Assume the only thing you know about an integrable monotonically increasing function $f:[0,1] \rightarrow [0,1]$ is that $f(0) = 0$ and $f(1) = 1$. Consider the family of all those functions.
You then know not more than
$0 \leq \int_0^1 f(x)\ dx \leq 1$
For $f(x) = x$ one has $\int_0^1 f(x)\ dx = \frac{1}{2}$
For each function $f(x)$ with $\int_0^1 f(x)\ dx = F$, the function $f^{-1}(x)$ has $\int_0^1 f^{-1}(x)\ dx = 1 - F$.
So the values of the integrals are somehow symmetrically distributed around $\frac{1}{2}$ probability-wise.
But how are they distributed?
Since $f(x) = x$ leaves more space for variations there will be more functions with $\int_0^1 f(x)\ dx = \frac{1}{2}$ than e.g. with $\int_0^1 f(x)\ dx = 1$, so the probability density of a given value of the integral will not be distributed uniformly around $\frac{1}{2}$.
But how exactly? How to calculate this distribution? Is it a Gauss-like curve, possibly?
Or are these questions ill-posed? How then to do better?