Answer
$$A = \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& x = {y^3} - y,{\text{ }}x = 0 \cr
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_{ - 1}^0 {\left( {{y^3} - y} \right)} dy + \int_0^1 {\left( {y - {y^3}} \right)} dy \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{1}{4}{y^4} - \frac{1}{2}{y^2}} \right]_{ - 1}^0 + \left[ {\frac{1}{2}{y^2} - \frac{1}{4}{y^4}} \right]_0^1 \cr
& {\text{Evaluating}} \cr
& A = - \frac{1}{4}{\left( { - 1} \right)^4} + \frac{1}{2}{\left( { - 1} \right)^2} + \frac{1}{2}{\left( 1 \right)^2} - \frac{1}{4}{\left( 1 \right)^4} \cr
& {\text{Simplifying}} \cr
& A = \frac{1}{4} + \frac{1}{4} \cr
& A = \frac{1}{2} \cr} $$
