Idon't know how to start this. this is for calc 1

corresponds to the area under the curve from 0 to 1. This region is a right triangle with base and height 1, so the area of the triangle, and thus the value of the integral, is .

is the area of one quadrant of the unit circle. Recall that is the equation of the circle with radius 1 centered at the origin. Solving for gives , where the positive root corresponds to the top half of the circle. The top half is defined over the interval , but you're only interested in the right half, . So this area is .

Put these together to get