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).
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).
Hmm..
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.