Solve these functional equations: $\int_0^1\!{f(x)^2\, \mathrm{dx}}= \int_0^1\!{f(x)^3\, \mathrm{dx}}= \int_0^1\!{f(x)^4\, \mathrm{dx}}$ 
Let $f : [0,1] \to \mathbb{R}$ be a continuous function such that
  $$\int_0^1\!{f(x)^2\, \mathrm{dx}}= \int_0^1\!{f(x)^3\, \mathrm{dx}}= \int_0^1\!{f(x)^4\, \mathrm{dx}}$$
  Determine all such functions $f$.

So far, I've managed to show that $\int_0^1\!{f(x)^t\, \mathrm{dx}}$ is constant for $t \geq 2$. Help would be appreciated. 

Calculate the difference of first and second integral then second and third.
$$\int_0^1f^2(x)-f^3(x)dx=0$$
$$\int_0^1f^3(x)-f^4(x)dx=0$$
Subtract both equations:
$$\int_0^1f^2(x)-2f^3(x)+f^4(x)dx=0$$
$$\int_0^1f^2(x)\left[1-2f(x)+f^2(x)\right]dx=0$$
$$\int_0^1f^2(x)(1-f(x))^2dx=0$$
Look at the integrand it consists of the product of two squares. It can only become $0$ if $f(x)=0$ or $f(x)=1$.
