Answer
$$A = \frac{1}{2}$$
Work Step by Step
$$\eqalign{
& y = x,{\text{ }}y = {x^3} \cr
& x > {x^3}{\text{ on the interval }}\left( {0,1} \right) \cr
& {\text{From the graph and using symmetry properties, we obtain }} \cr
& A = 2\int_0^1 {\left( {x - {x^3}} \right)} dx \cr
& {\text{Integrate and evaluate}} \cr
& A = 2\left[ {\frac{1}{2}{x^2} - \frac{1}{4}{x^4}} \right]_0^1 \cr
& A = 2\left[ {\frac{1}{2}{{\left( 1 \right)}^2} - \frac{1}{4}{{\left( 1 \right)}^4}} \right] - 2\left[ {\frac{1}{2}{{\left( 0 \right)}^2} - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& A = 2\left[ {\frac{1}{4}} \right] - \left[ 0 \right] \cr
& A = \frac{1}{2} \cr} $$
