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. 
 A: By Holder's Inequality,
$$\left(\int_{[0,1]}ff^2dx\right)^2\leq \int_{[0,1]}f^2dx \int_{[0,1]}f^4 dx=\left(\int_{[0,1]}f^3dx\right)^2$$
Equality occurs in this inequality iff $f=af^2$ for some $a\in\mathbb{R}$ i.e., Linearly dependence of $f,f^2$..
$f=af^2$ implies $f^2=af^3$. Considering the integrals and their equality, we see that $f=f^2$.
Which in turn implies that $f(x)=0$ or $1$ for $x\in [0,1]$
A: Note that 
$$\int_0^1 f(x)^2 (1 - f(x))^2 \; dx = \int_0^1 \left(f(x)^2 - 2 f(x)^3 + f(x)^4\right) \; dx = 0$$
so $f(x)(1-f(x))$ must always be $0$.  Since $f$ is continuous, the only solutions are $f(x)=0$ and $f(x)=1$.
A: 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 $f(x)=1$.
