Sketch the region in the first quadrant enclosed by the curves given below.

Question

y=arccos(x/2)

y=(pi/4)(2-x)

Decide whether to integrate with respect to x or y. Then find the area of the region.

In my previous unsuccesful attempts I have gotten -pi/2 and 4-pi. But they are all wrong.

Answer 1

arccos(x/2) = π/4(2 - x); x = 0 or x = 2.

So, area s = ∫₀²(cos^-1(x/2) - π/4(2 - x))dx = [2(cos^-1(x/2) - √1 - x²/4) + π/8(2 - x)²]₀² = 2 - π/2