Integral inequality 5 How can I prove that:
$$8\le \int _3^4\frac{x^2}{x-2}dx\le 9$$
My teacher advised me to find the asymptotes, why? what helps me if I find the asymptotes?
 A: The important point is that
$$\frac{d}{dx}\left(\frac{x^2}{x-2}\right)<0$$
for $x\in[3,4]$, so that $\frac{x^2}{x-2}$ is decreasing there. Hence,
$$\int_3^4\frac{4^2}{4-2}dx\leq \int_3^4\frac{x^2}{x-2}dx\leq\int_3^4\frac{3^2}{3-2}dx,$$
and the left and right hand sides evaluate to $8$ and $9$ respectively.
A: Hint 
$$(b-a) m \leq \int_a^b f(x) \, dx \leq (b-a) M$$
where $m, M$ are the minimum and maximum of $f$ on $[a,b]$.
The asymptotes of this function occur around $x=2$, so I don't see why your teacher asks you to use them.
A: Direct approach is not so painful:
$$\int_{3}^{4}\frac{x^2}{x-2}\,dx = \int_{1}^{2}\frac{(x+2)^2}{x}\,dx = \int_{1}^{2}\left(x+4+\frac{4}{x}\right)\,dx =\frac{11}{2}+4\log 2 $$
and by the Hermite-Hadamard inequality we have:
$$ \log 2 =\int_{1}^{2}\frac{dx}{x} \in \left(\frac{2}{3},\frac{3}{4}\right) $$
so:
$$\int_{3}^{4}\frac{x^2}{x-2}\,dx \in \left(8+\frac{1}{6},8+\frac{1}{2}\right).$$
A: Hint
Show that $f(x)=\frac{x^2}{x-2}$ satisfies: $f(3)=9,$ $f(4)=8$ and $f$ is decreasing in the interval $[3,4].$
