$$\lim\limits\_{n \to \infty}\frac{\sin(\frac{b}{n})+\sin(\frac{2b}{n})+...+\sin(\frac{nb}{n})}{n}$$

It is clear that sum in sigma notation is $$\sum\_{i=1}^n \frac{\sin(\frac{bi}{n})}{n} $$ while general formula is $$\sum\_{i=1}^n f(c\_i)\Delta{x} $$

In this problem $\Delta{x} = \frac{1}{n}$, so $b-a=1$

How do I get to know that $b = 1$ and $a=0$ to evaluate definite integral and why $f(x) = \sin(bx)$ rather than $\sin(\frac{bx}{n})$?