Integral of x² from 0 to b - using archimedes sum of squares - Apostol's I'm reading Apostol's Calculus book and in the first chapter is presented the way archimedes found the sum of the square and how it can be used to calculate the integral of $x²$. But I'm not able to follow some steps of the proof.
from the book:
We subdivided the base in $n$ parts each with length  $\ \frac{b}{n} $, a typical point corresponds to $\frac{kb}{n}$ where $k$ takes values from $k = 1, 2, 3, ..., n$
We can construct rectangles from for each $k th$  point:
$Base = \frac{b}{n}$
$Height = (\frac{kb}{n})^2$
$Area = Base * Height =  \frac{b}{n} . (\frac{kb}{n})^2$
$Area  =  \frac{b^3}{n^3}.k^2 $
If we sum all the rectangles, we get a bit more than the area under the curve $x^2$
$S_{big} = \frac{b^3}{n^3}.(1² + 2² + 3² + ... + (n-1)² + n²)$
If we can construct smaller rectangles, using $n-1$ points, we we get a bit less than the area under the curve $x^2$.
$S_{small} = \frac{b^3}{n^3}.(1² + 2² + 3² + ... + (n-1)²)$
So the real area under the curve $x^2$ is between the two areas:
$S_{small} < A < S_{big}$
After a bit of algebra we get that:
$S_{big} = \frac{b^3}{n^3} . (\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6})$
$S_{small} = \frac{b^3}{n^3} . (\frac{n^3}{3} - \frac{n²}{2} + \frac{n}{6})$
To prove that $A$ is $\frac{b^3}{3}$ he uses this inequalities:
$1² + 2² + 3² + ... + (n-1)² < \frac{n³}{3} < 1² + 2² + 3² + ... + n²$
But I don't understand where the $\frac{n³}{3}$ came from. And can't follow the proof
EDIT:
After taking the average of the two expression:
$\frac{(\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6}) + (\frac{n^3}{3} - \frac{n²}{2} + \frac{n}{6})}{2} = \frac{n^3}{3} + \frac{n}{6}$
So I understan where $\frac{n³}{3}$ came from, but why is $\frac{n}{6}$ is thrown away?
 A: The central idea is that you have to find a unique number (independent of $ n$) which lies between every $S_{\text{small}} $ and $S_{\text{big}} $ both of which depend on $n$. First thing to note here is that there can't be two distinct numbers say $A, A'$ lying between these $S$'s. Without any loss of generality let $A<A' $ and then if $$S_{\text{small}}\leq A<A'\leq S_{\text{big}} $$ then we have $$S_{\text{big}} - S_{\text{small}} \geq A'-A>0$$ ie $$\frac{b^{3}}{n}\geq A'-A$$ Clearly this is a contradiction because we can choseo the positive integer $n$ as large as we want and we can definitely choose $n>b^3/(A'-A)$.
Next we need to show that $b^3/3$ lies between between $S_{\text{small}} $ and $S_{\text{big}} $. This is easily done by noting that $$1-\frac{3}{2n}+\frac{1}{2n^2}<1<1+\frac{3}{2n}+\frac{1}{2n^2}$$ and thus $A=b^3/3$.
A: You may want to read this. It really is not an easy or intuitive inequality to find.

Alternatively, you can consider this:
Eventually, you would want $n$ to approach $+\infty$.
So let's consider
\begin{align}
S_{\text{big}} &= \frac{b^3}{n^3} \cdot \left(\frac{n^3}{3} + \frac{n²}{2} + \frac{n}{6}\right)\\
&=b^3\left(\frac{1}{3} + \frac{1}{2n} + \frac{1}{6n^2}\right)\\
\lim_{n\to\infty}S_{\text{big}} &=b^3\left(\frac{1}{3} +0+0\right)\\
&=\frac{b^3}{3}
\end{align}
Similarly, you will have 
$$\lim_{n\to\infty}S_{\text{small}} =\frac{b^3}{3}$$
too.
So you can actually just squeeze here already.
