0
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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, $$\frac{5}{8}<\frac{533}{840}=\sum_{n=1}^8\frac{(-1)^{n-1}}{n}<\log 2<\sum_{n=1}^3\frac{(-1)^{n-1}}{n}=\frac{5}{6}<\frac{7}{8}.$$

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.