# Darboux Integral Partition Question

I looking at a sample problem to use the Darboux Integral to find the integral of $f(x)=x^2$.

Textbook Solution:

For a partition P $={0=t_0<t_1<t_2<...<t_n=b}$ we have, $U(f,P) = \sum^n_{k=1}sup(x^2:x \in [t_{k-1},t_k])*(t_k-t_{k-1})$

(I get lost here) If we choose $t_k=\frac{kb}{n}$ then we have:

$U(f,P) = \sum^n_{k=1} \frac{k^2b^2}{n^2}(\frac{b}{n})$...

My question is how was $t_k=\frac{kb}{n}$ "picked". There is no explanation for it in the book.

What is being done is we are partitioning the interval $[0,b]$ into $n$ subintervals of equal length. So each subinterval is $1/n$ the size of the full interval, so each length is $b/n$. So the interval endpoints are $$0 < \frac{b}{n} < \frac{2b}{n} < \frac{3b}{n} < \cdots< \frac{(n-1)b}{n} < b$$
This is what is meant by the "picked" values for the partition subinterval endpoints. Rather than choosing an arbitrary partition, it is just the standard $n$-partition $\mathcal{P}^{(n)}$, the partition that splits the interval into $n$ subintervals of equal length.
Since $f(x) = x^2$ is a continuous function, the supremum on a subinterval is a maximum on the subinterval. Since $f(x) = x^2$ is increasing on $[0,b]$, the maximum value occurs at the right endpoint of the subinterval. So, denoting the partition endpoints as $x_k$ and the $k$-th subinterval as $[\Delta x_k] = [x_{k-1},x_k]$, each with length $\Delta x = b/n$ we see: