Evaluate the following definite integral as the limit of a sum:
$\int_{0}^{4} (x + e^{2x}) \, dx$

(N/A) We know that the definition of a definite integral as the limit of a sum is:
$\int_{a}^{b} f(x) \, dx = (b - a) \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(a + rh)$,where $h = \frac{b-a}{n}$.
Here,$a = 0$,$b = 4$,and $f(x) = x + e^{2x}$.
Thus,$h = \frac{4-0}{n} = \frac{4}{n}$.
$\int_{0}^{4} (x + e^{2x}) \, dx = 4 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} f(rh)$
$= 4 \lim_{n \to \infty} \frac{1}{n} \sum_{r=0}^{n-1} (rh + e^{2rh})$
$= 4 \lim_{n \to \infty} \frac{1}{n} \left[ h \sum_{r=0}^{n-1} r + \sum_{r=0}^{n-1} (e^{2h})^r \right]$
Using the formulas $\sum_{r=0}^{n-1} r = \frac{(n-1)n}{2}$ and the geometric series sum $\sum_{r=0}^{n-1} (e^{2h})^r = \frac{(e^{2h})^n - 1}{e^{2h} - 1} = \frac{e^{2nh} - 1}{e^{2h} - 1}$:
$= 4 \lim_{n \to \infty} \frac{1}{n} \left[ \frac{4}{n} \cdot \frac{n(n-1)}{2} + \frac{e^{8} - 1}{e^{8/n} - 1} \right]$
$= 4 \lim_{n \to \infty} \left[ 2 \cdot \frac{n-1}{n} + \frac{e^{8} - 1}{n(e^{8/n} - 1)} \right]$
$= 4 \left[ 2(1) + \frac{e^{8} - 1}{8 \lim_{n \to \infty} \frac{e^{8/n} - 1}{8/n}} \right]$
Since $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$,we have:
$= 4 \left[ 2 + \frac{e^{8} - 1}{8(1)} \right] = 8 + \frac{e^{8} - 1}{2} = \frac{16 + e^{8} - 1}{2} = \frac{15 + e^{8}}{2}$.