Consider a weakly increasing function $f : [0,1] \rightarrow [0,1]$ and define its generalized inverse (since an inverse does not necessariliy exist unless it increases strictly) as $$ f^{-1}(t):= \begin{cases} 1 & \text{if } f(1) <t,\\ \inf\{s \in [0,1] \mid f(s) \leq t\}& \text{otherwise}. \end{cases}$$
Note that, similar to the result on the integral of the inverse of a function, we have
$$\int_0^1 f(x) dx + \int_0^1 f^{-1}(y) dx = bd -ac$$ for $f(a) = c$ and $f(b) = d$.
Can we say anything similar about $$\int_0^1 (f^{-1}(x))^2dx$$ that is, express it somehow as a function of an integral of $f$ only?