Sketch the region enclosed by the given curves. Decide whether to integrate with respect to x or y. Then find the area of the region.

Question

2y=3 sqrt x,y=3 and 2y+1x=4

Answer 1

Since the region is roughly triangular, with a horizontal top but a vertex underneath, either way the region has to be broken into two parts. Either

∫[0,1] 3-(2-x/2) dx + ∫[1,4] 3-(3/2 √x) dx
or
∫[3/2,2] (4/9 y^2)-(4-2y) dy + ∫[2,3] 4/9 y^2 dy