Chapter 7, Problem 12RE

To determine

To calculate: The area of the region

Answer to Problem 12RE

The area of the region is $4$ .

Explanation of Solution

Given information:

  $y=2\sin x,$

  $y=\sin 2x,$

  $0\le x\le \pi$

Calculation:

To find the area of the region which is enclosed by the sinusoidal curves $y=2\sin x$ and $y=\sin 2x$ for each $x\in [0,\pi ]$ .

The sketch of these graphs will help us to more easily spot a given problem.

Let’s sketch the graphs into Desmos. Further by entering the code, which see in the red rectangle below, mark the required region with the color.

  

By observing the given picture, notice that our area is bounded above with the curve $y=2$ sin $x$ and below with the curve $y=\sin 2x$ from $x=0$ to $x=\pi$ .

Let’s find the area of the green region.

Area of the region is represented as

  $\begin{array}{c}\text{Area}={\displaystyle {\int }_0^{\pi }[2\sin x-\sin 2x]dx}\\ =2{\displaystyle {\int }_0^{\pi }\sin xdx}-{\displaystyle {\int }_0^{\pi }\sin 2xdx}\end{array}$

It is not difficult to solve those integrals but for the second integral it need the substitution.

Using the substitution $2x=t$ has that $\text{d}x=\frac{1}{2} \text{d}t$ and further more limits of integration become

  $\begin{array}{l}x=0\to t=0\\ x=\pi \to t=2\pi \end{array}$

Therefore,

  ${\displaystyle {\int }_0^{\pi }\sin 2x}dx =\frac{1}{2}{\displaystyle {\int }_0^{2\pi }\sin t}dt$

Substitute the above integral in the formula

  $\begin{array}{c}\text{Area}=2{\displaystyle {\int }_0^{\pi }\sin x}dx -\frac{1}{2}{\displaystyle {\int }_0^{2\pi }\sin t}dt \\ =-{\left.2\cos x\right|}_0^{\pi }+\frac{1}{2}{\left.\cos t\right|}_0^{2\pi }\\ =-2\cdot (-1-1)+\frac{1}{2}(1-1)\\ =(-2)\cdot (-2)\\ =4\end{array}$

Area of the region equals $4$ .