Evaluate the following definite integral as a limit of sums: $\int_{a}^{b} x \, dx$

It is known that,$\int_a^b f(x) \, dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f(a) + f(a + h) + \dots + f(a + (n - 1)h)]$ where $h = \frac{b - a}{n}$.
Here,$f(x) = x$.
$\therefore \int_a^b x \, dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [a + (a + h) + (a + 2h) + \dots + (a + (n - 1)h)]$.
$= (b - a) \lim_{n \to \infty} \frac{1}{n} [(a + a + \dots + a) + (h + 2h + \dots + (n - 1)h)]$.
$= (b - a) \lim_{n \to \infty} \frac{1}{n} [na + h(1 + 2 + \dots + (n - 1))]$.
$= (b - a) \lim_{n \to \infty} \frac{1}{n} \left[ na + h \frac{(n - 1)n}{2} \right]$.
$= (b - a) \lim_{n \to \infty} \left[ a + \frac{(n - 1)nh}{2n} \right] = (b - a) \lim_{n \to \infty} \left[ a + \frac{(n - 1)h}{2} \right]$.
Since $h = \frac{b - a}{n}$,we have $\lim_{n \to \infty} \frac{(n - 1)(b - a)}{2n} = \frac{b - a}{2}$.
$= (b - a) \left[ a + \frac{b - a}{2} \right] = (b - a) \left[ \frac{2a + b - a}{2} \right]$.
$= \frac{(b - a)(b + a)}{2} = \frac{b^2 - a^2}{2}$.