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Let $f$ is function with continuous derivatives, such that $f(\sqrt{2})=2$ and for any real numbers $x$, $$f(x)=\lim_{t \to 0}{\dfrac{1}{2t}\int_{x-t}^{x+t}sf'(s)ds}.$$

I have tried but not able able to get the correct solution by Leibniz's Rule but got correct answer by " Fundamental Theorem Of Calculus ".Thanks for help in advance.

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  • $\begingroup$ Please restate the first sentence. It doesn't make much sense as it stands. $\endgroup$
    – zhw.
    Commented Aug 18, 2017 at 15:54
  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ Commented Aug 21, 2017 at 10:20

2 Answers 2

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Note that by the Fundamental Theorem Of Calculus, \begin{align*} \lim_{t \to 0}\dfrac{1}{2t}\int_{x-t}^{x+t}sf'(s)ds&= \dfrac{1}{2}\lim_{t \to 0}\dfrac{1}{t}\int_{x}^{x+t}sf'(s)ds +\dfrac{1}{2}\lim_{t \to 0}\dfrac{1}{(-t)}\int_{x}^{x-t}sf'(s)ds\\ &=\dfrac{1}{2}xf'(x)+\dfrac{1}{2}xf'(x)=xf'(x). \end{align*} Hence $f$ solves the differential equation $f(x)=xf'(x)$ which implies $f(x)=cx$ with $c\in\mathbb{R}$. By using the initial condition $f(\sqrt{2})=2$ we obtain that $f(x)=\sqrt{2}x$.

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  • $\begingroup$ How can this help? $\endgroup$
    – Nosrati
    Commented Aug 18, 2017 at 17:02
  • $\begingroup$ @MyGlasses Is it better now? $\endgroup$
    – Robert Z
    Commented Aug 18, 2017 at 17:14
  • $\begingroup$ The last equality in the set of equations should just be $f(x)$, no? $\endgroup$
    – Harry
    Commented Aug 18, 2017 at 17:29
  • $\begingroup$ @Harry Corrected. Thanks. $\endgroup$
    – Robert Z
    Commented Aug 18, 2017 at 17:32
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Hint: Use Hospital rule and with Leibniz integral rule $$f(x)=\lim_{t \to 0}\dfrac{\int_{x-t}^{x+t}sf'(s)ds}{2t}=\dfrac{xf'(x)+xf'(x)}{2}=xf'(x)$$

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  • $\begingroup$ Nice approach +1. $\endgroup$
    – Robert Z
    Commented Aug 18, 2017 at 17:52

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