Answer
$$A = \frac{{128}}{{15}}$$
Work Step by Step
$$\eqalign{
& y = {x^4} - 2{x^2},{\text{ }}y = 2{x^2} \cr
& {\text{Let }}f\left( x \right) = {x^4} - 2{x^2}{\text{ and }}g\left( x \right) = 2{x^2} \cr
& {\text{From the graph shown below}} \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the interval }} - 2 \leqslant x \leqslant 2 \cr
& {\text{The enclosed area is given by }} \cr
& A = \int_{ - 2}^2 {\left[ {2{x^2} - \left( {{x^4} - 2{x^2}} \right)} \right]dx} \cr
& {\text{By symmetry}} \cr
& A = 2\int_0^2 {\left[ {2{x^2} - \left( {{x^4} - 2{x^2}} \right)} \right]dx} \cr
& A = 2\int_0^2 {\left( {4{x^2} - {x^4}} \right)dx} \cr
& {\text{Integrate}} \cr
& A = 2\left[ {\frac{{4{x^3}}}{3} - \frac{{{x^5}}}{5}} \right]_0^2 \cr
& A = 2\left[ {\frac{{4{{\left( 2 \right)}^3}}}{3} - \frac{{{{\left( 2 \right)}^5}}}{5}} \right] - 2\left[ {\frac{{4{{\left( 0 \right)}^3}}}{3} - \frac{{{{\left( 0 \right)}^5}}}{5}} \right] \cr
& {\text{Simplifying}} \cr
& A = \frac{{128}}{{15}} \cr} $$
