Evaluate the following definite integral as a limit of sums:
$\int_{0}^{5}(x+1) d x$

(D) Let $I = \int_{0}^{5}(x+1) d x$.
It is known that,$\int_a^b f (x)dx = (b - a) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a + rh)$,where $h = \frac{b-a}{n}$.
Here,$a=0, b=5$,and $f(x)=(x+1)$.
$\Rightarrow h = \frac{5-0}{n} = \frac{5}{n}$.
$\therefore \int_0^5 (x + 1) dx = 5 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f\left(r \cdot \frac{5}{n}\right)$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} \left( \frac{5r}{n} + 1 \right)$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \left[ \sum_{r=0}^{n-1} 1 + \frac{5}{n} \sum_{r=0}^{n-1} r \right]$.
$= 5 \lim_{n \to \infty} \frac{1}{n} \left[ n + \frac{5}{n} \cdot \frac{(n-1)n}{2} \right]$.
$= 5 \lim_{n \to \infty} \left[ 1 + \frac{5(n-1)}{2n} \right]$.
$= 5 \lim_{n \to \infty} \left[ 1 + \frac{5}{2} \left( 1 - \frac{1}{n} \right) \right]$.
$= 5 \left[ 1 + \frac{5}{2} \right] = 5 \left[ \frac{7}{2} \right] = \frac{35}{2}$.