Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This time there's another example from Riemann sums that I don't know how to approach I guess I have to use a different way of partitioning than even parts.

$$\int_a^b \frac{1}{x^2}\mathrm{d}x$$

where $0<a<b$. Any help would be great! Thanks in advance!

share|cite|improve this question
What are you asking? How to evaluate the integral using riemann sums? – gt6989b May 7 '13 at 15:30
Have you written out this integral using Riemann sums? If not, where are you stuck? If so, with what you are having difficulties? – JavaMan May 7 '13 at 15:31
@gt6989b exactly, I've written it as $\lim_{n\rightarrow\infty}\frac{b-a}{n}\sum_{k=1}^{n}\frac{1}{(a+\frac{k(b-a)}{n‌​})^2}$ and I'm kinda lost there – darenn May 7 '13 at 16:04
@darenn You're right, looks pretty nasty. – gt6989b May 7 '13 at 16:42
up vote 5 down vote accepted

You can use any subdivision; for instance, set $r=(b/a)^{1/n}$ and use $x_0=a$, $x_1=ar$, …, $x_n=ar^n=b$:

\begin{align} \sum_{i=1}^n f(x_i)(x_i-x_{i-1})&= \sum_{i=1}^n\biggl( \frac{1}{(ar^i)^2}(ar^i-ar^{i-1}) \biggr)\\[2ex] &= \frac{r-1}{ra}\sum_{i=1}^n\frac{1}{r^{i}}\\[2ex] &= \frac{r-1}{ra}\frac{1}{r}\frac{1-(1/r)^n}{1-(1/r)}\\[2ex] &= \frac{1}{ra}\frac{b-a}{b}\\[2ex] &=\sqrt[n]{\frac{a}{b}}\biggl(\frac{1}{a}-\frac{1}{b}\biggr) \end{align}

Since $\lim_{n\to\infty}\sqrt[n]{a/b}=1$ and so you have the limit of these Riemann sums is $$ \frac{1}{a}-\frac{1}{b} $$ as required.

Note that this method actually works for $x^k$, except when $k=-1$, and only requires computing the sum of terms of a geometric progression.

share|cite|improve this answer
Thanks, really appreciate your help! – darenn May 7 '13 at 17:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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