Answer
$\frac{4}{3}$
Work Step by Step
We can evaluate the integral to find the area of the region:
$\int_{0}^{2}(2y-y^2)~dy$
$=(y^2-\frac{y^3}{3})~\vert_{0}^{2}$
$=(2^2-\frac{2^3}{3})-(0^2-\frac{0^3}{3})$
$=(4-\frac{8}{3})-(0)$
$= \frac{4}{3}$
