Obtain $\displaystyle \int \limits_{0}^{1} e^{2 - 3 x} \; dx$ as the limit of a sum.
Let $f\left(x\right) = e^{2 - 3x}$ $\;\;\; \cdots \; (1)$
$f\left(x\right)$ is continuous on $\left[0,1\right]$.
Lower limit $= a = 0$; Upper limit $= b = 1$
Divide $\left[0,1\right]$ into n congruent sub-intervals.
Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{1 - 0}{n} = \dfrac{1}{n}$
$\implies$ $n h = 1$ $\;\;\; \cdots \; (2)$
Now, $f\left(a + kh\right) = f\left(kh\right) = e^{2 - 3 k h} = \dfrac{e^2}{e^{3 k h}}$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]
Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$
By definition, $\displaystyle \int \limits_{0}^{1} e^{2 - 3x} \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$
$\begin{aligned}
\therefore \; \int \limits_{0}^{1} e^{2 - 3x} \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} \dfrac{e^2}{e^{3kh}}\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\
& = \lim\limits_{h \to 0} e^2 h \left[\dfrac{1}{e^{3h}} + \dfrac{1}{e^{6h}} + \dfrac{1}{e^{9h}} + \cdots + \dfrac{1}{e^{nh}}\right] \\\\
& \left[\begin{aligned}
\text{Note: } & \dfrac{1}{e^{3h}} + \dfrac{1}{e^{6h}} + \dfrac{1}{e^{9h}} + \cdots + \dfrac{1}{e^{nh}} \text{ is a geometric series} \\\\
& \text{with first term} = \dfrac{1}{e^{3h}} \text{ and common ratio } = \dfrac{1}{e^{3h}} \\\\
& \text{Sum to n terms of a geometric series with first term A} \\\\
& \text{ and common ratio R } = \dfrac{A \left(1 - R^n\right)}{1 - R} \text{ when } R < 1
\end{aligned}\right] \\\\
& = \lim\limits_{h \to 0} e^2 h \left[\dfrac{\dfrac{1}{e^{3h}} \left(1 - \dfrac{1}{e^{3nh}}\right)}{1 - \dfrac{1}{e^{3h}}}\right] \\\\
& = \lim\limits_{h \to 0} e^2 h \left[\dfrac{1}{e^{3h}} \times \dfrac{\left(1 - \dfrac{1}{e^3}\right)}{e^{3h} - 1} \times e^{3h}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\
& = \dfrac{e^3 -1}{e} \; \lim\limits_{h \to 0} \dfrac{1}{\left(\dfrac{e^{3h} - 1}{3h}\right) \times 3} \\\\
& = \dfrac{1}{3} \left(\dfrac{e^3 - 1}{e}\right) \times \dfrac{1}{\log_{e} e} \\\\
& = \dfrac{1}{3} \left(\dfrac{e^3 - 1}{e}\right)
\end{aligned}$