find a curve inducing uniform areas below a given curve Let $f:\mathbb{R}_+\rightarrow \mathbb{R}_+$ be a given monotonically increasing, continuous function.  For example, consider $f(x) = x$.
How can I determine (if one exists) a monotonically increasing function $g:\mathbb{R}_+\rightarrow \mathbb{R}_+$ such that, for every $t\ge 0$,


*

*$g(t) \le f(t)$ and 

*the area of $\{(x,y) : x \ge t, g(x) \le y \le f(t)\}$ equals $f(t)$? 


The diagram below illustrates the area in question, labeled $A$, for a single $t$.  It is the intersection of an axis-parallel rectangle (with upper-left corner $(t, f(t))$, and upper and left boundaries shown with dashed lines in the figure) and the area above the curve $\{(x, g(x)) : x \ge 0\}$.

I would accept an answer for just the special case when $f(x) = x$.  In this case, surely $g(x) = f(x) - \Theta(\sqrt x)$ (for large $x$), but I'm looking for a more precise answer.
For the case when $f(x) = e^x$, the answer appears to be $g(x) = e^{x-c}$ for $c\approx 1.84$.
 A: The question is interesting. Here are some developments. At the end of the answer, I found a closed-form expression for $g$ in the case $f(t)=t$. 
Let us define the function
$$
h(t) = g^{-1}(f(t)),
$$
which defines the argument for which $g$ intersects with $f(t)$, see the figure below. (Just in case, $g^{-1}$ is the inverse of $g$. If $g$ is strictly monotone, then $g^{-1}$ truly exists. But the situation when $g$ has some "flat" parts is also ok, in fact.)

Let us denote the considered set as $A(t)$, i.e.,
$$
A(t) := \{(x,y): x \geq t,~ g(x) \leq y \leq f(t)\},
$$
and denote its area as $|A(t)|$.
By the assumption, we require $|A(t)| = f(t)$ for each $t \geq 0$. 
Let us explicitly calculate $|A(t)|$. We have
$$
f(t) = |A(t)| = f(t)(h(t)-t) - \int_t^{h(t)} g(s) \, ds,
$$
i.e. $|A(t)|$ is the area of the rectangle with sides $f(t)$ and $(h(t)-t)$, minus the area of the set which lies below $g$. 
Therefore,
$$
\tag{1}
\boxed{f(t)(h(t)-t-1) = \int_t^{h(t)} g(s) \, ds
\quad \text{for all } t \geq 0.}
$$
This equation already gives some information about $g$. Namely, we necessarily have
$$
h(t) = g^{-1}(f(t)) \geq t+1,
$$
which implies (by the monotonicity of $g$) that
$$
f(t) \geq g(t+1).
$$
In particular, $f(0) \geq g(1)$. For example, if $f(t)=t$, then $g(t)=0$ for all $t \in [0,1]$, and only for some $t >1$ it starts to grow.
A bit more developed bound can be obtained by estimating the integral on the right-hand side of (1) from below (by the monotonicity of $g$):
$$
f(t)(h(t)-t-1) \geq g(t) (h(t)-t).
$$
In particular, we see the following property: 

If $f(0)>0$, then $f(0) > g(0)$. Or, equivalently, if $f(0)=g(0)$, then $f(0)=g(0)=0$.


Suppose now that $f$ and $h$ are differentiable. Since the equality (1) holds for all $t \geq 0$, we can differentiate both sides  and obtain
$$
f'(t)(h(t)-t-1) + f(t)(h'(t)-1) = g(h(t)) h'(t) - g(t).
$$
Note that, by definition of $h$, we have $g(h(t))=f(t)$. That is,
$$
\tag{2}
\boxed{f'(t)(h(t)-t-1) - f(t) + g(t) = 0
\quad \text{for all } t \geq 0.}
$$
More or less, this is the equation for $g$, under the assumption that $f$ and $h$ are differentiable. But it is not explicit, since it contains the inverse function $g^{-1}$. 
Note that $h$ can be nondifferentiable. Indeed, if we assume that (2) holds all the time, then if $f'(0)=0$, we have $f(0)=g(0)$. However, if, additionally, $f(0)>0$, then we get a contradiction from the property stated above.
So, the nondifferentiability of $h$ occurs when $f'=0$. 
However, I conjecture that if $f'(t)>0$ for all $t$, then (2) is valid. Suppose, this is true. Then, in the particular case $f(t)=t$, we get the following elegant equation for $g$:
$$
g(t) + g^{-1}(t) = 2t+1,
$$
which gives the recurrence formula
$$
t = g(2t+1-g(t)).
$$
It is possible to resolve it step by step. For instance, here are first few steps (see the figure below):
$$
g(t) = 
\left\{
\begin{aligned}
&0, &&t \in [0,1];\\
&\frac{t-1}{2}, &&t \in [1,3];\\
&\frac{2t-3}{3}, &&t \in [3,6];\\
&...
\end{aligned}
\right.
$$
It can be seen that such $g$ is indeed what we need. That is, the application of (2) is ok.

