Answer
$A = 8$
Work Step by Step
$$\eqalign{
& {\text{From the graph we can see that }}y = x{\text{ is above }}y = {x^3} - 3x,{\text{ }} \cr
& {\text{on the interval }}\left[ {0,2} \right],{\text{and the area is equivalent to the interval}} \cr
& \left[ { - 2,0} \right]. \cr
& {\text{Let }}f\left( x \right) = x{\text{ and }}g\left( x \right) = {x^3} - 3x{\text{, on the interval }}\left[ {0,2} \right] \cr
& \cr
& {\text{Use }}A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx,{\text{ with }}f\left( x \right) \geqslant g\left( x \right) \cr
& x = a = 0{\text{ and }}x = b = 2 \cr
& {\text{Therefore}}{\text{,}} \cr
& \underbrace {A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx}_ \Downarrow \cr
& A = \int_0^2 {\left[ {x - \left( {{x^3} - 3x} \right)} \right]} dx \cr
& {\text{By symmetry on the interval }}\left[ { - 2,0} \right]{\text{ we can express the}} \cr
& {\text{enclosed area as}} \cr
& A = 2\int_0^2 {\left[ {x - \left( {{x^3} - 3x} \right)} \right]} dx \cr
& A = 2\int_0^2 {\left( {x - {x^3} + 3x} \right)} dx \cr
& A = 2\int_0^2 {\left( {4x - {x^3}} \right)} dx \cr
& {\text{Integrate}} \cr
& A = 2\left[ {2{x^2} - \frac{1}{4}{x^4}} \right]_0^2 \cr
& A = 2\left[ {2{{\left( 2 \right)}^2} - \frac{1}{4}{{\left( 2 \right)}^4}} \right] - 2\left[ {2{{\left( 0 \right)}^2} - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& {\text{Simplify}} \cr
& A = 2\left( 4 \right) - 2\left( 0 \right) \cr
& A = 8 \cr} $$
