Answer
\[
3 \sqrt{2} \leq \int_{0}^{3} \sqrt{2+x^{3}} d x \leq 3 \sqrt{29}
\]
Work Step by Step
Given:
For $0 \leq x \leq 3,$
\[
\begin{array}{c}
\Rightarrow 0 \leq x^{3} \leq 27 \\
\Rightarrow 2 \leq 2+x^{3} \leq 29 \\
\Rightarrow \sqrt{2} \leq \sqrt{2+x^{3}} \leq \sqrt{29} \\
\therefore ,\quad m(\min )=\sqrt{2}, M(\operatorname{mix})=\sqrt{29}
\end{array}
\]
\[
(-a+b)m \leq \int_{a}^{b} f(x) d x \leq (-a+b)M
\]
We then get:
\[
\begin{array}{l}
(-0+3)\sqrt{2} \leq \int_{0}^{3} \sqrt{2+x^{3}} d x \leq (-0+3)\sqrt{29} \\
\quad \Rightarrow \mathbb{3 \sqrt { 2 }}=\int_{0}^{3} \sqrt{2+x^{3}} d x \leq {3 \sqrt{29}}
\end{array}
\]
