Can the area between an arc of the curve $y = \sin kx$ and $x-$axis ever be $2/k$, where $k$ is any positive constant?

Question

Can the area between an arc of the curve $y = \sin kx$ and $x-$axis ever be $2/k$, where $k$ is any positive constant?

I guess it's true. So, I tried to fix values for the limits as $a=0$ and $b=\pi$, then $$\int^{b}_a \sin kx dx=\dfrac{1-\cos kx}{k}.$$

However, I'm not getting it. So, I think I'm missing it. Can anyone help out?

Answer 1

You are asking to solve $$\int^{b}_a \sin kx dx=-\left[\dfrac{\cos kx}{k}\right]_a^b=\frac1k\left({\cos ka}-{\cos kb}\right)=\frac2k$$I.e. we want solutions to $$\cos ka-\cos kb=2$$This is possible only if $\cos ka=1, \cos kb=-1$. That is $$ka=2n\pi, n\in\Bbb N\\kb=(2n+1)\pi,n\in\Bbb N$$Thus, it works for any region where $$a=\frac{2n\pi}k;\,\,\,\,\,b=\frac{(2m+1)\pi}k$$for integers $m,n$. In particular, you can adjust your example by using $a=0$ and say, $b=\pi/k$.