Talk:Riemann sum

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I think the Riemann sum and the Riemann integral have too much in common. I suggest merging information from them into Riemann integral, and make Riemann sum a redirect. Please see discussion at talk:Riemann integral.(Igny 21:51, 5 December 2005 (UTC))

Yes I was looking for this article but found that one instead. Very confusing. If you knew the difference between a Riemann sum and an integral why would you need to look it up?Circuitboardsushi (talk) 21:56, 7 April 2012 (UTC)


I think it would be worth clarifying that the distance between the points (x1, x2, xi-1, xi) have to be a uniform distance, and it is done for simplicities sake. The graphs would give that idea as well, which isn't really true of Riemann's original "Riemann sums." --AstoVidatu 04:47, 7 December 2006 (UTC)

 ==Error Estimation==

Are the error estimation formulas correct for the "middle sum" and "trapezoidal sum" methods? The "middle sum" error estimate is currently quoted as :\left \vert \int_{a}^{b} f(x) - A_\mathrm{mid} \right \vert \le \frac{M_2(b-a)^3}{(24n^2)},

...and the "trapezoidal sum" error estimate is :\left \vert \int_{a}^{b} f(x) - A_\mathrm{trap} \right \vert \le \frac{M_2(b-a)^3}{(12n^2)},

I don't have a calculus book handy, but it doesn't make intuitive sense that the "trapezoidal sum" error could be twice the size of the "middle sum" error. Is this right? Is there a handy reference online where these formulas are derived?

--Imperpay 22:30, 26 March 2007 (UTC)

The formulae are correct. Accuracy of Trapezoid and Simpson Approximations lists the formula for the trapezoidal error as \left \vert \int_{a}^{b} f(x) - A_\mathrm{trap} \right \vert \le \frac{M_2(b-a)^3}{(12n^2)},

It may not be intuitive, but oddly enough, the consideration of multiple derivatives and the fact that the middle sum method overlaps in both directions makes the error bound for the middle sum method smaller.

-- Icedemon —Preceding unsigned comment added by (talk) 09:35, 25 December 2007 (UTC)


how have you guys not added the limit of the Riemann sum, which solves the integral with no error? can someone please add this vital information? -- (talk) 23:49, 1 November 2008 (UTC)

I've added an example but I need some help with the formatting. If someone could clean it up for me, that would be great. Thanks. Dwees (talk) 04:53, 26 November 2008 (UTC)

Subsets or elements?[edit]


Because P is a partition with n elements of I, the Riemann sum of f over I with the partition P is defined as

Should that be "n subsets of I" instead of "n elements of I"? (talk) 17:06, 14 October 2009 (UTC)

Simpson's Rule is Blank![edit]

There are examples of using Simpson's Rule, but there is no theory under the blue heading like the other Reimann sum methods, just examples. —Preceding unsigned comment added by (talk) 03:14, 14 March 2010 (UTC)

This article should reference Archimedes' use of Reimann sums[edit]

The article located here states, "When rigorously proving theorems, Archimedes often used what are now called Riemann sums."

I think Archimedes' use of this method should be noted in this article so as to make it clear that Reimann did not originate Reimann Sums. — Preceding unsigned comment added by (talk) 11:34, 24 December 2011 (UTC)