Proving the Area under the curve is exactly $\frac{b^3}3$ using Method of Exhaustion and Law of Trichotomy

Question

I am reading Tom M Apostol's Calculus Vol 1, and was confused by how he used the method of exhaustion to determine the value of the area under the curve $ {b}^{2} $. He first shows a method of approximating the sum using two series, $S_n$ and $s_n$. He divides the area under $b^2$ into a sum of $n$ rectangles of width $ \frac {b}{n} $ and height ${(\frac {kb}{n})}^2$. This can be expressed as $ \frac {b^3}{n^3}k^2$ where $k^2$ is $$ \sum_{k=1}^n k^2 $$ for $S_n$ and $$ \sum_{k=1}^{n-1} k^2 $$ for $s_n$. This can be seen in pages 1 and 2: Page 1 Page 2

He then gives the identity in page 3: $$ \sum_{k=1}^n k^2 = \frac {n^3}{3} + \frac {n^2}{2} + \frac {n}{6}$$

Page 3

He also states an inequality: $$ \sum_{k=0}^n k^2> \frac {n^3}{3} > \sum_{k=0}^{n-1} k^2$$

Multiplying the inequality by $ \frac {b^3}{n^3}$, he obtains: $ S_n > \frac {b^3}3 > s_n$

He then tries to prove that it is this value: $ \frac {b^3}{3} $, that is the exact area under the curve where I am confused by his reasoning

Page 4

Thus, I tried looking online for more such solutions, and found one, in a site linked below, except it uses the curve $2b^2$, which should make no difference in the essence of reasoning why $ \frac {2b^3}{3} $ is the area under the curve.

This site uses the following way of proving that it is indeed $ \frac {2b^3}{3} $ which is the area under the curve: Proof In the proof: $$ \frac {2b^3}{3} < S_n (1) $$ $$ \frac {2b^3}{3} - s_n < S_n - s_n (2) $$ $$ \frac {2b^3}{3} - A < \frac {2b^3}{n} (3) $$ $$ n < \frac {2b^3}{\frac {2b^3}{3} - A} (4) $$

I don't understand how the $s_n$ in (2) magically changes into A in (3)

Proof Part 2 He does it, again, here, where $s_n$ seemingly magically changes into $\frac {2b^3}{3}$

https://www.stumblingrobot.com/2015/06/16/use-the-method-of-exhaustion-to-calculate-the-area-under-the-following-curves/

This confuses me greatly, and thus I am not able to understand the first part of the textbook, and proceed with reading ahead. If someone could help me understand Apostol's proof listed as Page 4, or explain this mysterious transformation of the inequalities, it would be much appreciated!

Answer 1

$\frac {b^3}{n^3} \sum_\limits{k=0}^{n-1}k^2 < A < \frac {b^3}{n^3}\sum_\limits{k=1}^{n}k^2$

The left side of the inequality represents the sum of the rectangles when the left corner of each rectangle is on the curve. The right side represents the area of the rectangle when the right corners of the rectangles are on the curve. Since the curve is strictly increasing, one is the upper sum and one is the lower sum and the area must be somewhere in between.

From the identity for the sum of squares

$\frac {b^3}{n^3}(\frac {n^3}{3} - \frac {n^2}{2} + \frac {n}{6}) < A < \frac {b^3}{n^3}(\frac {n^3}{3} + \frac {n^2}{2} + \frac {n}{6})$

$b^3(\frac 13 - \frac {1}{2n} + \frac {1}{6n^2}) < A < b^3(\frac {1}{3} + \frac {1}{2n} + \frac {1}{6n^2})$

Now consider the limit as $n$ approaches infinity. I am guessing that since this proof is on pages 4-7 of the book, that the subject of the limit has not yet been formally introduced, and this is the source of confusion.

In a few more pages you will have the mechanics of limits under your belt, and you can just make a statement of how at the limit the left side equals the right side and A gets squeezed in between.

Until that time, show that if $A \ne \frac {b^3}{3}$ for a sufficiently large value of $n$ one of the inequalities is going to break.

First Appostol makes a simplifying assumption.

