I'm really struggling to even make a start on this question and even just a hint to get me going would be much appreciated.

The function $f(x)$ is defined on $[0, \pi/2]$ as

$$f(x) = \int_0^x \sin^{1/2}{t} \, dt$$

Show that

$$\frac{\pi}{4} \left(\frac{1}{2}\right)^{1/4} \lt \int_{\pi/4}^{\pi/2} \sin^{1/2}{t} \, dt \lt \frac{\pi}{4} $$

My understanding is that for $g(x) = \frac{df(x)}{dx}$ there exists some $c$ for which $g(c)$ is equal to the average value of $g(x)$ however I'm not entirely sure how I would then apply the average value formula to the function, as I have no idea how to (and don't think I'm supposed to) integrate $\sqrt{\sin{x}}$. Thanks in advance for any help.

  • $\begingroup$ The length of the interval $I:=[\frac{\pi}{4},\frac{\pi}{2}]$ is $\frac{\pi}{4}$. The minimum of $\sin^{1/2}t$ on $I$ is $(\frac{\sqrt{2}}{2})^{1/2}$ and its maximum is $1$. $\endgroup$ – Guest Sep 25 '15 at 23:04

You should be aware that on $[\pi/4,\pi/2]$, $\sin{x}$ is increasing and has minimum value $1/\sqrt{2}$, maximum value $1$ (by evaluating at the endpoints). Then $$ \left(\frac{1}{2}\right)^{1/4} < \sqrt{\sin{t}} < 1 $$ for every point in the interior of the interval of integration. Integrating this inequality from $\pi/4$ to $\pi/2$ gives $$ \frac{\pi}{4}\left(\frac{1}{2}\right)^{1/4} < \int_{\pi/4}^{\pi/2} \sqrt{\sin{t}} \, dt < \frac{\pi}{4}; $$ the inequalities do not become equalities because the inequality is true for the entire interval of integration with only two exceptional points.

  • $\begingroup$ Thanks for clearing that up! This makes a lot more sense, I was confusing myself by making the problem more complicated than it needed to be. $\endgroup$ – nECS Sep 25 '15 at 23:33

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