How to find $\int_0^1f(x)dx$ if $f(f(x))=1-x$? A friend of mine gave this question to me : 
Find the value of $\int_0^1f(x)dx$ if $f$ is a real, non-constant, differentiable function satisfying $$f(f(x))=1-x  \space\space\space\space\space\space\forall x\in(0,1)$$ 
The only thing that came to my mind was to form a differential equation with the given equation and find a solution but that was a dead end. If someone could give me a hint on how to tackle this problem it would be great. 
EDIT: All this time wonghang and John Ma's answers had me convinced that this problem was not well thought-out. But like Sanket has pointed out, if we disregard the differentiable assumption for a while, does there now exist an $f$ satisfying the given conditions? Can someone find an example? I tried to produce one using some variants of the absolute value function but to no avail.

I think $f(x)$ does not exist at all. $f(f(x))$ is decreasing on $(0,1)$. 
$f(x)$ is either increasing or decreasing on $(0,1)$. (non-constant, real, differentiable)
If $f(x)$ is increasing, then $f(f(x))$ is increasing.
If $f(x)$ is decreasing, then $f(f(x))$ is increasing.
There is no chance for $f(f(x))$ to be a decreasing function like $1-x$.
