## Related questions with answers

Use the formula

$\sin h+\sin 2 h+\sin 3 h+\cdots+\sin m h =\frac{\cos (h / 2)-\cos ((m+(1 / 2)) h)}{2 \sin (h / 2)}$

to find the area under the curve y = sin x from x = 0 to $x=\pi / 2$ in two steps: Partition the interval $[0, \pi / 2]$ into n subintervals of equal length and calculate the corresponding upper sum U

Solution

Verified$y=\sin x, x\in[0,\pi/2]$

Partition of $[0,\pi/2]$ into n subintervals is $[\frac{i\pi}{2n},\frac{(i+1)\pi}{2n}]$ for $i=0,...,n,$ $x_i=a+\frac{\pi}{2n}i.$

Using $\underline{\text{Riemanns sum}}$,

$\begin{align*} \text{Area}&=\lim_{n\to\infty}\sum_{i=0}^{n} \frac{\pi}{2n}f(x_i) \\&=\lim_{n\to\infty}\frac{\pi}{2n} \sum_{i=0}^{n} f(\frac{\pi}{2n}i)=\{h=\frac{\pi}{2n}\}\\&=\lim_{h\to 0} h\sum_{i=0}^{n}f(ih)\\&=\lim_{h\to 0} h(\sin h+\sin 2h + ... +\sin nh)\\&=\lim_{h\to 0} h\cdot \frac{\cos(h/2)-\cos((n+1/2)h)}{2\sin(h/2)}\\&=\lim_{h\to 0} \frac{\cos h/2-\cos (\pi/2+h/2)}{\frac{sin(h/2)}{h/2}}\\& =\lim_{h\to 0} \frac{\cos h/2 - \sin h/2}{\frac{sin(h/2)}{h/2}}\\&=\frac{1-0}{1}=1\end{align*}$

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