Evaluate $\int_{0}^{2} e^{x} d x$ as a limit of sums.

(N/A) Let $I = \int_{0}^{2} e^{x} d x$.
Here,$a = 0$ and $b = 2$.
We know that $h = \frac{b-a}{n}$,so $nh = 2 - 0 = 2$.
By the definition of the definite integral as a limit of sums,$\int_{a}^{b} f(x) d x = \lim_{n \to \infty} h \sum_{r=0}^{n-1} f(a + rh)$.
Here,$f(x) = e^{x}$,so $f(0 + rh) = e^{rh}$.
Thus,$I = \lim_{n \to \infty} h \sum_{r=0}^{n-1} e^{rh} = \lim_{n \to \infty} h [1 + e^{h} + e^{2h} + \dots + e^{(n-1)h}]$.
This is a geometric progression with $n$ terms,first term $1$,and common ratio $e^{h}$.
$I = \lim_{n \to \infty} h \left[ \frac{1(e^{nh} - 1)}{e^{h} - 1} \right] = \lim_{n \to \infty} \left( \frac{e^{nh} - 1}{\frac{e^{h} - 1}{h}} \right)$.
Since $nh = 2$,as $n \to \infty$,$h \to 0$.
$I = \frac{e^{2} - 1}{\lim_{h \to 0} \frac{e^{h} - 1}{h}} = \frac{e^{2} - 1}{1} = e^{2} - 1$.