Chapter 8, Problem 12RE

To determine

To find the area of the region enclosed by the lines and curves.

Answer to Problem 12RE

The area of the region enclosed by the lines and curves $y=2\sin x,y=\sin 2x,0\le x\le \pi$ is

  $4$ square units.

Explanation of Solution

Given:

The equation of lines and curves are

  $y=2\sin x,y=\sin 2x,0\le x\le \pi$

Calculation:

Using graphing calculator to graph the given equations

  $y=2\sin x,y=\sin 2x,0\le x\le \pi$

Graph:

  

From graph it is observed that when $0\le x\le \pi$ the graph of $y=2\sin x$ lies above the graph of $y=\sin 2x$

The area of region is

  $A={\displaystyle \underset{0}{\overset{\pi }{\int }}\left(2\sin x-\sin 2x\right)dx}={\left|-2\cos x-\left(\frac{-\cos 2x}{2}\right)\right|}_0^{\pi }=\left(-2\cos \pi -\left(\frac{-\cos 2\pi }{2}\right)\right)-\left(-2\cos 0-\left(\frac{-\cos 0}{2}\right)\right)=4$

Therefore,

the area of the region enclosed by the lines and curves $y=2\sin x,y=\sin 2x,0\le x\le \pi$ is

  $4$ square units.