Answer
$$A = \frac{5}{2}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = \sin x{\text{ and }}y = \sin 2x \cr
& {\text{Find the intersection points between the graphs let }}y = y \cr
& \sin x = \sin 2x \cr
& \sin x = 2\sin x\cos x \cr
& 1 = 2\cos x \cr
& \cos x = \frac{1}{2} \cr
& {\text{On the interval }}\left( {0,\frac{\pi }{2}} \right){\text{ cos }}x = \frac{1}{2}{\text{ at }}x = \frac{\pi }{3} \cr
& {\text{The total area is given by}} \cr
& A = \int_0^{\pi /3} {\left( {\sin 2x - \sin x} \right)} dx + \int_{\pi /3}^\pi {\left( {\sin x - \sin 2x} \right)} dx \cr
& {\text{Integrating}} \cr
& A = \left[ { - \frac{1}{2}\cos 2x + \cos x} \right]_0^{\pi /3} + \left[ { - \cos x + \frac{1}{2}\cos 2x} \right]_{\pi /3}^\pi \cr
& {\text{Evaluating}} \cr
& A = - \frac{1}{2}\cos 2\left( {\frac{\pi }{3}} \right) + \cos \left( {\frac{\pi }{3}} \right) + \frac{1}{2}\cos 0 - \cos 0 \cr
& \,\,\,\,\,\,\,\,\, - \cos \left( \pi \right) + \frac{1}{2}\cos 2\left( \pi \right) + \cos \left( {\frac{\pi }{3}} \right) - \frac{1}{2}\cos 2\left( {\frac{\pi }{3}} \right) \cr
& A = \frac{1}{4} + \frac{9}{4} \cr
& A = \frac{5}{2} \cr} $$
