Prove that $\int^{1}_{0} f^{-1} = 1 - \int^1_0 f$ One more from hard to believe facts, which I'm curious why are true.
Let $f : [0,1] \rightarrow [0,1] $ is a continuous, monotonically increasing and surjective function
Then 
$$\int^{1}_{0} f^{-1} = 1 - \int^1_0 f$$
Sorry if it was asked, but I can't find. 
I've tried to use the theorem about derivative of reverse function, but I wasn't able today to connect it with this task.
Thanks in advance for help!

Because $f$ is continuous, monotonically increasing and surjective, $f(0) = 0$ and $f(1) = 1$.
Let $u = f^{-1}(x)$:
$$
\int^1_0 f^{-1}(x) \,dx = \int^1_0 u\, df(u) = uf(u)\bigg|^1_0 - \int^1_0 f(u)\,du = 1- \int^1_0 f(x)\,dx.
$$

Just drew a picture using tikz :

The code is as follows, feel free to take and use.
\begin{tikzpicture}[scale=4]
\shade[top color=blue,bottom color=gray!50] plot [smooth, tension=1] 
 coordinates { (0,0) (0.3,0.2) (0.6,0.8) (1,1)} |- (0,0);
\shade[bottom color=cyan,top color=gray!50] plot [smooth, tension=1] 
 coordinates { (0,0) (0.3,0.2) (0.6,0.8) (1,1)} |- (0,1);
\draw (0.7,0.2) node[above]   {$\displaystyle\int_0^{1} f(x)\,dx$};
\draw (0.3,0.7) node[above]   {$\displaystyle\int_0^{1} f^{-1}(y)\,dy$};

\draw[style=help lines] (0,0) grid (1.2,1.2);

\draw[->] (-0.2,0) -- (1.2,0) node[right] {$x$};
\draw[->] (0,-0.2) -- (0,1.2) node[above] {$y$};

\draw plot [smooth, tension=1] coordinates { (0,0) (0.3,0.2) (0.6,0.8) (1,1)};
\end{tikzpicture}

