Proving the Riemann Sum for the Integral of x^2 from 1 to 3

[NATURE.M]

So my textbook asks to show $\int^{3}_{1} x^{2}dx = \frac{26}{3}$.
They let the partition P = {$x_{0},...,x_{n}$}, and define the upper Riemann sum as U(P) = $\sum^{i=1}_{n} x_{i}Δx_{i}$ and lower sum as
L(P) = $\sum^{i=1}_{n} x_{i-1}Δx_{i}$

I understand this part, but the next part is where I'm confused.

For each index i, 1$\leq$i$\leq$n,
$3x^{2}_{i-1}\leq x^{2}_{i-1} + x_{i-1}x_{i}+x^{2}_{i}\leq3x^{2}_{i}$

It's probably something I'm overlooking by where does the middle term come from and the 3 ??

 

[Office_Shredder]

The inequality comes from the fact that $x_{i-1} \leq x_i$ for all i, and therefore
$$x_{i-1}^2 \leq x_{i-1}^2$$
and
$$x_{i-1}^2 \leq x_{i-1}x_i$$
and
$$x_{i-1}^2 \leq x_{i}^2$$
so adding these all together give
$$3 x_{i-1}^2 \leq x_{i-1}^2 + x_{i-1} x_i + x_{i}^2$$

 

[Newu]

I think they want to express this, because we have
$$3x_{i-1}^{2}\leq x_{i-1}^{2} + x_{i-1} x_{i}^{2} +x_{i}^{2}\leq 3 x_{i}^{2}, $$
so, $$ 3x_{i-1}^{2}(x_{i}-x_{i-1})\leq (x_{i-1}^{2} + x_{i-1} x_{i} +x_{i}^{2}) (x_{i}-x_{i-1}) \leq 3 x_{i}^{2} (x_{i}-x_{i-1}) ,$$
namely,$$3 x_{i-1}^{2}\Delta{x_{i}}\leq ( x_{i}^{3} - x_{i-1}^{3} )\leq 3x_{i}^{2}\Delta{x_{i}}.$$
Then, we can get
$$\sum_{i=1}^{n} x_{i-1}^{2} \Delta x_{i} \leq \frac{1}{3}( x_{n}^{3} - x_{0}^{3} ) \leq \sum_{i=1}^{n} x_{i}^{2} \Delta x_{i} ,$$
that is to say,
$$L(P)\leq \frac{26}{3} \leq U(P).$$
Because $f(x)=x^{2}$ is Riemann-integrable on $[1,3]$, let $I=\int _{1}^{3}f(x)dx\, , \lambda = \max \limits_{1\leq i \leq n}(\Delta{x_{i}})\rightarrow 0$, so
$$\lim_{\lambda\rightarrow 0}{U(P)}=L=I=l= \lim_{\lambda \rightarrow 0}{L(P)}.$$
According to the former reasoning, both of $L$ and $l$ equal $\frac{26}{3}$, so $I=\frac{26}{3}$.