Multiple-choice integration, picture included, include work

Recall that a circle of radius 2 centered at the origin has equation



where the positive root gives the top half of the circle in the x-y plane. The definite integral corresponds to the area of the right half of this top half. Since the area of a circle with radius is , it follows that the area of a quarter-circle would be .

You have , so the definite integral is equal to .

Another way to verify this is to actually compute the integral. Let , so that . Now



Recall the half-angle identity for cosine:



This means the integral is equivalent to

Hmm confusing don't understand