Quackulus Question

sum of squares 1..n = n(n+1)(2n+1)/2
sum of squares 1..10 = 385

integral of n^2 = (n^3)/3
intergral from 1 to 10 = 1000/3 - 1/3 = 333

shouldn't the integral give the sum or something?

integral = are.
i.e
Sx^2 1 to 10 = AREA OF parabola
from 1 to 10.

>>1

The first problem is a discrete problem of arithmetic, which gives its answer in linear units.  The second problem is a problem of the continuum, whose answer is obtained through a limiting process, and is given in SQUARE units.

>>2
Not neccessarily. For example, the integral of sin x over [0, 2pi]  produces a value of zero. But the area is actually 2. 

>>1
That would be the definition of a Reimann sum approximation.

>>4

What you meant to say is 4, dumbass.  The area of any of the finite regions described and bounded by sin(x) and the x-axis is two.  2+2=4; in other words, the absolute value of the integral from 0 to π of sin(x), plus the absolute value of the integral from π to 2π of sin(x), is four.

>>5

Right. My mistake. No need to call me a dumbass.

The point I was TRYING to make was that calling the integral of something the area of said function is misleading and mostly wrong.

>>6

Here's something you don't see on the message boards too much:

I was mean and I apologize.

>>7
Apology accepted. Feel the love.

>>8
buttsex?

>>9

Not a fan of the anal penetration. Reacharound is appreciated though.

>>9 is an imposter. Disregard, disregard, dilute, dilute, OK!

eew diluye

How did this get bumped?  The last post was from 2005...

This thread is ancient, but I'll contribute something nice anyways.

Define P_k(n) for non-negative integers k to be a polynomial which, when n is a non-negative integer, is equal to Sum[i=1,n](i^k).

The following holds:

P_k+1(n) = (k+1)*Integral[0,n](P_k(t) dt) + n*(1 - (k+t)*Integral[0,1](P_k(t) dt))

And of course P_0(n) = n, so you can construct a polynomial equal to Sum[i=1,n](i^k) for any nonnegative k.