Answer
Area = $\frac{9}{2}$
Work Step by Step
First find where the two functions intersect.
$ -y= 2y-y^2$
$y(-y+3)=0$ ,Factored form
$y=0,3$
Now Setup the integration using 0 and 3 as the limits and $f(y) \geq g(y) $
$\int^3_0 [y(2-y) +y] dy$
$\int^3_0 (-y^2 + 3y)dy$
$ [-\frac{1}{3}y^3 + \frac{3}{2}y^2]^3_0$
$(-9 + \frac{27}{2})-0$
$\frac{9}{2}$
