Question

\[\int\limits_0^{\pi/2} \sin x\ dx\]

Answer 1

\[\int_a^b f\left( x \right) d x = \lim_{h \to 0} h\left[ f\left( a \right) + f\left( a + h \right) + f\left( a + 2h \right) + . . . + f\left( a + \left( n - 1 \right)h \right) \right]\]
\[\text{where }h = \frac{b - a}{n}\]

\[\text{Here }a = 0, b = \frac{\pi}{2}, f\left( x \right) = \sin x, h = \frac{\frac{\pi}{2} - 0}{n} = \frac{\pi}{2n}\]
Therefore,
\[I = \int_0^\frac{\pi}{2} \sin x\ d\ x\]
\[ = \lim_{h \to 0} h\left[ f\left( 0 \right) + f\left( 0 + h \right) + . . . + f\left( 0 + \left( n - 1 \right)h \right) \right]\]
\[ = \lim_{h \to 0} h\left[ \sin0 + \sinh + \sin2h + . . . + \sin\left( n - 1 \right)h \right]\]
\[ = \lim_{h \to 0} h\left[ \frac{\sin\left( \left( n - 1 \right)\frac{h}{2} \right)\sin\frac{nh}{2}}{\sin\frac{h}{2}} \right]\]
\[ = \lim_{h \to 0} \left[ \frac{\frac{h}{2}}{\sin\frac{h}{2}} \times 2\sin\left( \frac{\pi}{4} - \frac{h}{2} \right)\sin\frac{\pi}{4} \right] \left( Using nh = \frac{\pi}{2} \right)\]
\[ = \lim_{h \to 0} \frac{\frac{h}{2}}{\sin\frac{h}{2}} \times \lim_{h \to 0} 2\sin\left( \frac{\pi}{4} - \frac{h}{2} \right)\sin\frac{\pi}{4}\]
\[ = 2\sin\frac{\pi}{4}\sin\frac{\pi}{4} = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = 1\]