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.
Best Answer
I still didn't find the proof in the book, but it seems it could be derived from the Additive and the Choice of scale properties, which are the axioms of an area function. Namely:
The definition of the integral of a non-negative step function is the sum of the areas of individual rectangles, which fits in the Property 2. Therefore, by being the part of the assumed area function, the integral of the non-negative step functions also satisfies the other properties too, including the exhaustion property.
Please let me know if I went wrong somewhere.