1
$\begingroup$

Let $f:[0,2] \rightarrow \mathbb{R}$ continuous and positive such that $\int_{0}^{1} f(x) dx=1$ 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 $F(1)-F(0)= \int_{0}^{1} f(x)dx$

$\endgroup$
1
$\begingroup$

Counterexample: Choose $f$ as above, with $$\int_{1/2}^1f=\int_1^2f=1/3.$$Then $$\int_{x}^{g(x)}f\le2/3 $$for every $x\in[1/2,1]$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.