The area of the region bounded by the curve \(y=\sin^{-1}(\sin x),0\leq x\leq \pi\) and \(x-\) axis is
Answer & explanation
Correct answer: option 3
\(\begin{aligned}\sin^{-1}(\sin x)&=\left\{\begin{aligned}x&:0\leq x\leq \frac{\pi}{2}\\ \pi-x &:\frac{\pi}{2}\leq x\leq \pi\end{aligned}\right.\\ \text{Area}&=\int_{0}^{\frac{\pi}{2}}xdx+\int_{\frac{\pi}{2}}^{\pi}(\pi-x)dx\\ &=\frac{\pi^{2}}{8}+\frac{\pi^{2}}{2}-\left[\frac{\pi^{2}}{2}-\frac{\pi^{2}}{8}\right]\\ &=\frac{\pi^{2}}{4}\end{aligned}\)