Answer
$$A = 8$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^4} - 4{x^2},{\text{ }}g\left( x \right) = {x^2} - 4 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval 0}} \leqslant x \leqslant 1 \cr
& g\left( x \right) \geqslant g\left( x \right){\text{ on the interval 1}} \leqslant x \leqslant 2 \cr
& {\text{From the graph shown below, we can define the area enclosed}} \cr
& {\text{as:}} \cr
& A = 2\int_0^1 {\left[ {{x^4} - 4{x^2} - \left( {{x^2} - 4} \right)} \right]} dx + 2\int_1^2 {\left[ {{x^2} - 4 - \left( {{x^4} - 4{x^2}} \right)} \right]} dx \cr
& A = 2\int_0^1 {\left( {{x^4} - 4{x^2} - {x^2} + 4} \right)} dx + 2\int_1^2 {\left( {{x^2} - 4 - {x^4} + 4{x^2}} \right)} dx \cr
& A = 2\int_0^1 {\left( {{x^4} - 5{x^2} + 4} \right)} dx + 2\int_1^2 {\left( {5{x^2} - 4 - {x^4}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = 2\left[ {\frac{{{x^5}}}{5} - \frac{{5{x^3}}}{3} + 4x} \right]_0^1 + 2\left[ {\frac{{5{x^3}}}{3} - 4x - \frac{{{x^5}}}{5}} \right]_1^2 \cr
& A = 2\left[ {\frac{{{{\left( 1 \right)}^5}}}{5} - \frac{{5{{\left( 1 \right)}^3}}}{3} + 4\left( 1 \right)} \right] + 2\left[ {\frac{{5{{\left( 2 \right)}^3}}}{3} - 4\left( 2 \right) - \frac{{{{\left( 2 \right)}^5}}}{5}} \right] \cr
& - 2\left[ {\frac{{5{{\left( 1 \right)}^3}}}{3} - 4\left( 1 \right) - \frac{{{{\left( 1 \right)}^5}}}{5}} \right] \cr
& A = 2\left( {\frac{{38}}{{15}}} \right) + 2\left( { - \frac{{16}}{{15}}} \right) - 2\left( { - \frac{{38}}{{15}}} \right) \cr
& A = 8 \cr} $$
