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Solution:

It is known that,

$\int_{a}^{b} f(x) d x=(b-a) \lim _{n \rightarrow \infty} \frac{1}{n}[f(a)+f(a+h)+\ldots+f(a+(n-1) h)]$, where $h=\frac{b-a}{n}$

Here, $a=0, b=4$, and $f(x)=x+e^{2 x}$

$\therefore h=\frac{4-0}{n}=\frac{4}{n}$

$\Rightarrow \int_{0}^{4}\left(x+e^{2 x}\right) d x=(4-0) \lim _{n \rightarrow \infty} \frac{1}{n}[f(0)+f(h)+f(2 h)+\ldots+f((n-1) h)]$

$=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\left(0+e^{0}\right)+\left(h+e^{2 h}\right)+\left(2 h+e^{22 h}\right)+\ldots+\left\{(n-1) h+e^{2(n-1) h}\right\}\right]$

$=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[1+\left(h+e^{2 h}\right)+\left(2 h+e^{4 h}\right)+\ldots+\left\{(n-1) h+e^{2(n-1) h}\right\}\right]$

$=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\{h+2 h+3 h+\ldots+(n-1) h\}+\left(1+e^{2 h}+e^{4 h}+\ldots+e^{2(n-1) h}\right)\right]$

$=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[h\{1+2+\ldots(n-1)\}+\left(\frac{e^{2 h m}-1}{e^{2 h}-1}\right)\right]$

$=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{(h(n-1) n)}{2}+\left(\frac{e^{2 h n}-1}{e^{2 h}-1}\right)\right]$

$=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{4}{n}, \frac{(n-1) n}{2}+\left(\frac{e^{8}-1}{e^{\frac{8}{n}}-1}\right)\right]$

$=4 \lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{4}{n} \cdot \frac{(n-1) n}{2}+\left(\frac{e^{8}-1}{e^{\frac{8}{n}}-1}\right)\right]$

$=4(2)+4 \lim _{n \rightarrow \infty} \frac{\left(e^{8}-1\right)}{\left(\frac{\frac{8}{n}-1}{\frac{8}{n}}\right)} 8$

$=8+\frac{4 \cdot\left(e^{8}-1\right)}{8}$       $\left(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}=1\right)$

$=8+\frac{e^{8}-1}{2}$

$=\frac{15+e^{8}}{2}$