Evaluate the following definite integral as the limit of a sum: $\int_{-1}^{1} e^{x} dx$

Let $I = \int_{-1}^{1} e^{x} dx$.
It is known that,
$\int_{a}^{b} f(x) dx = (b - a) \lim_{n \to \infty} \frac{1}{n} [f(a) + f(a + h) + \dots + f(a + (n - 1)h)]$,where $h = \frac{b - a}{n}$.
Here,$a = -1$,$b = 1$,and $f(x) = e^{x}$.
$\therefore h = \frac{1 - (-1)}{n} = \frac{2}{n}$.
$\therefore I = 2 \lim_{n \to \infty} \frac{1}{n} [f(-1) + f(-1 + \frac{2}{n}) + f(-1 + 2 \cdot \frac{2}{n}) + \dots + f(-1 + (n - 1)\frac{2}{n})]$.
$I = 2 \lim_{n \to \infty} \frac{1}{n} [e^{-1} + e^{-1 + \frac{2}{n}} + e^{-1 + \frac{4}{n}} + \dots + e^{-1 + (n - 1)\frac{2}{n}}]$.
$I = 2 \lim_{n \to \infty} \frac{e^{-1}}{n} [1 + e^{\frac{2}{n}} + e^{\frac{4}{n}} + \dots + e^{(n - 1)\frac{2}{n}}]$.
This is a geometric progression with $n$ terms,where the first term $a = 1$ and common ratio $r = e^{\frac{2}{n}}$.
Using the sum formula $S_n = \frac{a(r^n - 1)}{r - 1}$,we get:
$I = 2 \lim_{n \to \infty} \frac{e^{-1}}{n} \left[ \frac{e^{\frac{2n}{n}} - 1}{e^{\frac{2}{n}} - 1} \right] = 2 \lim_{n \to \infty} \frac{e^{-1}}{n} \left[ \frac{e^{2} - 1}{e^{\frac{2}{n}} - 1} \right]$.
$I = e^{-1} (e^{2} - 1) \lim_{n \to \infty} \frac{2/n}{e^{2/n} - 1}$.
Since $\lim_{h \to 0} \frac{e^{h} - 1}{h} = 1$,where $h = \frac{2}{n}$,we have:
$I = (e - e^{-1}) \cdot 1 = e - \frac{1}{e} = \frac{e^{2} - 1}{e}$.