<s>[INST] 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
 [/INST] Slightly Sharper Bounds
For $x\geq 0$, we have by the AM-GM Inequality that
$$x^3+2\cdot1\geq 3\sqrt[3]{x^3\cdot1^2}=3x\,.$$
That is,
$$x^3+4\geq 3x+2\,.$$
When $0\leq x\leq 2$, we also get
$$x^3+4\leq \left(x^{\frac{3}{2}}+2\right)^2\leq \big(\sqrt{2}x+2\big)^2\,.$$
Therefore,
$$\frac{1}{\sqrt{2}x+2}\leq \frac{1}{\sqrt{x^3+4}}\leq \frac{1}{\sqrt{3x+2}}$$
for every $x\in[0,2]$.  Consequently,
$$\frac{1}{\sqrt{2}}\,\ln\left(\sqrt{2}+1\right)=\int_0^2\,\frac{1}{\sqrt{2}x+2}\,\text{d}x\leq \int_0^2\,\frac{1}{\sqrt{x^3+4}}\,\text{d}x\leq \int_0^2\,\frac{1}{\sqrt{3x+2}}\,\text{d}x=\frac{2\sqrt{2}}{3}\,.$$
Hence, we have a slight improvement to the original bounds:
$$\frac1{\sqrt{3}}<0.623< \int_0^2\,\frac{1}{\sqrt{x^3+4}}\,\text{d}x<0.943<1\,.$$
The improvement is not significant, nonetheless, and the actual value of $\displaystyle\int_0^2\,\frac{1}{\sqrt{x^3+4}}\,\text{d}x$ is roughly $0.854$.
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