MHB How Did Archimedes Achieve Quadrature of the Parabola Without Calculus?

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Archimedes achieved the quadrature of the parabola by dividing it into a series of triangles, demonstrating that their areas create a geometric sequence. He then applied the formula for the sum of an infinite geometric series to derive the area of the parabola. This method illustrates his innovative approach to geometry without the use of calculus. References to this technique can be found in major calculus texts and works by Apostle. The discussion highlights Archimedes' significant contributions to mathematics and geometry.
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How did Archimedes discover the Quadrature of the parabola without the use of calculus?

If someone could please explain, I would be eternally grateful.
 

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I recall an explanation of this in one of Apostle's Analysis books. Don't have it with me and can't remember how detailed it was, but could be useful. It's likely mentioned in any major book on calculus.

Also @HallsofIvy, Bing? :)
 

Thread 'Proving that convexity implies second order derivative being positive'

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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