I do not understand how
$\displaystyle \lim_{n \rightarrow \infty} \sum_{i=1}^n \, c \frac{b-a}{n} = c(b-a) \, .$
Can somebody show the steps?
Note that $\displaystyle \sum_{i=1}^{n} \frac{1}{n} = 1$, thus we obtain
$\displaystyle \lim_{n \to \infty}\sum_{i=1}^{n} c \frac{b-a}{n} = c(b-a) \lim_{n \to \infty} \sum_{i=1}^{n} \frac{1}{n} = c(b-a) \lim_{n \to \infty} 1 = c(b-a)$