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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!

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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

1 Answer 1

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.

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Thanks, really appreciate your help! – darenn May 7 '13 at 17:14

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