0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Please restate the first sentence. It doesn't make much sense as it stands. $\endgroup$ – zhw. Aug 18 '17 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?. $\endgroup$ – projectilemotion Aug 21 '17 at 10:20
4
$\begingroup$

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 $f(x)=\sqrt{2}x$.

$\endgroup$
4
  • $\begingroup$ How can this help? $\endgroup$ – Nosrati Aug 18 '17 at 17:02
  • $\begingroup$ @MyGlasses Is it better now? $\endgroup$ – Robert Z Aug 18 '17 at 17:14
  • $\begingroup$ The last equality in the set of equations should just be $f(x)$, no? $\endgroup$ – Harry Aug 18 '17 at 17:29
  • $\begingroup$ @Harry Corrected. Thanks. $\endgroup$ – Robert Z Aug 18 '17 at 17:32
2
$\begingroup$

Hint: Use Hospital rule and with Leibniz integral rule $$f(x)=\lim_{t \to 0}\dfrac{\int_{x-t}^{x+t}sf'(s)ds}{2t}=\dfrac{xf'(x)+xf'(x)}{2}=xf'(x)$$

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

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.