Evaluate the following definite integral as a limit of sums:
$\int_{2}^{3} x^{2} d x$

(N/A) It is known that,
$\int_{a}^{b} f(x) d x = (b-a) \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a+rh)$,where $h = \frac{b-a}{n}$.
Here,$a=2, b=3$,and $f(x)=x^{2}$.
$\Rightarrow h = \frac{3-2}{n} = \frac{1}{n}$.
$\therefore \int_{2}^{3} x^{2} d x = (3-2) \lim _{n \rightarrow \infty} \frac{1}{n} \left[ f(2) + f\left(2+\frac{1}{n}\right) + f\left(2+\frac{2}{n}\right) + \dots + f\left(2+\frac{n-1}{n}\right) \right]$
$= \lim _{n \rightarrow \infty} \frac{1}{n} \left[ 2^{2} + \left(2+\frac{1}{n}\right)^{2} + \left(2+\frac{2}{n}\right)^{2} + \dots + \left(2+\frac{n-1}{n}\right)^{2} \right]$
$= \lim _{n \rightarrow \infty} \frac{1}{n} \left[ n(2^{2}) + \frac{1}{n^{2}}(1^{2} + 2^{2} + \dots + (n-1)^{2}) + \frac{4}{n}(1 + 2 + \dots + (n-1)) \right]$
Using the formulas $\sum_{k=1}^{m} k = \frac{m(m+1)}{2}$ and $\sum_{k=1}^{m} k^{2} = \frac{m(m+1)(2m+1)}{6}$ with $m = n-1$:
$= \lim _{n \rightarrow \infty} \frac{1}{n} \left[ 4n + \frac{1}{n^{2}} \cdot \frac{(n-1)(n)(2n-1)}{6} + \frac{4}{n} \cdot \frac{(n-1)n}{2} \right]$
$= \lim _{n \rightarrow \infty} \left[ 4 + \frac{(n-1)(2n-1)}{6n^{2}} + \frac{2(n-1)}{n} \right]$
$= \lim _{n \rightarrow \infty} \left[ 4 + \frac{1}{6}\left(1-\frac{1}{n}\right)\left(2-\frac{1}{n}\right) + 2\left(1-\frac{1}{n}\right) \right]$
$= 4 + \frac{1}{6}(1)(2) + 2(1) = 4 + \frac{1}{3} + 2 = 6 + \frac{1}{3} = \frac{19}{3}$.