Evaluate $\int_{0}^{1} e^{2-3 x} d x$ as a limit of a sum.

(N/A) Let $I = \int_{0}^{1} e^{2-3 x} d x$.
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,$a = 0, b = 1,$ and $f(x) = e^{2-3x}$.
$\Rightarrow h = \frac{1 - 0}{n} = \frac{1}{n}$.
$\therefore \int_0^1 e^{2 - 3x} dx = (1 - 0) \lim_{n \to \infty} \frac{1}{n} [f(0) + f(h) + \dots + f((n - 1)h)]$
$= \lim_{n \to \infty} \frac{1}{n} [e^2 + e^{2 - 3h} + e^{2 - 6h} + \dots + e^{2 - 3(n - 1)h}]$
$= \lim_{n \to \infty} \frac{e^2}{n} [1 + e^{-3h} + e^{-6h} + \dots + e^{-3(n - 1)h}]$
Using the sum of a geometric series $S_n = \frac{a(1 - r^n)}{1 - r}$,where $r = e^{-3h}$:
$= \lim_{n \to \infty} \frac{e^2}{n} \left[ \frac{1 - (e^{-3h})^n}{1 - e^{-3h}} \right] = \lim_{n \to \infty} \frac{e^2}{n} \left[ \frac{1 - e^{-3}}{1 - e^{-3/n}} \right]$
$= e^2 (1 - e^{-3}) \lim_{n \to \infty} \frac{1}{n(1 - e^{-3/n})} = e^2 (1 - e^{-3}) \lim_{n \to \infty} \frac{1/n}{1 - e^{-3/n}}$
Let $x = -3/n$. As $n \to \infty, x \to 0$. Then $\frac{1}{n} = -x/3$.
$= e^2 (1 - e^{-3}) \lim_{x \to 0} \frac{-x/3}{1 - e^x} = e^2 (1 - e^{-3}) \cdot \frac{1}{3} \lim_{x \to 0} \frac{x}{e^x - 1}$
Since $\lim_{x \to 0} \frac{x}{e^x - 1} = 1$,we get:
$= \frac{e^2 (1 - e^{-3})}{3} = \frac{e^2 - e^{-1}}{3} = \frac{1}{3} \left( e^2 - \frac{1}{e} \right)$.