Proving bounds of an integral

Question

$\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$

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

Answer 1

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

Answer 2

Hint: $$\int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{(2)^3 +4}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{(0)^3 +4}} $$

Answer 3

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 $0.854$.