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Let $f:[-1,1]\to\mathbb{R}$ be an integrable real valued function such that $\left| f(x) \right| \le 2$.

Prove that there exist a $y\in [-1,1]$ such that $\int _{ -1 }^{ y }{ f(t)\,{\rm d}t=4y } $

(Please any hint. I tried using the mean value theorem and mean value theorem for integrals without success)

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Let $F(y) = \int_{-1}^y f(t)dt - 4y$. Note that $F$ is a continuous function.

Note that $F(-1) = 0-(-4) = 4>0$, and $F(1) = \int_{-1}^1 f(t)dt - 4 \leq 2(1-(-1)) - 4 \leq 0$

Hence, using the intermediate value theorem, it follows that there is some $x$ such that $F(x) = 0$, which is to say, $\int_{-1}^x f(t)dt = 4x$.

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  • $\begingroup$ What does absolute continuity have to do with it? $\endgroup$ – zhw. Sep 25 '16 at 23:08
  • $\begingroup$ Oh it doesn't have anything. My apologies. I've edited it to avoid the complication. I am just saying that the stronger condition can be used, namely that the integral is absolutely continuous. $\endgroup$ – астон вілла олоф мэллбэрг Sep 25 '16 at 23:09

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