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.
 A: Another point of view: 
For all $C > 1$, let $f$ be a differentiable function on $(0,C+1)$ such that 
$$f(x) = \begin{cases} C+x & \text{if }x \in (0,1) \\ C+1 - x & \text{if }x \in (C, C+1).\end{cases}$$
Then for all $x\in (0,1)$, $f(f(x)) = f(C+x) = C+1 - (C+x) = 1-x$. But 
$$\int_0^1 f(x) dx = C + \frac  12$$
can be any number bigger than $\frac 32$. 
A: The given function is not possible, but if we assume the given function is not differentiable which should have been mentioned then answer would have been $\frac 12$.
$$f(f(x)) =1-x\implies f(1-x)=1-f(x)$$
$$\int_0^1 f(x)dx=1-\int_0^1f(1-x)dx=1-\int_0^1 f(x)dx\implies \int_0^1 f(x)dx=\frac 12$$
A: 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 $1-x$.
