Question
Evaluate : $\int_0^{\pi / 2} \sin x \cdot d x$
$\begin{aligned} & \therefore \quad \sum_{r=1}^n \sin r h=\frac{\cos \frac{h}{2}+\sin \frac{h}{2}}{2 \sin \frac{h}{2}} \\ & \text { Now from I, } \\ & \int_0^{\pi / 2} \sin x \cdot d x=\lim _{n \rightarrow \infty} \sum_{r=1}^n h \cdot \sin r h \\ & =\lim _{n \rightarrow \infty} h \cdot\left[\frac{\cos \frac{h}{2}+\sin \frac{h}{2}}{2 \sin \frac{h}{2}}\right] \\ & \because \quad n h=\frac{\pi}{4} \text { as } n \rightarrow \infty \Rightarrow h \rightarrow 0\left(\frac{1}{n} \rightarrow 0\right) \\ & =\lim _{\substack{h \rightarrow \infty \\ h \rightarrow 0}}\left[\frac{\cos \frac{h}{2}+\sin \frac{h}{2}}{\frac{2 \cdot \sin \frac{h}{2}}{h}}\right] \\ & =\frac{\cos 0+\sin 0}{\left(\frac{1}{2}\right)} \\ & =\frac{1+0}{2 \cdot \frac{1}{2}}=1 \\ & \therefore \quad \int_0^{\pi / 2} \sin x \cdot d x=1 \\ & \end{aligned}$
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$\frac{1}{a}-\frac{1}{b}=\frac{1}{a^{\prime}}-\frac{1}{b^{\prime}}$
| $\text{X}=\text{x}_\text{i}:$ | $-2$ | $-1$ | $0$ | $1$ |
| $\text{P}(\text{X}=\text{x}_\text{i}):$ | $\frac{1-\text{a}}{4}$ | $\frac{1+2\text{a}}{4}$ | $\frac{1-2\text{a}}{4}$ | $\frac{1+\text{a}}{4}$ |