I'm trying to recover a function based on the area under its curve. The domain of the function itself is in the $[0; 1]$ interval, and all I know is that the area between $0$ and any $x$ value between $0$ and $1$ equals to $1-x$ (eg. $\int_0^{0.1}f(t)dt = 0.9$, $\int_0^{0.3}f(t)dt = 0.7$ etc. $\int_0^1 f(t)dt = 0$. It seems to have a hyperbolic shape and negative function values for some points, can anyone please give me some hints on how I can recover the exact function for this?
-- UPDATE
I've already got an awesome suggestion from @draks with the Heaviside Step Function and I know that as the value of the integral decreases with x increasing, the function I'm looking for must have negative values. However, I wonder whether there exists a function with positive values.
Basically I'm trying to visualize the generalized 80-20 rule (in the form of the x - (100 - x) rule): the set of points in the first 20% (hence $\int_0^{0.2}$) have a collective utility of 0.8, the set consisting of the first 10% will have a collective utility of 0.9 and so on. The main reason I wonder whether such function with positive values exists is whether a specific utility value can be mapped to the specific point lying on the very boundary between the first 20% and the remaining 80% (ie. what is each member's utility in a group or a set based on their ordering).
I think $$ \int_0^x \lim_{\epsilon \to 0^+} \biggr(\delta(y-\epsilon)-1\biggr) dy =\\ \lim_{\epsilon \to 0^+}\int_0^x \biggr(\delta(y-\epsilon)-1\biggr) dy = \lim_{\epsilon \to 0^+} \big(H(y-\epsilon)-y\big)\Biggr|_0^x\\=\lim_{\epsilon \to 0^+} \biggr(H(x-\epsilon)-H(0-\epsilon)-x-0\biggr)=1-x, $$ where $H(x)$ is the Heaviside Step Function, would do. I additionally assume $x>\epsilon$.
So avoid the problems, concerning the half-mean convention, your "function" would look like $$ \lim_{\epsilon \to 0^+} \biggr(\delta(y-\epsilon)-1\biggr) $$ and would actually be a distribution.