Answer
$$A = \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph}} \cr
& {\text{We let: }}f\left( x \right) = x{\text{ and }}g\left( x \right) = {x^3} \cr
& {\text{ }}g\left( x \right) \geqslant {\text{ }}f\left( x \right)\,{\text{ on the interval }}\left[ { - 1,0} \right] \cr
& {\text{ }}f\left( x \right) \geqslant {\text{ }}g\left( x \right)\,{\text{ on the interval }}\left[ {0,1} \right] \cr
& {\text{Using the symmetry of the graph we can define the area as}} \cr
& A = 2\int_0^1 {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx \cr
& A = 2\int_0^1 {\left( {x - {x^3}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = 2\left[ {\frac{{{x^2}}}{2} - \frac{{{x^4}}}{4}} \right]_0^1 \cr
& A = 2\left[ {\frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( 1 \right)}^4}}}{4}} \right] - 2\left[ {\frac{{{{\left( 0 \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^4}}}{4}} \right] \cr
& A = 2\left( {\frac{1}{4}} \right) - 2\left( 0 \right) \cr
& A = \frac{1}{2} \cr} $$
