Question

Sketch the region enclosed by the curves given below. Decide whether to integrate with respect to x or y. Then find the area of the region.
y=4cos(x),y=4−8x/π.

I thought it was the integral of 4cos(x)- the integral of 4- 8x/π on the interval 0 to π/2.

This gave me 4- π which isn't right..

Answer 1

From 0 to π/2, 4cos(x) is above 8-4x/π, so the area would be

∫[0,π/2] (8-4x/π)-4cos(x) dx = 7π/2 - 4

But there is a second symmetric region between π/2 and π, so the area is double that, or 7π-8

see

http://www.wolframalpha.com/input/?i=plot+y+%3D+4cos%28x%29,+y+%3D+4-8x%2Fpi

Evidently you integrated wrong, and didn't see fit to show your work...