find a $\int_{x}^{g(x)} f(t)\,dt = 1$

Let $f : [0, 2] → \mathbb{R}$ be a continuous, positive function and $$\int_{0}^1 f(x)\,dx = \int_{1}^2 f(x)\,dx= 1$$ For each $x \in [0, 1]$, prove that there exists a unique function $g(x) \in [1, 2]$ such that $$\int_{x}^{g(x)} f(t)\,dt = 1$$ and $g \in C^1$. Well, I think in $g(x) = x +1$, but I'm not sure. Any hint is appreciated.

Thank you.

Solution 1:

Hint: let $F(x) = \int_0^x f(x) dx$. $F'\ge 0$ so $F$ increases strictly, so it is a bijection. You look for a $g(x)$ such as $F(g(x)) - F(x)= 1$, which has a unique solution: $$ g(x) = F^{-1}(1+F(x)) $$and it proves that $g$ is $C^1$ with derivative $$ g'(x) = \frac{f(x)}{f(F^{-1}(1+F(x)))} = \frac{f(x)}{f\circ g(x)} $$