Chapter 7.2, Problem 28E

To determine

To find: The area of the region enclosed by the lines and curve.

Answer to Problem 28E

The area is: $A=4$

Explanation of Solution

Given information:

The curves are:

  $\begin{array}{c}y=2\sin x\\ y=\sin 2x,0\le x\le \pi \end{array}$

Calculation:

Let find the intersection of the graphs of the functions

  $\begin{array}{c}y=2\sin x\\ y=\sin 2x\end{array}$

on the interval $[0,\pi ]$ from the equation:

  $\begin{array}{c}2\sin (x)=\sin (2x)\\ \Rightarrow \sin (2x)=2\sin (x)\cos (x)\\ \Rightarrow \sin (x)=\sin (x)\cos (x)\end{array}$

Consider that,

  $x\in [0,\pi ]$

Therefore, from the fact that,

  $2\sin (x)\ge 2\sin (x)\cos (x)=\sin (2x)$

The integral represents the area of the specified area.

  

Put,

  $t=1-\cos (x)$

Where is

  $\sin x\text{d}x=\text{d}t$

And

Boundaries are $0$ and $2$ .

So get:

  

Therefore, the required area of the region enclosed by the lines and curve is $4$ .