Given that $f(x) = \frac{x^2}{x-2}$ show that $8 \le \int_3^4 f(x)\,dx \le 9$. Consider the function:
$$f : \mathbb{R} \setminus \{ 2\} \rightarrow \mathbb{R} \hspace{2cm} f(x) = \dfrac{x^2}{x - 2}$$
I have to show that the following statement is true:
$$8 \le \int_3^4 f(x) \, dx \le 9$$
The first thing I did was to find:
$$\int_3^4 f(x) \, dx = \frac{11}{2} + 4 \ln 2$$
I think this is correct. The problem I have now is to show the following:
$$8 \le \dfrac{11}{2} + 4 \ln 2 \le 9$$
What I did in order to accomplish this was to first show that the result of the integral is first $\ge 8$ and then that it is $\le 9$. So first I tried to show:
$$\frac{11}{2} + 4 \ln 2 \ge 8$$
$$\frac{11}{2} - 8 + 4 \ln 2 \ge 0$$
After a few calculations I got to the point where I have to show:
$$8 \ln 2 \ge 5$$
$$\ln 2^8 \ge \ln e^5$$
$$ 256 \ge e^5 \tag 1$$
So I have to show that $256 \ge e ^5$. In a similar manner, working with the other part of the inequality (to trying to show that the result of the integral is $\le 9$) I have to show:
$$ 256 \le e^7 \tag 2$$
So I am stuck at proving $(1)$ and $(2)$. If I approximate $e \approx 2.71$ there is no problem, the inequalities are clearly true, but this feels a bit sloppy. Is there another way to prove $(1)$ and $(2)$ (or even the whole given integral inequality from the start) a little more rigorously, not relying on approximations?
 A: Hint: What are the minimum and the maximum of $f(x)$ for $x\in[3,4]$?
This is a whole different route, but should be much simpler.
A: notice that:


*

*$f(3)=9$

*$f(4)=8$
Also by quickly looking at the function in this range it is decreasing, meaning the maximum is 9 and the minimum is 8. Using this, if the function was continuously 8 over the range of one the integral would be 8. using the same logic for 9 you can show the statement is true.

A: Since $f(x)=x+2+\frac{4}{x-2}$, $f^\prime(x)=1-\frac{4}{(x-2)^2}$, so $f^\prime(x)\le0$ on $[3,\,4]$. Since $f$ decreases on $[3,\,4]$ from $f(3)=9$ at $x=3$ to $f(4)=8$ at $x=4$, we can integrate $8\le f(x)\le 9$ on $[3,\,4]$ to get $8\le\int_3^4f(x)dx\le9$.
A: $$f(x)=\frac{x^2}{x-2} \implies f'(x)=\frac{x(x-4)}{(x-2)^2}\le 0, x \in[3,4].$$
$f(x)$ is a decreasing function in the interval. The global maximum is $M=f(3)=9$, its
global minimumm is $m=f(4)=8$. So by MVT, we have
$$m(b-a) \le \int_{a}^{b} f(x) dx\le M(b-a) \implies  8 \le \int_{3}^{4} \frac{x^2}{x-2} dx \le 9$$
A: While the analysis approach as given by others is obviously the "best" method, just following on from what you were doing:
(1) is obviously true because $$3>e\implies 3^5=243>e^5$$
so $$e^5<256$$
(2) is also true because $e>2$ and $$e^7>2^3\times e^4$$
and since $e>2.5$. $$e^4>(2+\frac12)^4>16+4\times8\times\frac12=32$$
Hence $$e^7>8\times32=256$$
A: $$8 \le \int_3^4 \frac{x^2}{x-2} \, dx \le 9 \text{ ?}$$
\begin{align}
& \overbrace{ \, \frac{x^2}{x-2} = (x+2) + \frac 4 {x-2} \, }^\text{This can be done by long division.} \\[12pt]
& \int_3^4 (x+2)\,dx + \int_3^4 \frac 4 {x-2} \, dx \\[10pt]
= {} & \frac {11} 2 + \int_3^4 \big( \text{something between 2 and 4} \big)\,dx \\[10pt]
= {} & \frac {11} 2 + \big( \text{something between 2 and 4} \big)
\end{align}
This narrows it down to the interval from $7.5$ to $9.5.$
Now let's get a bit subtler: The curve $y = \dfrac 4{x-2}$ is conxex (i.e. concave upward) on the interval $[3,4].$ Therefore the graph lies below the chord that connects the points on the graph at $x=3$ and $x=4.$ The area below the chord is merely the area of a trapezoid, so you don't even need antiderivatives. It is $3.$ So the integral we seek is less than $8.5.$ But covexity also implies the graph is above the two tangent lines at $x=3$ and $x=4.$ Look at where those two tangent lines meet and you need the areas of two trapezoids. This gives you a lower bound.
A: A very quick way to show this is by finding the area under the curve in the interval from 3 to 4
The plot of $ f(x) $ looks like this- here is the plot.
From the plot it's very clear that area under the curve in $ [3, 4] $ is bounded between 8 and 9
Also, this function can be re-written as:
$$
f(x)=\frac{x^{2}}{x-2} = \frac{x^{2}-4+4}{x-2}= \frac{x^{2}-4}{x-2} + \frac{4}{x-2} \
=x+2+ \frac{4}{x-2}$$
For the values of $ f(x) $ in $ [3,4], f(3)=5+4=9 $ and $ f(4)= 6+2=8$,  function integral in the range $[3,4]$ is calculated as
$$\begin{aligned}
f(x)&=\frac{x^{2}}{x-2}\\
&= \frac{x^{2}-4+4}{x-2}\\
&= \frac{x^{2}-4}{x-2} + \frac{4}{x-2}\\
&=x+2+ \frac{4}{x-2}\end{aligned}$$
$$\begin{aligned}
&\int_{3}^{4} x+2+ \frac{4}{x-2}\\
=& \left[\frac{x^{2}}{2}+2x+4\ln(x-2)\right]_3^4\\
=& \left[\frac{4^{2}}{2}+8+4\ln2\right]-\left[\frac{3^{2}}{2}+6+0\right] \\=& 16+ 4\ln2-4.5-6 \\
=& 5.5+4\ln2 (\because \ln2\approx .69)\\ 
\approx& 8.3
\end{aligned}$$
A: use the fact that $m(b-a)\leq \int_a^b f(x)dx\leq M(b-a$).
 where $m,M $ be the global maximum and minimum of of $f$ on $[a,b]$, in this case $m=f(4), M=f(3)$, since the funtion is decrasing on the specified interval, which can easily be seen by first derivative test.
