Find $x>0$ for which the integral value of $$\int_0^{\sqrt {x}} \sin \left(\frac{2\pi t}{t+2}\right) dt$$ is the largest.

My try:
Let: $$f(x)=\int_0^{\sqrt {x}} \sin \left(\frac{2\pi t}{t+2}\right) dt$$ and $$F'(t)=\sin \left(\frac{2\pi t}{t+2}\right) dt$$ Then: $$f(x)=F(\sqrt {x})-F(0)$$ $$f'(x)=\frac{F'(\sqrt {x})}{2\sqrt {x}}-F'(0)\cdot 0=\frac{F'(x)}{2\sqrt {x}}=\frac{\sin \left(\frac{2\pi \sqrt {x}}{\sqrt {x}+2}\right)}{2\sqrt {x}}$$In this moment I am wanting to find $x$ for which $f'(x)=0$ and $f'$ changes the sign because it is a maximum for $f$. However I have a sine and I don't know how to do it in the simplest way, because the study of all ranges of sine variation is a terrible calculation.

Can you help me?

  • $\begingroup$ In the last expression, replace $\sqrt{x}$ by $y$, say, and analyse the corresponding function of $y$ in $(0,\infty)$. It looks doable. $\endgroup$ Jun 20 '19 at 10:21
  • $\begingroup$ @ViktorGlombik yes, but when I calculate only $f'(x)=0$ I don't know if in this $x$ is maximum or minimum $\endgroup$
    – MP3129
    Jun 20 '19 at 10:37
  • $\begingroup$ I am getting infinite values. $\endgroup$
    – u_sre
    Jun 20 '19 at 10:41
  • $\begingroup$ @AjayMishra WolframAlpha says otherwise. $\endgroup$
    – Ramanujan
    Jun 20 '19 at 10:43
  • $\begingroup$ I didn't get that at all. $\endgroup$
    – u_sre
    Jun 20 '19 at 10:46

Given function, $$ f(x) = \int_0^{\sqrt{x}} \sin\bigg( \cfrac{2 \pi t}{t + 2} \bigg) dt $$

On differentiating, $$f'(x) = \cfrac{1}{2 \sqrt{x} } \sin\bigg( \cfrac{2 \pi \sqrt{x}}{\sqrt{x} + 2} \bigg)$$For stationary points, $f'(x) = 0$. Here, in order to satisfy the condition, $$x = \bigg( \cfrac{2n}{2-n} \bigg)^2 $$

Where $ n \ne 0 $ and $ \in \mathbb I$ . But, remember at these values of $x$, there is stationary points, in order to find the maxima you can plug these values in $f'(x)$ and use first derivative test to find that. Can you do this from now?


You did almost half of the solution After using $f'(x)=0$ we get $$\sin \bigg(\frac{2\pi \sqrt x}{\sqrt x +2}\bigg)=0$$ Which yields $$\frac{2\pi \sqrt x}{\sqrt x +2}=n\pi$$ or $$2\sqrt x=n(\sqrt x+2)$$ By rearranging one can see that $x$ must be an integer in order to satisfy above equality because $n$ is integer And this equation satisfies only when $$x=0,4$$ for some $n\in Z$ But as $x>0$ Thus, The only solution is $$x=\boxed{4}$$

  • $\begingroup$ Where it was stated that $x$ must be integer? $\endgroup$
    – u_sre
    Jun 20 '19 at 10:57
  • $\begingroup$ It is not stated it's obtained in the solution because n is integer , so if you take any integral values of X which is not perfect square then equations won't satisfy $\endgroup$ Jun 20 '19 at 10:59

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