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  1. #1
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    riemann integral

    i want to calculate the integral of sine(x) using riemann sums and got the stuck with the series.

    the process scan is attached and shows the problem clearly.
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  2. #2
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    Krizalid's Avatar
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    Make it more generally:

    Let's compute \int_{a}^{b}{\sin x\,dx}. Let's use the following partition \Delta x_{i}=\frac{b-a}{n}, thus, our limit becomes \underset{n\to \infty }{\mathop{\lim }}\,\frac{b-a}{n}\sum\limits_{i=1}^{n}{\sin \left( a+\frac{b-a}{n}\cdot i \right)}. Now, use the formula \cos u-\cos v=-2\sin \left( \frac{u+v}{2} \right)\sin \left( \frac{u-v}{2} \right) and the limit is

    -\frac1{2}\underset{n\to \infty }{\mathop{\lim }}\,\frac{b-a}{n\sin \left( \frac{b-a}{n} \right)}=\sum\limits_{i=1}^{n}{\left\{ \cos \left( a+\frac{b-a}{n}\cdot (i+1) \right)-\cos \left( a+\frac{b-a}{n}\cdot (i-1) \right) \right\}},

    now let's turn this sum into a telescoping one,

    \sum\limits_{i=1}^{n}{\left\{ \cos \left( a+\frac{b-a}{n}\cdot (i+1) \right)-\cos \left( a+\frac{b-a}{n}\cdot i \right) \right\}}\,(^*) equals \cos \left( a+\frac{b-a}{n}\cdot (n+1) \right)-\cos \left( a+\frac{b-a}{n} \right); and \sum\limits_{i=1}^{n}{\left\{ \cos \left( a+\frac{b-a}{n}\cdot i \right)-\cos \left( a+\frac{b-a}{n}\cdot (i-1) \right) \right\}}\,(^{**}) equals \cos (b)-\cos (a). (Note that by adding (^*,\,^{**}) you'll get the sum of the limit.)

    Finally, by taking the limit, we have

    \frac{b-a}{n\sin \left( \frac{b-a}{n} \right)}\to 1,

    and

    \cos \left( a+\frac{b-a}{n}\cdot (n+1) \right)\to\cos(b), \cos \left( a+\frac{b-a}{n} \right)\to\cos(a) as n\to\infty, thus the limit equals -\frac{2\cos (b)-2\cos (a)}{2}=\cos (a)-\cos (b), and we're done.

    (You just need to put (a,b)=(0,\pi) and there's your sum.)
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