Showing that $\int^1_0g(x)dx + \int^1_0g^{-1}(x)dx=1$ Let $g:[0,1]\mapsto [0,1]$ be strictly increasing and onto, and let $g$ be continuously differentiable on $(a,b)$ with
$$0<\inf_{0<x<1}h'(x)<\sup_{0<x<1}h'(x)<\infty$$
Let $g^{-1}$ denote its inverse.  Show that 
$$\int^1_0g(x)dx + \int^1_0g^{-1}(x)dx=1$$
Thanks for the help!
 A: This follows from Integration by Parts.
Substituting $g(u)=x$, we get
$$
\begin{align}
\int_0^1g(x)\,\mathrm{d}x+\int_0^1g^{-1}(x)\,\mathrm{d}x
&=\int_0^1g(x)\,\mathrm{d}x+\int_0^1u\,\mathrm{d}g(u)\\
&=\int_0^1g(x)\,\mathrm{d}x+ug(u)\Big]_0^1-\int_0^1g(u)\,\mathrm{d}u\\
&=1\cdot1-0\cdot0\\[9pt]
&=1
\end{align}
$$
A: Hint (Informal proof): $(x,g(x))$ should start with the point $(0,0)$ and end at $(1,1)$. Draw such a $g(x)$, together with the square $[0,1]^2$. $g(x)$ separates this square into two sections. The first integral is the right section, and the second integral is the left section.
A: This is even true if you only assume $g$ continuous and bijective from $[0,1]$ onto $[0,1]$. The argument is given by Lord Soth. 
But that's probably not what the asker has in mind. The assumptions are here for you to make the change of variable $x=g(u)$, $dx=g'(u)du$. This gives
$$
\int_0^1g^{-1}(x)dx=\int_{g^{-1}(0)}^{g^{-1}(1)}g^{-1}(g(u))g'(u)du=\int_0^1ug'(u)du.
$$
Now you can integrate by parts
$$
=ug(u)\Big|_0^1-\int_0^1g(u)du=1-\int_0^1g(u)du.
$$
