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