$\frac 13 - \frac {1}{n}<\frac 13 - \frac {1}{2n} + \frac {1}{6n^2}$

and

$\frac 13 + \frac {1}{n}>\frac 13 + \frac {1}{2n} + \frac {1}{6n^2}$

$b^3(\frac 13 + \frac {1}{n})<\frac {b^3}{n^3}(\frac {n^3}{3} - \frac {n^2}{2} + \frac {n}{6}) < A < \frac {b^3}{n^3}(\frac {n^3}{3} + \frac {n^2}{2} + \frac {n}{6})<b^3(\frac 13 + \frac {1}{n})$

$\frac {b^3}{3} - \frac {b^3}{n} < A < \frac {b^3}3 + \frac {b^3}{n}$

Claim: $A = \frac {b^3}{3}$

Proof by contradiction.

Suppose $A\ne\frac {b^3}{3}.$

In the case that $A > \frac {b^3}{3}$

The inequality above says:
$A < \frac {b^3}{3} + \frac {b^3}{n}$
$A-\frac {b^3}{n} < \frac {b^3}{n}$
Our assumption that $A > \frac {b^3}{3}$ implies $A-\frac {b^3}{3}>0,$ but when $n$ is big enough $\frac {b^3}{n}$ is going to be smaller than $A-\frac {b^3}{n}$

More specifically, when $n>\frac {b^3}{A-\frac {b^3}{3}}$ (which we can do because we have assumed $A-\frac {b^3}{3}>0$) the inequality breaks.

Simillarly, in the case that $A<\frac {b^3}{3}$ the inequality $\frac {b^3}{3} - \frac {b^3}{n} < A$ is going to break when $n> \frac {b^3}{\frac {b^3}{3} - A}$.

Providing the contradiction we need to prove our claim.

Answer 2

In the solution by https://www.stumblingrobot.com/2015/06/16/use-the-method-of-exhaustion-to-calculate-the-area-under-the-following-curves/, we see that this is the inequality obtained, where we are trying to prove that $\frac {2b^3}{3}$ is the area under the curve.

$$ \frac {2b^3}{3} < S_n (1) $$ $$ \frac {2b^3}{3} - s_n < S_n - s_n (2) $$ $$ \frac {2b^3}{3} - A < \frac {2b^3}{n} (3) $$ $$ n < \frac {2b^3}{\frac {2b^3}{3} - A} (4) $$ From (2) to (3), we see $s_n$ change into $A$. I was initially confused by this, but this is just indeed a clever manipulation of inequalities, which can be explained as: $$ s_n = \sum_{k=0}^{n-1} k^2 $$ $$ \sum_{k=0}^{n-1} k^2 < \frac {n^3}{3} <\sum_{k=0}^{n} k^2 $$ $$ \sum_{k=0}^{n-1} k^2 < \frac {n^3}{3} $$ Adding $n^2$ $$ \sum_{k=0}^{n} k^2 < \frac {n^3}{3} + n^2 $$ Multiplying by $\frac {2b^3}{n^3}$ $$ S_n < \frac {2b^3}{3} + \frac {2b^3}{n} $$ Similarly, we also get $$ s_n > \frac {2b^3}{3} - \frac {2b^3}{n} $$ From this we also conclude: $$ \frac {2b^3}{3} - \frac {2b^3}{n} < A < \frac {2b^3}{3} + \frac {2b^3}{n} $$ And from this we get: $$ \frac {2b^3}{3} - \frac {2b^3}{n} < A $$ $$ \frac {2b^3}{3} - A < \frac {2b^3}{n} $$ And finally transposing: $$ n < \frac {2b^3}{\frac {2b^3}{3} - A} (4) $$ The reason this proof works is because, $\frac {2b^3}{\frac {2b^3}{3} - A}$ is a constant, and we can always choose an arbitrary integer $n$ greater than it. By this very clever proof, we can see that $ A < \frac {2b^3}{3} $ is obviously false. The same follows for $ A > \frac {2b^3}{3} $ Assuming this: $$ A - \frac {2b^3}{3} < \frac {2b^3}{n} $$ This works, as this time we assume $A$ is larger, and therefore the inequality remains positive. Transposing: $$ n < \frac {2b^3}{A - \frac {2b^3}{3}} $$ Again, we can just choose $n$ such that it is bigger than this constant. This proof was quite hard for me to grasp initially but makes perfect sense now.