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__TRON__
01-10-2007, 12:17 PM
howdy, I'm a real novice with a simple question relating to method of exhaustion and area of parabolas (in this example using outer rectangles).

Given a parabola with base b divided into n sections, where x = kb/n, and the curve is found at (kb/n)², the area under the curve is described by
L x W = (b/n) • (kb/n)² = (b³/n³)k. Clear so far.

And so to approximate the sum of all these sections, would be described by:
Sn = (b³/n³) • (1² + 2² +...+ n²)
Again, clear so far.

Okay, so the text from which I'm reading is called "Calculus, Volume I" by a Tom M. Apostol, and at this point he says "there is an interesting identity which makes it poss. to evaluate this sum [the squares of all values for k] in a simpler way, namely

1² + 2² +...+ n² = n³/3 + n²/2 + n/6"

And he goes on to talk about a proof of this, and says it's valid for every value n such that n >= 1.

What confuses me is this following text:

"Start with the formula (k + 1)³ = k³ + 3k² + 3k + 1 and rewrite it in the form

3k² + 3k + 1 = (k + 1)³ - k³"

What confuses me is not the formula itself, it's a common enough product, but what confuses me is WHY this formula was selected. My background in math is weak enough that I'm not familiar with most of the formulas related to parabolas and such, does (k + 1)³ have some special significance? Thanks in advance.

HallsofIvy
01-17-2007, 02:03 PM
I'm not certain myself but I suspect that it is due to a result you will later see: that the "integral" defined in terms of those "Riemann sums" is the same as the "anti-derivative". Since the derivative of x^3 is 3x^2, the anti-derivative of 3x^2 is x^3. That's why "integrating" a quadratic, as you are doing, gives rise to a cubic and that's why they are using that cubic formula.