The integral from 0 to 1 of x^2 is equivalent to the integral from 0 to 1 of the sum of the first n squares formula (x(2x+1)(x+1)/6) does anyone know if this is some special occurrence (it's only true for 0 to 1 as far as I can tell).

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- Mar 6th 2008, 08:55 PMpDeCrunchSum of squares from 0 to 1
The integral from 0 to 1 of x^2 is equivalent to the integral from 0 to 1 of the sum of the first n squares formula (x(2x+1)(x+1)/6) does anyone know if this is some special occurrence (it's only true for 0 to 1 as far as I can tell).

- Mar 7th 2008, 05:20 AMwingless
Hmm..

http://img528.imageshack.us/img528/4585/graphze8.png

I guess this is only a coincidence.

$\displaystyle \int \frac{ x (x+1) (2x+1)}{6} ~dx = \frac{x^2}{12}+\frac{x^3}{6}+\frac{x^4}{12}$

$\displaystyle \int x^2 ~dx = \frac{x^3}{3}$

$\displaystyle \frac{1^2}{12}+\frac{1^3}{6}+\frac{1^4}{12}=\frac{ 1^3}{3} = \frac{1}{3}$

$\displaystyle \frac{0^2}{12}+\frac{0^3}{6}+\frac{0^4}{12}=\frac{ 0^3}{3} = 0$

I don't know, such things are likely to happen with numbers 0 and 1. - Mar 7th 2008, 05:26 AMcolby2152
Coincidence at best.