Answer
$$A = \frac{{{\pi ^2}}}{8} - 1$$
Work Step by Step
$$\eqalign{
& x \geqslant x\sin x{\text{ on the interval }}\left[ {0,\frac{\pi }{2}} \right],{\text{ then}} \cr
& {\text{The area is given by}} \cr
& A = \int_0^{\pi /2} {\left( {x - x\sin x} \right)dx} \cr
& {\text{Integrating}} \cr
& A = \int_0^{\pi /2} {xdx} - \underbrace {\int_0^{\pi /2} {x\sin xdx} }_{\operatorname{int} {\text{ by parts}}} \cr
& A = \left[ {\frac{{{x^2}}}{2}} \right]_0^{\pi /2} - \left[ {\sin x - x\cos x} \right]_0^{\pi /2} \cr
& A = \left[ {\frac{{{x^2}}}{2} + x\cos x - \sin x} \right]_0^{\pi /2} \cr
& {\text{Evaluate}} \cr
& A = \left[ {\frac{{{{\left( {\pi /2} \right)}^2}}}{2} + \frac{\pi }{2}\cos \left( {\frac{\pi }{2}} \right) - \sin \left( {\frac{\pi }{2}} \right)} \right] - \left[ 0 \right] \cr
& A = \frac{{{\pi ^2}}}{8} - 1 \cr} $$
