Answer
$$A = \frac{{677}}{{10}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^4} - 4{x^2},{\text{ }}g\left( x \right) = {x^2} - 4 \cr
& f\left( x \right) \geqslant g\left( x \right){\text{ on the interval 0}} \leqslant x \leqslant 1 \cr
& g\left( x \right) \geqslant f\left( x \right){\text{ on the intervals }} - {\text{3}} \leqslant x \leqslant 0{\text{ and }}1 \leqslant x \leqslant 3 \cr
& {\text{From the graph shown below, we can define the area enclosed}} \cr
& {\text{as:}} \cr
& A = \int_{ - 3}^0 {\left( {{x^3} - 9x - {x^4} + 9{x^2}} \right)} dx + \int_0^1 {\left( {{x^4} - 9{x^2} - {x^3} + 9x} \right)} dx \cr
& + \int_1^3 {\left( {{x^3} - 9x - {x^4} + 9{x^2}} \right)dx} \cr
& {\text{Integrating}} \cr
& A = \left[ {\frac{{{x^4}}}{4} - \frac{{9{x^2}}}{2} - \frac{{{x^5}}}{5} + 3{x^3}} \right]_{ - 3}^0 + \left[ {\frac{{{x^5}}}{5} - 3{x^3} - \frac{{{x^4}}}{4} + \frac{{9{x^2}}}{2}} \right]_0^1 \cr
& + \left[ {\frac{{{x^4}}}{4} - \frac{{9{x^2}}}{2} - \frac{{{x^5}}}{5} + 3{x^3}} \right]_1^3 \cr
& {\text{Evaluating}} \cr
& A = - \left[ {\frac{{{{\left( { - 3} \right)}^4}}}{4} - \frac{{9{{\left( { - 3} \right)}^2}}}{2} - \frac{{{{\left( { - 3} \right)}^5}}}{5} + 3{{\left( { - 3} \right)}^3}} \right] \cr
& + \left[ {\frac{{{{\left( 1 \right)}^5}}}{5} - 3{{\left( 1 \right)}^3} - \frac{{{{\left( 1 \right)}^4}}}{4} + \frac{{9{{\left( 1 \right)}^2}}}{2}} \right] + \left[ {\frac{{{{\left( 3 \right)}^4}}}{4} - \frac{{9{{\left( 3 \right)}^2}}}{2} - \frac{{{{\left( 3 \right)}^5}}}{5} + 3{{\left( 3 \right)}^3}} \right] \cr
& - \left[ {\frac{{{{\left( 1 \right)}^4}}}{4} - \frac{{9{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( 1 \right)}^5}}}{5} + 3{{\left( 1 \right)}^3}} \right] \cr
& A = \frac{{1053}}{{20}} + \frac{{29}}{{20}} + \frac{{243}}{{20}} + \frac{{29}}{{20}} \cr
& A = \frac{{677}}{{10}} \cr} $$
