Answer
$\frac{1}{33}\leq \int_2^4\frac{1}{x^3+2}dx\leq \frac{1}{5}$
Work Step by Step
Property 8
If $m\leq f(x)\leq M$ for $a\leq x\leq b$, then $m(b-a)\leq \int_a^bf(x)dx\leq M(b-a)$
Given $f(x)=\frac{1}{x^3+2}$
Find $m$ and $M$:
Notice that
$2\leq x\leq 4$
$\Rightarrow 2^3\leq x^3\leq 4^3$
$\Rightarrow 2^3+2\leq x^3+2\leq 4^3+2$
$\Rightarrow \frac{1}{4^3+2}\leq \frac{1}{x^3+2}\leq \frac{1}{2^3+2}$
$\Rightarrow \frac{1}{66}\leq f(x)\leq \frac{1}{10}$
$\therefore m=\frac{1}{66}$ and $M=\frac{1}{10}$
Now, estimate the integral using the Property 8:
$m(b-a)\leq \int_a^bf(x)dx\leq M(b-a)$
$\frac{1}{66}(4-2)\leq \int_2^4\frac{1}{x^3+2}dx\leq \frac{1}{10}(4-2)$
$\frac{1}{33}\leq \int_2^4\frac{1}{x^3+2}dx\leq \frac{1}{5}$