Answer
$\frac{4}{\pi}-1$
Work Step by Step
Step 1. Find the intersection between the two functions; let $sin(\frac{\pi x}{2})=x$ and we have $x=\pm1$ as shown in the figure.
Step 2. The total enclosed area contains two integrals: $A=\int_{-1}^0(x-sin(\frac{\pi x}{2}))dx+\int_0^1(sin(\frac{\pi x}{2})-x)dx=\int_{-1}^0xdx- \int_{-1}^0\frac{2}{\pi}sin(\frac{\pi x}{2}))d((\frac{\pi x}{2}))+\int_0^1\frac{2}{\pi}sin(\frac{\pi x}{2}))d((\frac{\pi x}{2}))-\int_0^1xdx=\frac{1}{2}x^2|_{-1}^0+\frac{2}{\pi}cos(\frac{\pi x}{2})|_{-1}^0-\frac{2}{\pi}cos(\frac{\pi x}{2})|_{0}^1-\frac{1}{2}x^2|_{0}^1=-\frac{1}{2}(-1)^2++(\frac{2}{\pi}cos(\frac{0}{2})-\frac{2}{\pi}cos(\frac{-\pi}{2}))-(\frac{2}{\pi}cos(\frac{\pi}{2})-\frac{2}{\pi}cos(\frac{0}{2}))-\frac{1}{2}(1)^2=\frac{4}{\pi}-1$
