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Math Help - finding area using the limit of a riemann sum

  1. #1
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    finding area using the limit of a riemann sum



    The sum is ∑ i=1 to n of (2\frac{2i}{n} +1)*\frac{2}{n}

    multiplying it out:
    (*formatting is weird, there should be an n and i=1 around the
    ∑ and the limit goes all the way down)

    lim
    n-> infinity ∑ \frac{8i}{n^2} +\frac{2}{n}

    n-> infinity ∑ \frac{8}{n^2}∑i + \frac{2}{n}

    n-> infinity ∑  \frac{8}{n^2}[\frac{n(n+1)}{2}] + \frac{2}{n}

    = \frac{8}{2}

    =4

    Answer should be 6.
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  2. #2
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    Quote Originally Posted by hazecraze View Post


    The sum is ∑ i=1 to n of (2\frac{2i}{n} +1)*\frac{2}{n}

    multiplying it out:
    (*formatting is weird, there should be an n and i=1 around the ∑ and the limit goes all the way down)

    lim
    n-> infinity ∑ \frac{8i}{n^2} +\frac{2}{n}

    n-> infinity ∑ \frac{8}{n^2}∑i + \frac{2}{n}

    n-> infinity ∑  \frac{8}{n^2}[\frac{n(n+1)}{2}] + \frac{2}{n}

    = \frac{8}{2}

    =4

    Answer should be 6.
    What happend to the sum

    \sum_{i=1}^{n}\frac{2}{n}


    You left this out of your calculation!
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  3. #3
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    So when it says ∑ 1= n, I plug that in to get \frac{2}{1}=2? That + the 4 does make 6.
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by hazecraze View Post
    So when it says ∑ 1= n, I plug that in to get \frac{2}{1}=2? That + the 4 does make 6.

    That is correct.
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