Proving bounds of an integral $\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$
I found $f'(x)=\displaystyle\frac{-3x^2}{2(x^3 + 4)^{\frac{3}{2}}}.$
Set $f'(x) = 0$. Got $x = 0$ as a local max.
$f(0) = 1/\sqrt{5}$
$f(2) = 1/\sqrt{12}$
So $1/\sqrt{3} \leq 1/\sqrt{12} \leq$ the integral $\leq 1/\sqrt{5} \leq 1$.
Correct?
Unsure how to progress from here

Hint: $$\frac{1}{\sqrt{a^3+4}} \le \frac{1}{\sqrt{x^3+4}} \le \frac{1}{\sqrt{0^3+4}}$$
for $0 \le x \le a$.
