Thread: Limit of Riemannian Sum of definite integral cos(x) dx from pi to 2pi

1. Limit of Riemannian Sum of definite integral cos(x) dx from pi to 2pi

$\displaystyle \text{Riemannian sum, limit, and integration:}$

$\displaystyle \int_{\pi}^{2\pi}\cos(x)dx$

\displaystyle \begin{align*}\displaystyle \Delta x &= \frac{b -a}{n}\\ &= \frac{2\pi - \pi}{n}\\ &= \frac{\pi}{n} \end{align*}

\displaystyle \begin{align*}x_{i} &= a + \Delta x . i\\ &= \pi + \frac{\pi i}{n} \end{align*}

\displaystyle \begin{align*} &\lim_{n \to \infty} \sum_{i = 1}^{n} \Delta x . \cos(x_{i})dx\\ &=\lim_{n \to \infty} \sum_{i = 1}^{n} \frac{\pi}{n} . \cos(\pi + \frac{\pi i}{n})dx \end{align*}

How can I evaluate the above expression and get 0? I know how to get this value using integration. But I want to know practically how to arrive at that value?

2. Re: Limit of Riemannian Sum of definite integral cos(x) dx from pi to 2pi

I think my strategy would be to use the identity:

$\displaystyle \cos(2\pi-\theta)=-\cos(\pi+\theta)$

And break the sum into two sums that add to zero.

3. Re: Limit of Riemannian Sum of definite integral cos(x) dx from pi to 2pi

Prove

$\displaystyle \sum _{i=1}^n \cos \left(\pi +\frac{\pi i}{n}\right)=1$

or equivalently show that

$\displaystyle \sum _{i=1}^{n-1} \cos \left(\frac{\pi i}{n}\right)=0$

This can be done using a change of index $\displaystyle j=n-i$

Therefore the Riemann sum $\displaystyle = \frac{\pi }{n}$ which goes to zero

4. Re: Limit of Riemannian Sum of definite integral cos(x) dx from pi to 2pi

Originally Posted by Idea
Prove

$\displaystyle \sum _{i=1}^n \cos \left(\pi +\frac{\pi i}{n}\right)=1$

or equivalently show that

$\displaystyle \sum _{i=1}^{n-1} \cos \left(\frac{\pi i}{n}\right)=0$

This can be done using a change of index $\displaystyle j=n-i$

Therefore the Riemann sum $\displaystyle = \frac{\pi }{n}$ which goes to zero

Sorry for not understanding and for my ignorance.

Is it possible to kindly tell me how did you come to your solution?