Answer
$$A = 8$$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can note that the area is given by}} \cr
& A = \int_{ - 2}^0 {\left[ {\left( {{x^3} - 4x} \right) - 0} \right]} dx + \int_0^2 {\left[ {0 - \left( {{x^3} - 4x} \right)} \right]} dx \cr
& A = \int_{ - 2}^0 {\left( {{x^3} - 4x} \right)} dx + \int_0^2 {\left( {4x - {x^3}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = \left[ {\frac{1}{4}{x^4} - 2{x^2}} \right]_{ - 2}^0 + \left[ {2{x^2} - \frac{1}{4}{x^4}} \right]_0^2 \cr
& {\text{Evaluating}} \cr
& A = - \left[ {\frac{1}{4}{{\left( { - 2} \right)}^4} - 2{{\left( { - 2} \right)}^2}} \right] + \left[ {2{{\left( 2 \right)}^2} - \frac{1}{4}{{\left( 2 \right)}^4}} \right] \cr
& {\text{Simplifying}} \cr
& A = 4 + 4 \cr
& A = 8 \cr} $$
