Answer
$$ \frac{3\pi^{2}}{8} -1 $$
Work Step by Step
The area enclosed between two curves $y=f(x)$ and $y = g(x)$ over $[ a,b]$ such that $f(x) \geq g(x)$ is $$ A = \int_{a}^{b} f(x) - g(x)dx$$
From the graph, we can see that $x\gt \sin x$ on the interval $\bigg[\frac{\pi}{2}, \pi \bigg]$
Therefore, the Area between the two curve is $\int_{\frac{\pi}{2}}^{\pi} x - \sin (x) dx$
$$=\left[\frac{x^{2}}{2}+ \cos (x) \right]^{\pi}_\frac{\pi}{2}$$
$$= \bigg[ \frac{\pi^{2}}{2} + \cos (\pi) \bigg] - \bigg[\frac{\pi^{2}}{8} + \cos (\frac{\pi}{2})\bigg]$$
$$ = \bigg[ \frac{\pi^{2}}{2} -1 \bigg] - \bigg[\frac{\pi^{2}}{8} + 0 \bigg]$$
$$ = \frac{\pi^{2}}{2} - \frac{\pi^{2}}{8} -1 $$
$$= \frac{3\pi^{2}}{8} -1 $$