Trouble with Riemann sums

1. Sep 8, 2008

guynoone

Alright, I started doing Riemann sums and I am ripping my hair out in frustration. I just can't wrap my head around how I'm supposed to do it, and my woefully vague textbook isn't helping either. I'm wondering how I'm supposed to solve a Riemann sum question with sigma notation (no limits), and with only the width and height of the rectangles provided (no curve function). I used the image as an example. I have absolutely no clue how/why they get the 30 and 27. If anyone could provide a step by step explanation, it would be great. Thanks

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2. Sep 9, 2008

tiny-tim

Welcome to PF!

Hi guynoone! Welcome to PF!

ok … for some reason, the 0th Riemann sum, from 1 to n, is defined as being 1/n times the ∑, and the 1st Riemann sum as 1/n² times ∑.

In this case, you're expecting 10/3 times the 0th sum, and 1/3 times the 1st, but this particular definition means that you multiply the 0th one by 9, = 10/3 * 9 = 30, and the 1st one by 81, = 81 * 1/3 = 27.

(it isn't the only way of working out this problem … it's only really useful if you happen to have tables of Riemann sums

personally, i prefer to remember ∑i = n(n+1)/2)

3. Sep 9, 2008

guynoone

Thanks for the answer. To be honest I still don't really get what is happening for the solution I posted, but I just plan on sticking to the ∑i = n(n+1)/2) method which is way better (they must be trying to confuse us on purpose with these identities!)

4. Sep 20, 2008

Re: Welcome to PF!

''ok … for some reason, the 0th Riemann sum, from 1 to n, is defined as being 1/n times the ∑, and the 1st Riemann sum as 1/n² times ∑.''

Little abuse of notation here, even though I'm sure tiny-tim understands the entire problem.
The 0th order term refers to a sum which involves $$i^0$$, or $$1$$, a first order term involves $$i^1 = i$$, and so on. Writing $$\sum$$ without any explicit is common, but writing it without any arguments (summands) is not.
Actually, all that is done in this work a (very poor, in my opinion) condensation of the basic steps.

\begin{align*} \sum_{i=1}^9 {\frac{10-i} 3} & = \frac 1 3 \sum_{i=1}^9 {(10-i)}\\ & = \frac 1 3 \left(\sum_{i=1}^9 10 - \sum_{i=1}^9 i \right) \\ & = \frac 1 3 \left( 10 \sum_{i=1}^9 1 - \sum_{i=1}^9 i \right)\\ & = \frac 1 3 \left(10 \cdot 9 - \frac{10\cdot 9}{2} \right) \\ & = \frac 1 3 \left(90 - 45\right) = \frac{45}{3} = 15 \end{align*}

Note that the formula

$$\sum_{i=1}^n =\frac{n(n+1)} 2$$

was used, with $$n = 9$$, was used.

As I indicated at the top of my post, I believe the solution for this problem (the one posted by the OP) is very poorly presented and typeset