Let's have some generalization

Let $I=\displaystyle \int_0^1 (x-f(x))^n \, dx$ where $n$ is an even positive integer. Then, using integration by parts, $\begin{aligned} I&=\left[x( x-f(x))^n \right]_0^1 - n \times J \\ Where \quad J&=\int_0^1 x(1-f'(x))(x-f(x))^{n-1}\, dx\\ \end{aligned}$

Note that $f(f(0))=0 \implies f(1)= 0$ .

Now, $J= \int_0^1 x(x-f(x))^{n-1} \, dx - \int_0^1 x(x-f(x))^{n-1} f'(x) \, dx\\ \text{ For the second integral, make the substitution } \\ t=f(x) \implies x=f(t) \quad dt=f'(x)dx$ A little simplification, keeping in mind that $n$ is even, gives the following: $\begin{aligned} J&=I\\ \implies I&=1-nI\\ \implies I&=\frac{1}{n+1} \end{aligned}$

Note: The conditions stated in the question after the second one are irrelevant. It was given so that people who do it (without computational aids) can't cheat ;)