Integral inequality 5

Question

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?

Answer 1

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.

Answer 2

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.

Answer 3

Hint

Show that $f(x)=\frac{x^2}{x-2}$ satisfies: $f(3)=9,$ $f(4)=8$ and $f$ is decreasing in the interval $[3,4].$

Answer 4

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: $$\int_{3}^{4}\frac{x^2}{x-2}\,dx \in \left(8+\frac{1}{6},8+\frac{1}{2}\right).$$