Answer
$$A = \frac{8}{{15}}$$
Work Step by Step
$$\eqalign{
& {\text{From the graph}} \cr
& {\text{We let: }}f\left( x \right) = {x^4} - {x^2}{\text{ and }}g\left( x \right) = {x^2} - {x^4} \cr
& {\text{ }}g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }}\left[ { - 1,1} \right] \cr
& {\text{Using the symmetry of the graph we can define the area as}} \cr
& A = 2\int_0^1 {\left[ {g\left( x \right) - f\left( x \right)} \right]} dx \cr
& A = 2\int_0^1 {\left[ {\left( {{x^2} - {x^4}} \right) - \left( {{x^4} - {x^2}} \right)} \right]} dx \cr
& A = 2\int_0^1 {\left( {{x^2} - {x^4} - {x^4} + {x^2}} \right)} dx \cr
& A = 2\int_0^1 {\left( {2{x^2} - 2{x^4}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = 2\left[ {\frac{{2{x^3}}}{3} - \frac{{2{x^5}}}{5}} \right]_0^1 \cr
& A = 2\left[ {\frac{{2{{\left( 1 \right)}^3}}}{3} - \frac{{2{{\left( 1 \right)}^5}}}{5}} \right] - 2\left[ {\frac{{2{{\left( 0 \right)}^3}}}{3} - \frac{{2{{\left( 0 \right)}^5}}}{5}} \right] \cr
& A = 2\left( {\frac{4}{{15}}} \right) - 2\left( 0 \right) \cr
& A = \frac{8}{{15}} \cr} $$
