integration of sum of $f$ and its inverse If $a,b$ are fixed points of $f$, then $\int_a^b (f(x)+f^{-1}(x))dx =b^2-a^2$.
I tried to sub. $u=f(x)$ and $y=f^{-1}$, but it didnt bring me to anywhere. I knew that if $a$ is fixed point of $f$, then it is also a fixed point of $f^{-1}$. Please helps. 
 A: Just notice that 
$$
\int_a^b f
$$
is the area between the graph of the function $f$ and the $x$ axis. Similarly
$$
\int_a^b f^{-1}
$$
is the area between the graph of the function $f$ and the $y$ axis. So if you make the union of the two areas you get the square of side $b$ with a square of side $a$ removed.
See this picture:
http://upload.wikimedia.org/wikipedia/commons/5/59/FunktionUmkehrIntegral2.svg
A: As an alternative to Emanuele's 'graph-ical' solution, you can do the following:
Since, $f(b)=b$ and $f(a)=a$ and $f$ is invertible, it follows that $f^{-1}(b)=b$ and $f^{-1}(a)=a.$
Now, we need $$I=\int_a^b f(x) + f^{-1}(x) \, dx$$
$$I=\int_a^b f(x) \, dx +\int_a^b f^{-1}(x) \, dx$$
In the second integral, set $x=f(t).$ Thus, $t \in [a,b].$ We have,
$$I=\int_a^b f(x) \, dx + \int_a^b t f'(t) \, dt$$
Changing the variable of integration in the second integral from $t$ to $x$, we have
$$I=\int_a^b \left( f(x) + x f'(x) \, \right) dx$$
However, $$ f(x) + x f'(x) dx = d\left(xf(x)\right)$$
Thus,
$$I=\left[xf(x) \right] ^ b_a$$
Finally,
$$I=bf(b)-af(a)=b^2-a^2$$ and we are done.
