Obtain $\displaystyle \int \limits_{0}^{\pi} \sin x \; dx$ $\;$ as the limit of a sum.
Let $f\left(x\right) = \sin x$ $\;\;\; \cdots \; (1)$
$f\left(x\right)$ is continuous on $\left[0, \pi\right]$.
Lower limit $= a = 0$; Upper limit $= b = \pi$
Divide $\left[0, \pi\right]$ into n congruent sub-intervals.
Length of each sub-interval $= h = \dfrac{b - a}{n} = \dfrac{\pi - 0}{n} = \dfrac{\pi}{n}$
$\implies$ $n h = \pi$ $\;\;\; \cdots \; (2)$
Now, $f\left(a + kh\right) = f\left(kh\right) = \sin \left(k h\right)$ $\;\;\; \cdots \; (3)$ $\;\;\;$ [from equation $(1)$]
Since $h = \dfrac{1}{n}$ $\implies$ when $n \rightarrow \infty, \; h \rightarrow 0$ $\;\;\; \cdots \; (4)$
By definition, $\displaystyle \int \limits_{0}^{\pi} \sin x \; dx = \lim\limits_{n \to \infty} \left[h \sum \limits_{k = 1}^{n} f \left(a + k h\right)\right]$
$\begin{aligned}
\therefore \; \int \limits_{0}^{\pi} \sin x \; dx & = \lim\limits_{h \to 0} \left[h \sum \limits_{k = 1}^{n} \sin\left(k h\right)\right] \;\;\; \left[\text{by equations }(3) \text{ and }(4)\right] \\\\
& = \lim\limits_{h \to 0} h \left[\sin \left(h\right) + \sin \left(2 h\right) + \sin \left(3 h\right) + \cdots + \sin \left(n h\right)\right] \;\;\; \cdots \; (5)
\end{aligned}$
Let $S_n = \sin \left(h\right) + \sin \left(2 h\right) + \sin \left(3 h\right) + \cdots + \sin \left(n h\right)$ $\;\;\; \cdots \; (6)$
Multiplying equation $(6)$ with $2 \sin \left(\dfrac{h}{2}\right)$ gives
$\begin{aligned}
2 \sin \left(\dfrac{h}{2}\right) S_n & = 2 \sin \left(h\right) \sin \left(\dfrac{h}{2}\right) + 2 \sin \left(2 h\right) \sin \left(\dfrac{h}{2}\right) \\\\
& \hspace{3em} + 2 \sin \left(3 h\right) \sin \left(\dfrac{h}{2}\right) + \cdots + 2 \sin \left(n h\right) \sin \left(\dfrac{h}{2}\right) \\\\
& \left[\text{Note: } 2 \sin A \; \sin B = \cos \left(A - B\right) - \cos \left(A + B\right)\right] \\\\
& = \cos \left(h - \dfrac{h}{2}\right) - \cos \left(h + \dfrac{h}{2}\right) + \cos \left(2 h - \dfrac{h}{2}\right) - \cos \left(2 h + \dfrac{h}{2}\right) \\\\
& \hspace{3em}+ \cos \left(3 h - \dfrac{h}{2}\right) - \cos \left(3 h + \dfrac{h}{2}\right) + \\\\
& \hspace{7em} \cdots + \cos \left(n h - \dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right) \\\\
& = \cos \left(\dfrac{h}{2}\right) - \cos \left(\dfrac{3 h}{2}\right) + \cos \left(\dfrac{3 h}{2}\right) - \cos \left(\dfrac{5 h}{2}\right) \\\\
& \hspace{3em} + \cos \left(\dfrac{5 h}{2}\right) + \cdots + \cos \left(n h - \dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right)
\end{aligned}$
$\begin{aligned}
\therefore \; S_n & = \dfrac{\cos \left(\dfrac{h}{2}\right) - \cos \left(n h + \dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \\\\
& = \dfrac{\cos \left(\dfrac{h}{2}\right) - \cos \left(\pi + \dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \;\;\; \left[\text{from equation }(2)\right] \\\\
& = \dfrac{\cos \left(\dfrac{h}{2}\right) + \cos \left(\dfrac{h}{2}\right)}{2 \sin \left(\dfrac{h}{2}\right)} \;\;\; \left[\text{Note: } \cos \left(\pi + \theta\right) = - \cos \theta\right] \\\\
& = \dfrac{2 \cos \left(h / 2\right)}{2 \sin \left(h / 2\right)} \\\\
& = \dfrac{\cos \left(h / 2\right)}{\sin \left(h / 2\right)} \;\;\; \cdots \; (7)
\end{aligned}$
$\therefore$ $\;$ In view of equation $(7)$, equation $(5)$ becomes
$\begin{aligned}
\int \limits_{0}^{\pi} \sin x \; dx & = \lim\limits_{h \to 0} \; h \times \dfrac{\cos \left(h / 2\right)}{\sin \left(h / 2\right)} \\\\
& = \lim\limits_{\frac{h}{2} \to 0} \; \cos \left(\dfrac{h}{2}\right) \times 2 \times \dfrac{1}{\lim\limits_{\frac{h}{2} \to 0} \; \dfrac{\sin \left(h / 2\right)}{h / 2}} \;\;\; \left[\text{Note: As } h \to 0, \; \dfrac{h}{2} \to 0\right] \\\\
& = 1 \times 2 \times 1 = 2
\end{aligned}$