# Solving this inequality with integral

We have function $f:\mathbb{R}-\{2 \}\to\mathbb{R}$ $$f(x)=\frac{x^2}{x-2}$$

Show that $8\le\int\limits _3^4f\left(x\right)dx\le9$

I solved the definite integral and got $\int\limits _3^4f\left(x\right)dx = \frac{11}{2}+ln16$. I tried solving the inequality but I get something like $e^{\frac{5}{2}}\le 16\le e^{\frac{7}{2}}$ which I don't know how to prove.

How to solve this?

## 2 Answers

Here's a fancy way to prove the explicit inequality $$\frac{5}{2}\le \log 16 \le \frac{7}{2} \\ \frac{5}{8}\le\log 2\le\frac{7}{8}.$$We have $$\log 2=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n},$$ therefore truncating the series at odd or even $n$ respectively provides upper and lower bounds for $\log 2$. In particular, $$\frac{5}{8}<\frac{533}{840}=\sum_{n=1}^8\frac{(-1)^{n-1}}{n}<\log 2<\sum_{n=1}^3\frac{(-1)^{n-1}}{n}=\frac{5}{6}<\frac{7}{8}.$$

This is a Riemann sum question. It's easy to see that a local minimum occurs at x = 4. Thus if you approximate the area under the curve by the rectangle with height f(4) you get a lower bound. Similarly. x = 3 gives a maximum (at least on the relevant interval [3,4]). Thus, approximating the area with the rectangle of height f(3) gives an upper bound.