Answer
$\displaystyle \frac{37}{12}$
Work Step by Step
Graphing the given equations, we find the intersections
or we solve the equation $f(y)=g(y), $ where
$ f(y)=2y \quad$ and $ g(y)=y^{3}-y^{2}\quad$
The solutions are $y=-1,0$ and $2$
When $y\in[c,d]=[-1,0],\ \quad g(y)$ is to the right of $f(y)$ the area between the graphs over this interval is
$A_{1}=\displaystyle \int_{c}^{d} [g(y)-f(y)]dy=\displaystyle \int_{-1}^{0}[y^{3}-y^{2}-2y]dy $
$=\displaystyle \left[\frac{y^{4}}{4}-\frac{y^{3}}{3}-y^{2} \right]_{-1}^{0}=\frac{(0-1)}{4}-\frac{0+1}{3}-(0-1)$
$=\displaystyle \frac{5}{12}$
Over $[c,d]=[0,2],$ the curve $f(y)$ is to the right of $g(y)$
$A_{2}=\displaystyle \int_{c}^{d} [f(y)-g(y)]dy=-\displaystyle \int_{0}^{2}[y^{3}-y^{2}-2y]dy $
$=-\displaystyle \left[\frac{y^{4}}{4}-\frac{y^{3}}{3}-y^{2} \right]_{0}^{2}=-[\frac{(16-0)}{4}-\frac{8-0}{3}-(4-0)]$
$=\displaystyle \frac{8}{3}$
Total area = $A_{1}+A_{2}=\displaystyle \frac{5}{12}+\frac{8}{3}=\frac{37}{12}$
