# Thread: riemann integral

1. ## 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.

2. 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.)