Obtain $\displaystyle \int \limits_{1}^{3} x^3 \; dx$ as the limits of a sum.
Let $f\left(x\right) = x^3$ $\;\;\; \cdots \; (1)$
$f\left(x\right)$ is continuous on $\left[1, 3\right]$.
Lower limit $= a = 1$; Upper limit $= b = 3$
Divide $\left[1, 3\right]$ into n congruent sub-intervals.
Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{3 - 1}{n} = \dfrac{2}{n}$ $\;\;\; \cdots \; (2)$
$\begin{aligned}
\text{Now, } f\left(a + kh\right) & = f\left(1 + kh\right) \\\\
& = \left(1 + k h\right)^3 \;\;\; \left[\text{by equation } (1)\right] \\\\
& = 1 + 3 k h + 3 k^2 h^2 + k^3 h^3 \;\;\; \cdots \; (3)
\end{aligned}$
By definition, $\displaystyle \int \limits_{1}^{3} x^3 \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$
$\begin{aligned}
\therefore \; \int \limits_{1}^{3} x^3 \; dx & = \lim \limits_{n \to \infty} \left[h \sum_{k = 1}^{n} \left(1 + 3 k h + 3 k^2 h^2 + k^3 h^3\right)\right] \;\;\; \left[\text{by equation } (3)\right] \\\\
& = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} 1 + 3 h \sum \limits_{k = 1}^{n} k + 3 h^2 \sum \limits_{k = 1}^{n} k^2 + h^3 \sum \limits_{k = 1}^{n} k^3\right] \\\\
& = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[n + 3 \times \dfrac{2}{n} \times \dfrac{n \left(n + 1\right)}{2} + 3 \times \left(\dfrac{2}{n}\right)^2 \times \dfrac{n \left(n + 1\right) \left(2 n + 1\right)}{6} \right. \\\\
& \hspace{11em} \left. + \left(\dfrac{2}{n}\right)^3 \times \dfrac{n^2 \left(n + 1\right)^2}{4}\right] \;\;\; \left[\text{by equation } (2)\right] \\\\
& = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[n + 3 \left(n + 1\right) + \dfrac{2 \left(n + 1\right) \left(2 n + 1\right)}{n} + \dfrac{2 \left(n + 1\right)^2}{n}\right] \\\\
& = \lim\limits_{n \to \infty} \dfrac{2}{n} \left[4 n + 3 + \dfrac{2 \left(2 n^2 + 3 n + 1\right)}{n} + \dfrac{2 \left(n^2 + 2 n + 1\right)}{n}\right] \\\\
& = \lim\limits_{n \to \infty} \left[8 + \dfrac{6}{n} + 8 + \dfrac{12}{n} + \dfrac{4}{n^2} + 4 + \dfrac{8}{n} + \dfrac{4}{n^2}\right] \\\\
& \left[\text{As } n \to \infty, \; \dfrac{1}{n} \to 0 \right] \\\\
& = 8 + 8 + 4 = 20
\end{aligned}$