Question

Use the formula

sinh+sin2h+sin3h++sinmh=cos(h/2)cos((m+(1/2))h)2sin(h/2)\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=π/2x=\pi / 2 in two steps: Partition the interval [0,π/2][0, \pi / 2] into n subintervals of equal length and calculate the corresponding upper sum U

Solution

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Answered 8 months ago
Answered 8 months ago
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y=sinx,x[0,π/2]y=\sin x, x\in[0,\pi/2]

Partition of [0,π/2][0,\pi/2] into n subintervals is [iπ2n,(i+1)π2n][\frac{i\pi}{2n},\frac{(i+1)\pi}{2n}] for i=0,...,n,i=0,...,n, xi=a+π2ni.x_i=a+\frac{\pi}{2n}i.
Using Riemanns sum\underline{\text{Riemanns sum}},

Area=limni=0nπ2nf(xi)=limnπ2ni=0nf(π2ni)={h=π2n}=limh0hi=0nf(ih)=limh0h(sinh+sin2h+...+sinnh)=limh0hcos(h/2)cos((n+1/2)h)2sin(h/2)=limh0cosh/2cos(π/2+h/2)sin(h/2)h/2=limh0cosh/2sinh/2sin(h/2)h/2=101=1\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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