Answer
$\frac{4\pi +12}{3\pi }$
Work Step by Step
The area between two curves $y=f(x)$ and $y=g(x)$ between $x=a$ and $x=b$ is given by the formula:
$$A=\int_a^b |f(x)-g(x)|dx.\quad\quad\quad(1)$$
Given
$$f(x)=\sin (\pi x / 2), \quad g(x)=x^2-2 x$$
Using graph to find the intersection points
$$(0,0),\ \ (2,0) $$ ,
Since $$\sin (\pi x / 2), \geq x^2-2 x,\ \ \ \ 0\leq x\leq 2 $$, then use $a=0$ and $b=2$ in $(1)$:
\begin{aligned}
\text{Area}
& =\int_0^2\left[\sin \left(\frac{\pi x}{2}\right)-\left(x^2-2 x\right)\right] d x \\
& =\left[-\frac{2}{\pi} \cos \left(\frac{\pi x}{2}\right)-\frac{1}{3} x^3+x^2\right]_0^2 \\
& =\left(\frac{2}{\pi}-\frac{8}{3}+4\right)-\left(-\frac{2}{\pi}-0+0\right)\\
&=\frac{4}{3}+\frac{4}{\pi} =\frac{4\pi +12}{3\pi }
\end{aligned}
