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

  1. #1
    Senior Member x3bnm's Avatar
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    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?
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  2. #2
    MHF Contributor MarkFL's Avatar
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    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.
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  3. #3
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    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
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  4. #4
    Senior Member x3bnm's Avatar
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    Re: Limit of Riemannian Sum of definite integral cos(x) dx from pi to 2pi

    Quote Originally Posted by Idea View Post
    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?
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