Answer
$A = \frac{4}{\pi } - \frac{1}{2}$
Work Step by Step
$$\eqalign{
& y = \sin \left( {\frac{{\pi x}}{2}} \right),{\text{ }}y = {x^3} \cr
& {\text{Graph the curves using Geogebra }}\left( {{\text{Shown Below}}} \right) \cr
& {\text{Find the intersection points using the graph}} \cr
& x = - 1{\text{ and }}x = 1 \cr
& {\text{We have the intervals }}\left[ { - 1,0} \right]{\text{, and }}\left[ {0,1} \right] \cr
& \cr
& {\text{We can find the area integrating with respect to }}x \cr
& A = \int_a^b {\left[ {f\left( x \right) - g\left( x \right)} \right]} dx{\text{ }}\left( {\bf{1}} \right){\text{ }}\left( {{\text{see page 439}}} \right) \cr
& {\text{From the graph }} \cr
& {x^3} \geqslant \sin \left( {\frac{{\pi x}}{2}} \right){\text{ for the interval }}\left[ { - 1,0} \right] \cr
& \sin \left( {\frac{{\pi x}}{2}} \right) \geqslant {x^3}{\text{ for the interval }}\left[ {0,1} \right] \cr
& {\text{Therefore by symmetry}} \cr
& A = 2\int_0^1 {\left( {\sin \left( {\frac{{\pi x}}{2}} \right) - {x^3}} \right)} dx \cr
& {\text{Integrating}} \cr
& A = 2\left[ { - \frac{2}{\pi }\cos \left( {\frac{{\pi x}}{2}} \right) - \frac{1}{4}{x^4}} \right]_0^1 \cr
& {\text{Evaluate the limits}} \cr
& A = 2\left[ { - \frac{2}{\pi }\cos \left( {\frac{\pi }{2}} \right) - \frac{{{{\left( 1 \right)}^4}}}{4}} \right] - 2\left[ { - \frac{2}{\pi }\cos \left( 0 \right) - \frac{1}{4}{{\left( 0 \right)}^4}} \right] \cr
& {\text{Simplifying}} \cr
& A = 2\left[ { - \frac{2}{\pi }\left( 0 \right) - \frac{1}{4}} \right] - 2\left[ { - \frac{2}{\pi } - 0} \right] \cr
& A = - \frac{1}{2} + \frac{4}{\pi } \cr
& A = \frac{4}{\pi } - \frac{1}{2} \cr} $$
