, you would have to break this up into two separate integrals to treat them separately. For the first one, cosines at top, sines on the bottom. For the second one, sines on the top, cosines on the bottom. But again, by symmetry, I can actually get away with just one of them. Because the area under the curve will be two times the integral from zero to pi fourths. Because like I said, this area is equal to this area right here. But for the first region, like I said, cosine is the one on top and sine is the one on the bottom. So as we integrate this, an antiderivative of cosine would be sine of x. The antiderivative of negative sine is a positive cosine. So we have to evaluate this at zero and at pi fourths. Now, at pi fourths, sine and cosine are both the same. They're both equal to root two over two. For the second one, though, sine of zero is itself equal to zero and cosine of zero is equal to one. So simplifying this, we get two times the square root of two minus one. If you distribute that two, you would get two root two minus two. And so we see that the correct answer is going to be choice C indicated right here.