The relationship between the energy and amplitude of a wave? Derivation?

From multiple online sources I read that

$$E \propto A^2$$ but when I mentioned this in class, my teacher told me I was wrong and that it was directly proportional to amplitude instead.

As far as I know, every website I stumbled upon concerning this said that is the case. My teacher has a Ph.D and seems pretty experienced, so I don't see why he would make a mistake, are there cases where $E \propto A$?

I also saw this derivation:

$$\int_0^A {F(x)dx} = \int_0^A {kx dx} = \frac{1}{2} kA^2$$

located here, does anyone mind explaining it in a bit more detail? I have a basic understanding of what an integral is but I'm not sure what the poster in the link was saying. I know there is a pretty good explanation here, but it seems way too advanced for me (gave up once I saw partial derivatives, but I see that they're basically the same later on). The first one I linked seems like something I could understand.

Answer

The poster from that link is saying that the work done by the spring (that's Hooke's law there: $F=-kx$) is equal to the potential energy (PE) at maximum displacement, $A$; this PE comes from the kinetic energy (KE) and is equal to the integral of Hooke's law over the range 0 (minimum displacement) to $A$ (maximum displacement).

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Anyway, your professor is wrong. The total energy in a wave comes from the sum of the the changes in potential energy,

$$\Delta U=\frac12\left(\Delta m\right)\omega^2y^2,\tag{PE}$$ and in kinetic energy, $$\Delta K=\frac12\left(\Delta m\right)v^2\tag{KE}$$ where $\Delta m$ is the change in mass. If we assume that the density of the wave is uniform, then $\Delta m=\mu\Delta x$ where $\mu$ is the linear density. Thus the total energy is $$E=\Delta U+\Delta K=\frac12\omega^2y^2\,\mu\Delta x+\frac12v^2\,\mu \Delta x$$ As $y=A\sin\left(kx-\omega t\right)$ and $v=A\omega\cos(kx-\omega t)$, then the energy is proportional to the square of the amplitude: $$E\propto\omega^2 A^2$$