Chapter 6.4, Problem 63E

To determine

Area between x − axis and one arch of the curvey is always $2/k$ .

Answer to Problem 63E

Between x − axis and one arch of the curve y,

  $\text{Area}=\frac{2}{k}$

Explanation of Solution

Given information:

  $y=\sin kx$

Where,

k is a positive constant.

The period of $\sin kx$ is $\frac{2\pi }{k}$ .

Such that

One arch of $\sin kx$ is from $x=0$ and $x=\frac{\pi }{k}$ .

Then

The area of one arch,

  $\begin{array}{l}\text{Area}={\displaystyle {\int }_0^{\frac{\pi }{k}}\sin kxdx}\\ ={-\frac{1}{k}\cos kx|}_0^{\frac{\pi }{k}}\\ =-\frac{1}{k}\cos \left(k\cdot \frac{\pi }{k}\right)+\frac{1}{k}\cos \left(k\cdot 0\right)\\ =-\frac{1}{k}\cos \pi +\frac{1}{k}\cos 0\\ =-\frac{1}{k}\left(-1\right)+\frac{1}{k}\left(1\right)\\ =\frac{1}{k}+\frac{1}{k}\\ =\frac{2}{k}\end{array}$