One Arch Area of Sine

Problem 191: Show that if $k$ is a positive constant, then the area between the $x$ -axis and one arch of the curve $y = \sin kx$ is $2/k$ .

Solution: One arch of the curve $y = \sin kx$ is from 0 to $\pi/k$ . That is,

(1) $\begin{align*} A & = \int_{0}^{\frac{\pi}{k}} \sin(kx) \ dx \\ & = -\frac{\cos kx}{k} \bigg|_{0}^{\pi/k} \\ & = -\frac{1}{k}\left[\cos\left( k \cdot \frac{\pi}{k}\right) - \cos\left(k \cdot 0 \right)\right] \\ & = -\frac{1}{k} \left( \cos(\pi) - \cos(0) \right) \\ & = -\frac{1}{k}(-1-1) \\ & = -\frac{1}{k}(-2) \\ & = \frac{2}{k}.\end{align*}$

That is, the area between the $x$ -axis and one arch of the curve $y = \sin kx$ is $2/k$ as desired.

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