When finding the shaded region in the area, we're going to look left to right. We're going to figure out our left sided bound for our X, which is going to be right here where the two intersect at 0 to our right sided bound for X, which is going to occur at this π thirds. So zero to π thirds. We're going to take the top equation, in this case 3 minus the bottom equation, so 3 minus the quantity 2 cosine X + 1. That's all going to be times DX, so 3 -, 1 is going to give us 2 minus cosine XI. Messed up my sines through here. Hold on a second while I fix them all. OK, so 3 minus the quantity 2 cosine X + 1 is going to give us 3 -, 1 or two and -2 cosine X the derivative. The anti derivative of two is going to be 2X2 cosine X -2 cosine X would be -2 sine X. So then when we stick in our upper bound π thirds we get 2 * π thirds -2 sine of π thirds. Sine of π thirds is root 3 / 2. So when we have root 3 / 2 here, two times root 3 / 2, the twos canceled, leaving us a negative root three 2 * 0 -, 2 times sine of 0. Sine of 0 is 0, so we end up with 2π Thirds minus root 3 is a final answer.