An evaluation of integral with unknown $f$

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function in $[0, 1]$, strictly monotonic and $f(0)=1$ . If forall $x \in \mathbb{R}$ holds $f\left( f(x) \right)=x$ then evaluate the integral

$$\mathcal{J} = \int_0^1 \left(x - f(x) \right)^{2016} \, {\rm d}x$$

Solution

Well,

\begin{align*}
\int_{0}^{1} \left ( x-f(x) \right )^{2016} \, {\rm d}x &\overset{x=f(u)}{=\! =\! =\! =\! =\!} \int_{1}^{0} \left ( f(u) - u  \right )^{2016} f'(u) \, {\rm d}u \\
 &= - \int_{0}^{1} \left ( u - f(u) \right )^{2016} f'(u) \, {\rm d}u \\
 &= \frac{1}{2}\int_{0}^{1}\left ( u - f(u) \right )^{2016} \left ( 1-f'(u) \right ) \, {\rm d}u \\
 &\!\!\!\!\!\!\!\overset{y=u-f(u)}{=\! =\! =\! =\! =\! =\!} \;  \frac{1}{2} \int_{-1}^{1} y^{2016} \, {\rm d}y \\
 &= \int_{0}^{1} y^{2016} \, {\rm d}y \\
 &= \frac{1}{2017}
\end{align*}

Welcome to 2017!

Remarks: 

1. We might not know what all those functions are but we were able to compute the integral for all those functions. 
2. It is of quite interesting to actually find all continuous functions $f:\mathbb{R} \rightarrow \mathbb{R}$ satisfying:
   
   $$f(f(x))=x \; , \;\; x  \in \mathbb{R}$$