Franklin’s Blog

Area Under a Sine Arch

2017 Mar 9

I noticed something funny the other day.

Consider a sine graph with equation \(y=c*\sin(cx)\), where \(c\) is a constant. The period of such a sine wave is \(\frac{2\pi}{c}\), so the area under a single one of its arches is
\[A=\int_0^{\frac{\pi}{c}} c\sin cx \,dx\]

Which is
\[A=-\cos \pi + \cos 0\]

That struck me as a little odd. Regardless of the value of \(c\), the area underneath a single arch was always \(2\)? But I suppose it makes sense, because as \(c\) increases, each arch is stretched vertically but compressed horizontally. However, it gets weirder. If \(y=f'(x)\sin f(x)\), the area is still \(2\), because the x-coordinates of the start and endpoints of the nth arch are \(f^{-1}(\pi n-\pi)\) and \(f^{-1}(\pi n)\) for integral \(n\), the area under the nth arch is
\[A=\int_{f^{-1}(\pi n-\pi)}^{f^{-1}(\pi n)} c\sin cx \,dx\]

Which simplifies to
\[-\cos f(f^{-1}(\pi n))+\cos f(f^{-1}(\pi n-\pi))\]
\[-\cos \pi n+\cos (\pi n-\pi)\]

We are interested in only positive areas, and the positive difference between \(-\cos \pi n\) and \(\cos (\pi n-\pi)\) is the difference between two consecutive peaks of the cosine arch, which is also always \(2\).

That means that all of the areas underneath all of the arches in the following pictures is equal to \(2\):

Fig 1
Fig 2
Fig 3