So, as the title describes, I'm trying to find a way to express all $y=f(x)$ differentiable on $[0,1]$ such that the center of mass of the function, assuming it has uniform density, will be a point on the graph of the function. Some obvious examples are $y=0.5$ or $y=x$.
What I tried to do was say that
$$x_{c.m.} = \frac{1}{L} \int_0^1 x\sqrt{1+f'(x)^{2}}dx$$
and
$$y_{c.m.} = \frac{1}{L} \int_0^1 f(x)\sqrt{1+f'(x)^{2}}dx$$
where
$$L= \int_0^1 \sqrt{1+f'(x)^{2}}$$
so I guess my question could also be stated:
Describe all $f(x)$ such that
$$f(\frac{1}{L} \int_0^1 x\sqrt{1+f'(x)^{2}}dx) = \frac{1}{L} \int_0^1 f(x)\sqrt{1+f'(x)^{2}}dx$$
or
$$f(\frac{1}{\int_0^1 \sqrt{1+f'(x)^{2}}dx} \int_0^1 x\sqrt{1+f'(x)^{2}}dx) = \frac{1}{\int_0^1 \sqrt{1+f'(x)^{2}}dx} \int_0^1 f(x)\sqrt{1+f'(x)^{2}}dx$$
There are going to be infinitely many lines that satisfy the condition. For example, as said above, $f(x)=\frac{1}{2}+g(x-\frac{1}{2})$, where g is an odd function.