Apostol's Archimedes area proof for a parabola

[gom2]

Homework Statement

My question is on the last bit of Apostol's proof, in his book Calculus Vol I 2nd ed, where he shows that the area under the parabola = b$$^{3}$$/3 where b is the base of the rectangle enclosing the parabola.

The bit I am confused about is where his contradiction n$$\geq$$b$$^{3}$$/(A-b$$^{3}$$/n) came from?

I know this is impossible for anyone to solve unless they own the book... However, please please could someone understand my frustration and also the lack of knowledge of knowing Latex enough to replicate the problem. If you own the book, please could you help me?

I would greatly appreciate help!

Homework Equations

The Attempt at a Solution

I have tried various approaches in trying to attain the n$$\geq$$b$$^{3}$$/(A-b$$^{3}$$/n), however, I have not been successful either way. Sorry, it might not seem like proper attempts have been made, but I have tried, and I have constantly failed to gain it.

 

[Feldoh]

It's just from the fact that you can pick a number of partitions n that is greater than or equal to $$\frac{b^3}{A-\frac{b^3}{n}}$$ so A < $$\frac{b^3}{3}$$ is false. The same logic works for the case A > $$\frac{b^3}{3}$$, just in reverse

 

[gom2]

It's just from the fact that you can pick a number of partitions n that is greater than or equal to $$\frac{b^3}{A-\frac{b^3}{n}}$$ so A < $$\frac{b^3}{3}$$ is false. The same logic works for the case A > $$\frac{b^3}{3}$$, just in reverse

Sorry and thank you for replying even though I may have been vague!

However, I do not understand how you get n that is greater than or equal to $$\frac{b^3}{A-\frac{b^3}{n}}$$

 

[ideasrule]

To the OP: if you want help from people who don't own the book, you can scan or photograph the book's proof and upload it here.

 

[gom2]

The book can be viewed here: http://www.scribd.com/doc/5874133/Calculus-Volume-1-686pp67

The proof starts from page 3 and ends at page 8. However, I am confused about the last bit, which is from pages 7-8.

If its hard to view on that, and you have bandwidth, please view the PDF file on the attachment.

 

[Subdot]

The bit I am confused about is where his contradiction n$$\geq$$b$$^{3}$$/(A-b$$^{3}$$/n) came from?

However, I do not understand how you get n that is greater than or equal to $\frac{b^3}{A-\frac{b^3}{n}}$

Well, firstly it is not n greater than or equal to what you wrote but rather n greater than or equal to (1) $$\frac{b^3}{A-\frac{b^3}{3}}$$. To figure out how to obtain this, remember that the inequality you obtained this from was valid for integer n ≥ 1.

Therefore, the inequality (2) n < $$\frac{b^3}{A-\frac{b^3}{3}}$$ is also valid for integer n ≥ 1.

However, the information on the right side of the inequality (2) is a constant. Thus, n can be made to be larger than the expression on the right. n can increase but the right side of (2) doesn't. n approaches infinity while the right side stays put. Obviously, n violates this inequality when (3) n ≥ $$\frac{b^3}{A-\frac{b^3}{3}}$$ (since the inequality (2) had n less than but not equal to the expression of (1)).
So there's a contradiction because the inequality (2) should be valid for all integer n ≥ 1, but as has been shown by (3), it is not valid for certain n. That's where the expression came from and is basically what Feldoh was getting at.