Examples of interesting integrable functions with at least 2 fixed points and an explicit inverse What are some interesting functions I can use to demonstrate this integration trick:
$$\int_a^b [f(x)+f^{-1}(x)]=b^2-a^2$$
I would like to know of some interesting functions where this trick is not obvious. EDIT: The functions I am receiving are the obvious ones like $f(x)=x$, which I don't want. :)
 A: Take $$f(x)=x+\sin x$$
Now, if you are asked to find $$\int_{0}^{\pi}f^{-1}(x)dx^{**}$$Do you know $f^{-1*}$,No!
Puzzled! Use the identity:
$$\int_{0}^{\pi}f^{-1}(x)dx=\pi^2-\int_{0}^{\pi}f(x)dx=\frac{\pi^2}2-2$$ 
Otherwise put you can use it anywhere you can't get $f^{-1}$ in explicit forms.
*in explicit form
**$\small f(0)=0\wedge f(\pi)=\pi$  
A: Well, you can take $f(x)=\sin(\pi x/2)^n$, so that $f^{-1}(x)=2 \arcsin(x^{1/n})/\pi$ and 
$$\int_0^1 f(x)+f^{-1}(x) = 1$$
If you know how to integrate $f$, this allows you to compute the definite integral
$$\int_0^1 \arcsin(x^{1/n})$$ 
But this might not be what you're looking for, since I suspect you want a case where the function $f+f^{-1}$, as opposed to the function $f^{-1}$, looks like something that might occur naturally.  
A: The first obvious one that satisfies your condition is $f(x)=x$.
A: I'll throw one in for kicks. This is the first one that came to mind that isn't directly given in the form you stated. It's nothing mind blowing but here we go.
$\int 2xdx = x^2$, obviously. Now let $f(x) = x$. Then we get
\begin{align*}\int_a^b 2xdx & = \int_a^b 2f(x)dx \\
& = \int_a^b (f(x) + f(x))dx \\
& = \int_a^b (f(x) + f^{-1}(x))dx \\
& = b^2 - a^2 \text{,}
\end{align*}
and of course this simply coincides with exactly what we'd expect from $\int_a^b 2xdx$.
A: Let $a$ and $b$ be greater than or equal to 1. $\int_a^b{e^{x^2}+\sqrt{\ln{x}}dx}=b^2-a^2$
