The definition for the Mean Value Theorem I am working with:
If a function $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, Then there exists at least one point $c$, where $a<c<b$ such that $f'(c) = \frac{f(b)-f(a)}{b-a}$
Now I am given this problem to look at, and I have no idea how the definition relates to the integral. I am not after an explicit answer, but just some ideas on how to understand what is being asked so I can attempt it.
Prove that for $0 \le x \le \frac{\pi}{3}$
$$\frac{x}{2} \le \sin(x) \le x$$
Hence, prove that $$\int_0^{\pi/3} \frac{\sin(x)}{x^{3/2}}\text{d}x$$ is improperly integrable
Thanks for looking.
Hint: MVT implies that if $f'\geq 0$ on $(a,b)$, then $f$ is increasing on $[a,b]$.
Apply this to $h(x) = x - \sin x $ and $k(x) = \sin x - x/2$ for all $x\in [0,\pi/3]$.
Second part of the problem, the integral converges because, using first part, we have:
$$ 2^{-1}x^{-1/2} \leq x^{-3/2}\sin x \leq x^{-1/2}$$ for all $x\in (0,\pi/3)$, $\int_0^{\pi/3}x^{-1/2} dx=\lim\limits_{\epsilon\to 0^+}\int_\epsilon^{\pi/3}x^{-1/2} dx$ is finite, and the integral operator monotony gives us bounds for integral:
$$ \int_0^{\pi/3} 2^{-1}x^{-1/2}dx \leq \int_0^{\pi/3}x^{-3/2}\sin x \; dx \leq \int_0^{\pi/3}x^{-1/2} dx.$$