Let $\mathcal{F}$ be the set of continuous and strictly increasing functions from $[0,1]$ to $[0,1]$ with $f(0)=0$. Is there a closed form for $$\sup_{f\in\mathcal{F},0\le x\le 1}\frac{(1-x)f(x)}{\int_0^1f(t)\,\mathrm{d}t}?$$
That is, the ratio of the following dark area to the integral area.
Consider $f(x)=x^{a}$, for $a\in (0,1)$. Then $$\sup_{x\in [0,1]}\frac{(1-x)f(x)}{\int_0^1f(t)\,dt}=(a+1)\sup_{x\in [0,1]}(1-x)x^a=\left(\frac{a}{a+1}\right)^a.$$ which goes to $1$ as $a\to 0^+$.
Moreover, for all $x\in [0,1]$, and for any strictly increasing functions from $[0,1]$ to $[0,1]$, we have that $$\int_0^1f(t)\,dt\geq \int_{x}^1f(t)dt\geq f(x)\int_x^1dt=f(x)(1-x).$$ So the desired $\sup$ is $1$. Note that the condition $f(0)=0$ is irrelevant.