Question

$\underset{0}{\overset{\mathrm{\pi }/2}{\int }}\sin xdx$

Answer 1

$$\begin{gathered} {\int }_a^{b}f\left(x\right)dx=\underset{h\to 0}{\lim }h\left[f\left(a\right)+f\left(a+h\right)+f\left(a+2h\right)+...+f\left(a+\left(n-1\right)h\right)\right] \\ \mathrm{where}h=\frac{b-a}{n} \end{gathered}$$
$$\begin{gathered} \mathrm{Here}a=0,b=\frac{\mathrm{\pi }}{2},f\left(x\right)=\sin x,h=\frac{{\displaystyle \frac{\mathrm{\pi }}{2}-0}}{n}=\frac{\mathrm{\pi }}{2n} \\ \mathrm{Therefore}, \\ I={\int }_0^{\frac{\mathrm{\pi }}{2}}\sin xdx \\ =\underset{h\to 0}{\lim }h\left[f\left(0\right)+f\left(0+h\right)+...+f\left(0+\left(n-1\right)h\right)\right] \\ =\underset{h\to 0}{\lim }h\left[\sin 0+\sin h+\sin 2h+...+\sin \left(n-1\right)h\right] \\ =\underset{h\to 0}{\lim }h\left[\frac{\sin \left(\left(n-1\right){\displaystyle \frac{h}{2}}\right)\sin {\displaystyle \frac{nh}{2}}}{\sin \frac{h}{2}}\right] \\ =\underset{h\to 0}{\lim }\left[\frac{{\displaystyle \frac{h}{2}}}{\sin {\displaystyle \frac{h}{2}}}\times 2\sin \left(\frac{\mathrm{\pi }}{4}-\frac{h}{2}\right)\sin \frac{\mathrm{\pi }}{4}\right]\left(\mathrm{Using}nh=\frac{\mathrm{\pi }}{2}\right) \\ =\underset{h\to 0}{\lim }\frac{{\displaystyle \frac{h}{2}}}{\sin {\displaystyle \frac{h}{2}}}\times \underset{h\to 0}{\lim }2\sin \left(\frac{\mathrm{\pi }}{4}-\frac{h}{2}\right)\sin \frac{\mathrm{\pi }}{4} \\ =2\sin \frac{\mathrm{\pi }}{4}\sin \frac{\mathrm{\pi }}{4}=2\times \frac{1}{\sqrt{2}}\times \frac{1}{\sqrt{2}}=1 \end{gathered}$$