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:
\begin{align*} U(\mathcal{P}^{(n)}) &= \sum_{k=1}^n \left( \sup_{x \in [\Delta x_k]} f(x)\right) \Delta x \\ &= \sum_{k=1}^n \left( \max_{x \in [\Delta x_k]} x^2\right) \frac{b}{n} \\ &= \frac{b}{n} \sum_{k=1}^n \left( k \cdot \frac{b}{n} \right)^2 \\ &= \frac{b^3}{n^3} \sum_{k=1}^n k^2 \\ &= \frac{b^3}{n^3} \cdot \frac{n(n+1)(2n+1)}{6} \\ &= b^3 \cdot \frac{(n+1)(2n+1)}{6n^2} \\ \lim_{n \to \infty} U(\mathcal{P}^{(n)}) &= \lim_{n \to \infty} \left[b^3 \cdot \frac{(n+1)(2n+1)}{6n^2} \right] \\ \int_0^b x^2 \, dx &= b^3 \cdot \lim_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} \\ &= b^3 \cdot \frac{2}{6} \\ &= \frac{b^3}{3} \end{align*}