SOLUTION 15: Compute the area of the region enclosed by the graphs of the equations $y=\sin x$ and $y=\sin 2x$ on the interval $[0, \pi]$. Begin by finding the points of intersection of the two graphs. From $y= \sin x$ and $y=\sin 2x$ we get that $$\sin x = \sin 2x \ \ \longrightarrow$$ $$\sin x - \sin 2x = 0 \ \ \longrightarrow$$ (Recall the double-angle trigonometry identity $\ \sin 2x = 2 \sin x \cos x$.) $$\sin x - 2 \sin x \cos x = 0 \ \ \longrightarrow$$ $$\sin x (1- 2 \cos x ) = 0 \ \ \longrightarrow$$ $$\sin x =0 \ \ or \ \ 1- 2 \cos x = 0 \ \ \longrightarrow$$ $$\sin x =0 \ \ or \ \ \cos x = \displaystyle\frac{1}{2} \ \ \longrightarrow$$ $$x = 0, x= \pi \ \ or \ \ x=\displaystyle\frac{\pi}{3}$$ Now see the given graph of the enclosed region.

Using vertical cross-sections to describe this region, which is made up of two smaller regions, we get that $$0 \le x \le \displaystyle\frac{\pi}{3} \ \ and \ \ \sin x \le y \le \sin 2x$$ in addition to $$\displaystyle\frac{\pi}{3} \le x \le \pi \ \ and \ \ \sin 2x \le y \le \sin x \ ,$$ so that the area of this region is $$AREA = \displaystyle{ \int_{0}^{\pi/3} (Top \ - \ Bottom) \ dx + \int_{\pi/3}^{\pi} (Top \ - \ Bottom) \ dx }$$ $$= \displaystyle { \int_{0}^{\pi/3} \Big( \sin 2x - \sin x \Big) \ dx + \int_{\pi/3}^{\pi} \Big( \sin x - \sin 2x \Big) \ dx }$$ $$= \displaystyle { \Big( - \frac{1}{2} \cos 2x - \Big( - \cos x \Big) \Big) \Big\vert_{0}^{\pi/3} + \Big( - \cos x - \Big( -\frac{1}{2} \cos 2x \Big) \Big) \Big\vert_{\pi/3}^{\pi} }$$ $$= \displaystyle { \Big( \Big( - \frac{1}{2} \cos \frac{2\pi}{3} + \cos \frac{\pi}{3} \Big) - \Big( - \frac{1}{2} \cos 0 + \cos 0 \Big) \Big) + \Big( \Big( - \cos \pi + \frac{1}{2} \cos 2\pi \Big) - \Big( - \cos \frac{\pi}{3} + \frac{1}{2} \cos \frac{\pi}{3} \Big) \Big) }$$ $$= \displaystyle { \Big( \Big( - \frac{1}{2} \Big( -\frac{1}{2} \Big) + \frac{1}{2} \Big) - \Big( - \frac{1}{2} \Big( 1 \Big) + 1 \Big) \Big) + \Big( \Big( - \Big( -1 \Big) + \frac{1}{2} \Big( 1 \Big) \Big) - \Big( - \frac{1}{2} + \frac{1}{2} \Big( \frac{1}{2} \Big) \Big) \Big) }$$ $$= \displaystyle { \Big( \frac{1}{4} + \frac{1}{2} + \frac{1}{2} - 1 \Big) + \Big( 1 + \frac{1}{2} + \frac{1}{2} - \frac{1}{4} \Big) }$$ $$= 2$$