# $\int_0^1 f(x) \,dx=\int_1^2 f(x) \, dx=1$ [closed]

Let $$f:[0,2] \rightarrow \mathbb{R}$$ continuous and positive such that $$\int_0^1 f(x) \, dx=\int_1^2 f(x) \, dx=1$$ for each $$x \in [0,1]$$ prove that there is a unique $$g(x) \in[1,2]$$ such that $$\int_x^{g(x)} f(t) \, dt=1$$ prove that the function $$g:[0,1] \rightarrow \mathbb{R}$$ is the class $$\mathbb{C}^1$$

by the fundamental theorem of calculus

$$F(1)-F(0)= \int_0^1 f(x) \, dx= \int_1^2 f(x)=F(2)-F(1)=1$$ then $$2F(1)=F(2)+F(0)$$.

• Consider the function $F(y) = \int_x^{y} f(t)dt$ for any fixed $x\in[0,1]$. What can you say about $F(y=x)$ and what can you say about $F(y=2)$? You want to compare these two values to $1$ and apply the intermediate value theorem. – Winther Sep 16 at 23:51

If $$x \in [0, 1]$$ and $$g(x) \in [1, 2]$$, then $$\int_x^{g(x)}f(t)\,dt = \int_x^1 f(t)\,dt + \int_1^2 f(t)\,dt -\int_{g(x)}^2 f(t)\,dt = 1$$

This simplifies to showing that $$\int_x^1 f(t)\,dt = \int_{g(x)}^2 f(t)\,dt \tag 1$$ has a solution $$g(x) \in [1, 2]$$. Since the LHS ranges over $$[0, 1]$$, it is sufficient to show that a $$g(x)$$ exists such that $$\int_{g(x)}^2 f(t) \, dt$$ ranges over $$[0, 1]$$. This is easy to show using the intermediate value theorem since $$\int_1^2 f(t) \, dt = 1$$, $$\int_2^2 f(t)\,dt = 0$$, and $$f(x)$$ is continuous.

Now for continuity and differentiability: Let $$F(x) = \int_0^x f(t)dt$$. Then you are trying to show that there exists a continuous and differentiable $$g(x) \in [1, 2]$$ such that $$F(g(x))-F(x)=1$$

The solution in $$g(x)$$ would be $$g(x) = F^{-1}(1+F(x))$$ where $$F^{-1}(x)$$ denotes the inverse of $$F(x)$$. Note that $$F(x)$$ is strictly increasing (since $$f(x)>0$$) so $$F^{-1}(x)$$ would also be continuous. Then since the composite of two continuous functions is also continuous, $$g(x)$$ would also be continuous. The derivative of $$g(x)$$ would be $$g'(x) = \frac{f(x)}{f(F^{-1}(1+F(x)))} = \frac{f(x)}{f(g(x))}$$

which would be defined for $$x \in [0, 1]$$.

• Please, could you clarify the LHS ranges over $[0,1]$, it is sufficient to show that a continuous $g(x)$ exists such that $\int_{g(x)}^{2} f(t) dt$ ranges over $[0,1]$. what does it mean LHS ranges? – Mayra Isabel Ferreira Ortiz Sep 16 at 21:58
• @MayraIsabelFerreiraOrtiz I was saying that the left hand side (LHS) of $(1)$ would be in $[0, 1]$. – Varun Vejalla Sep 16 at 22:06
• why is it easy to show that $g$ exists, I still don't understand why using $\int_{1}^{2} f(t)dt=1, \int_{2}^{2} f(t) dt=0$ and $f(x)$ is continuous, Please could you explain it to me? – Mayra Isabel Ferreira Ortiz Sep 16 at 22:13
• Let $h(x)=\int_x^2 f(t) dt$. Then $h(x)$ must be continuous since the integral of a continuous function would also be continuous and since $h(1) = 1$ and $h(2) = 0$, by the intermediate value theorem, $h(x) = y$ would have some solution $x$ for all $y \in [0, 1]$ – Varun Vejalla Sep 17 at 0:48
• It is not so clear why $g$ is continuous and differentiable. Can you include the proof? – Arctic Char Sep 17 at 3:54