Question on Tom Apostol's Proof for Method of Exhaustion

Question

Currently, I am going through Tom Apostol's "Calculus: Volume 1 Second Edition" as I heard this was a good book to get a solid understanding of calculus and I wasn't satisfied with James Stewart's 9th edition calculus book.

I am on page 6 of the book and I am having some trouble. On pages 5 and 6, Apostol is proving that the area under the curve of $y = x^2$ between $x = 0$ and $x = b$ is equal to $\frac{b^3}3$, or $\frac{1}3$ of a rectangle with a base of length $b$ and a height of $b^2$. After proving that $1^2 + 2^2 +...+(n-1)^2 = $ $\frac{n^3}3 +$ $\frac{n^2}2 + $ $\frac{n}6$ (although I didn't fully understand how Apostol proved that either, this video by blackpenredpen https://www.youtube.com/watch?v=OpA7oNmHobM&ab_channel=blackpenredpen cleared that up for me so if anyone has trouble with that, check the link out), Apostol goes on to say:

For our purposes, we do not need the exact expressions given in the right-hand members of (I.3) and (I.4), or they can be proved directly by induction. (A proof by induction is given in Section I 4.1.)

All we need are the two inequalities

$1^2 + 2^2 + ... + (n-1)^2$ < $\frac{n^3}3$ < $1^2 + 2^2 + ... + n^2$

which are valid for every integer n $\geq$ 1. These inequalities can be deduced easily as consequences of (I.3) and (I.4), or they can be proved by induction. (A proof by induction is given in Section I 4.1.)

If we multiply both inequalities in (I.5.) by $\frac{b^3}{n^3}$ and make use of (I.1) and (I.2), we obtain

$s_n$ < $\frac{b^3}3$ < $S_n$

for every n. The inequalities in (1.6) tell us that $\frac{b^3}3$ is a number which lies between $s_n$ and $S_n$ for every n. We will now prove that $\frac{b^3}3$ is the only number which has this property. In other words, we assert that if A is any number which satisfies the inequalities

$s_n$ < A < $S_n$

for every positive integer n, then A = $\frac{b^3}3$. It is because of this fact that Archimedes concluded that the area of the parabolic segment is $\frac{b^3}3$.

My questions arises at the inequality: $1^2 + 2^2 + ... + (n-1)^2$ < $\frac{n^3}3$ < $1^2 + 2^2 + ... + n^2$

I understand that the values less than $\frac{n^3}3$ are the sum of squares of the heights of rectangles under the parabolic segment where $y=x^2$ and the values greater than $\frac{n^3}3$ are the sum of squares of the heights of rectangles that extend beyond the parabolic segment where $y=x^2$.

My question is, how did Apostol come up with the idea to use $\frac{n^3}3$ in the inequality?

I have looked to this link Method of Exhaustion applied to Parabolic Segment in Apostol's Calculus before I asked this question as this was the closest I found to someone asking a similar question but I could not find the answer.

For context, I have provided images of pages 1 through 8 so I can show you all that Apostol is doing for this proof.

Also, sorry for any formatting errors as this is my first time posting on this site.

Images:

Pages 1 and 2

Pages 3 and 4

Pages 5 and 6

Pages 7 and 8

Answer 1

In the early 1700's, Bernoulli proved (without Calculus) that for any positive integer $k$, there exists a unique set of rational numbers $c_0, c_1, \cdots, c_k$ such that

for any positive integer $n$, $$\sum_{i=1}^n i^k = c_0n^{(k+1)} + c_1n^k + \cdots + c_kn^1.$$

Actually, Bernoulli took it one step further, and originated the idea of Bernoulli numbers. For the purposes of this discussion, that is irrelevant.

Bernoulli's demonstration is (partially) recreated here.

So, it was well known, for centuries that

$$\sum_{i = 1}^n i^3 = \frac{1}{3}n^3 + \frac{1}{2}n^2 + \frac{1}{6}n.$$

What Apostol is doing in these first few pages is giving you an informal (but valid) glimpse of estimating the area under the curve through a series of step functions (e.g. rectangles). His algebra in the pages that you cited is valid.

Whoever originated the idea that Apostol is merely repeating had the vision that the area of the inner rectangles would always be less than $\frac{b^2}{3}$ but would get closer and closer to $\frac{b^2}{3}.$

The analyis is that the $k$-th rectangle has width $\frac{b}{n}$ and height $\left(\frac{kb}{n}\right)^2.$ Therefore, the $k$-th rectangle has area $\frac{b^3}{n^3}k^2.$

Therefore, the sum of the area of the inner rectangles is

$$\frac{b^3}{n^3} \times \sum_{k=0}^{(n-1)}k^2$$

and the sum of the area of the outer rectangles is

$$\frac{b^3}{n^3} \times \sum_{k=1}^{n}k^2.$$

From the above algebra, it is immediate that as $n \to \infty$, the sum of the area of the inner triangles will converge to $\frac{b^3}{3}$ from below, and the sum of the area of the outer triangles will converge to $\frac{b^3}{3}$ from above.

Therefore, Apostol is (through the informal use of the squeeze theorem of limits) repeating the idea that some Calculus pioneer envisioned: the exact area under the curve has to be $\frac{b^3}{3}.$

What may be confusing is that Apostol's (algebra-only, no Calculus) derivation of the formula for $\sum_{i=1}^n i^3$, on page 6, though valid, is somewhat non-standard. The non-Calculus derivation normally uses induction, as in the mathSE link that I cited.