High School How did Cavalieri get his formula for the area underneath a parabola?

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Cavalieri derived his formula for the area underneath a parabola by utilizing a specific ratio related to the sum of squares. The key formula involved is the sum of the first m squares, expressed as $$\sum_{k=1}^m k^2 = \frac{m(m + 1)(2m + 1)}6$$. By substituting this sum into his initial equation, he was able to arrive at the desired result. This mathematical approach highlights the connection between geometric areas and algebraic summation. Understanding this derivation is crucial for grasping Cavalieri's contributions to calculus and geometry.
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I know he had this ratio:
1604325468203.png

But how did he get this:
1604325593307.png
?
 
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1604325899893.png
 
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erocored said:
I know he had this ratio:
View attachment 272042
But how did he get this:
View attachment 272043 ?
The result uses a formula for the sum of squares.
$$\sum_{k=1}^m k^2 = \frac{m(m + 1)(2m + 1)}6$$
Replace ##1^2 + 2^2 + \dots + m^2## in your first equation by the above, and you will get the result shown in the second equation.
 
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There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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