area

\hspace{-16}$Here $\mathbf{y=f(x)=x\sin x}$ and Line $\mathbf{y=x}$\\\\ Now If Function $\mathbf{f(x)}$ is Invertable, Then It is Symmetrical about $\mathbf{y=x}$\\\\ So Coordinate of Point $\mathbf{A}$ is $\mathbf{x\sin x= x\Leftrightarrow x.(1-\sin x)=0\Leftrightarrow x=0,\frac{\pi}{2}}$\\\\ So Coordinate of $\mathbf{O(0,0)}$ and $\mathbf{A\left(\frac{\pi}{2},\frac{\pi}{2}\right)}$\\\\ So Inverse of $\mathbf{f(x)}$ is $\mathbf{f^{-1}(x)=g(x)},$ Which Lie above The line $\mathbf{y=x}$ and\\\\ Symmetrical about $\mathbf{y=x}$\\\\ So Required Area is $\mathbf{=2\int_{0}^{\frac{\pi}{2}}\left(x-x.\sin x\right)dx}$\\\\\\ $\mathbf{=2.\left[\frac{x^2}{2}+x\cos x-\sin x\right]_{0}^{\frac{\pi}{2}}=2\left(\frac{\pi^2}{8}-1\right)}$\\\\\\ So Required Area is $\mathbf{=\left(\frac{\pi^2}{4}-2\right)}$ Sq. -unit

Yes Nishant Sir is saying Right.

Here I am also calculating for [0,pi/2]

in which function f(x) is Invertable

yes. I know it is not invertible. they forgot to give the interval in the question. probably it is [0,pi2] . this came in fiitjee aits ft-3.
@man111 thank you.