Evaluate $\int_{0}^{2}(x^{2}+3) dx$ as a limit of sums.

Let $I = \int_{0}^{2}(x^{2}+3) dx$.
Here,$a = 0$,$b = 2$,and $h = \frac{b-a}{n} = \frac{2-0}{n} = \frac{2}{n}$.
Thus,$nh = 2$ and $f(x) = x^{2}+3$.
By the definition of the definite integral as a limit of sums:
$\int_{a}^{b} f(x) dx = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a+rh)$.
Substituting the values:
$I = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} f(0+rh) = \lim_{n \to \infty} \frac{2}{n} \sum_{r=0}^{n-1} ((rh)^{2}+3)$.
$I = \lim_{n \to \infty} \frac{2}{n} [\sum_{r=0}^{n-1} r^{2}h^{2} + \sum_{r=0}^{n-1} 3]$.
Since $h = \frac{2}{n}$,$h^{2} = \frac{4}{n^{2}}$.
$I = \lim_{n \to \infty} \frac{2}{n} [\frac{4}{n^{2}} \sum_{r=0}^{n-1} r^{2} + 3n]$.
Using the formula $\sum_{r=0}^{n-1} r^{2} = \frac{(n-1)n(2n-1)}{6}$:
$I = \lim_{n \to \infty} \frac{2}{n} [\frac{4}{n^{2}} \cdot \frac{(n-1)n(2n-1)}{6} + 3n]$.
$I = \lim_{n \to \infty} [\frac{8}{n^{3}} \cdot \frac{(n-1)n(2n-1)}{6} + 6]$.
$I = \lim_{n \to \infty} [\frac{4}{3} \cdot \frac{n(n-1)(2n-1)}{n^{3}} + 6]$.
$I = \lim_{n \to \infty} [\frac{4}{3} \cdot (1)(1 - \frac{1}{n})(2 - \frac{1}{n}) + 6]$.
As $n \to \infty$,$\frac{1}{n} \to 0$:
$I = \frac{4}{3} \cdot (1)(1)(2) + 6 = \frac{8}{3} + 6 = \frac{8+18}{3} = \frac{26}{3}$.