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Math Help - Riemann Sum

  1. #1
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    Riemann Sum

    I was wondering if someone could show me the Riemann Sum for the following function:

    \sqrt{3}x^2 - x + 1 on [-1,2]

    I need to find the exact area under the curve, so I'd assume n->INFINITY

    I keep trying this, and I come in the realm of Mathematica's answer (If I split up the interval).. but not close enough.

    I keep getting: -1 + 3i/n

    Whenever I multiply everything out after substituting into the function, substituting in for i after factoring out the constant/n I come nowhere close.
    Last edited by chrisf; December 7th 2008 at 08:07 AM.
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  2. #2
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    First, we know that the  {\Delta}x = \frac {b-a}{n} = \frac {2-(-1)}{n}= \frac {3}{n}

    Use the right end point rule, we have C_k = -1 + \frac {3}{n}k ,

    Then we would have C_1= -1 + \frac {3}{n} \rightarrow -1
    and C_n = -1 + \frac {3}{n}n = 2

    Now, <br />
f(x) = \sqrt{3}x^2 - x + 1<br />
, implies that f(C_k)=<br />
\sqrt{3}(-1+ \frac {3}{n}k)^2 - (-1+ \frac {3}{n}k) + 1<br />

    so the Riemann Sum is  \frac {3}{n} \sum ^n _{k=1} \sqrt{3}(-1+ \frac {3}{n}k)^2 - (-1+ \frac {3}{n}k) + 1

    = \frac {3}{n} \sum ^n_{k=1} \sqrt{3}- \frac {3 \sqrt {3}}{n}k + \frac {9 \sqrt {3}}{n^2}k^2
    Last edited by tttcomrader; December 7th 2008 at 08:36 AM.
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  3. #3
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    That's exactly what I've got. Maybe my algebra sucks...

    I appreciate your reply. Let me re-work it again and I'll get back here!
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    = \sum ^n_{k=1} \sqrt{3}- \frac {3 \sqrt {3}}{n}k + \frac {9 \sqrt {3}}{n^2}k^2
    This is exactly what I have - but, I originally pull the \frac {3}{n} out in front of the summation.

    \frac {3}{n} [ \sum ^n_{k=1} \sqrt{3} - \frac {3 \sqrt {3}}{n} \sum ^n_{k=1} k + \frac {9 \sqrt {3}}{n^2} \sum ^n_{k=1} k^2]

    \frac {3 \sqrt{3}n}{n} - \frac {9 \sqrt{3}n(n+1)}{2n^2} + \frac {27 \sqrt{3}n(n+1)(2n+1)}{6n^3}

    When I take the limit of that as n->INFINITY, I come nowhere close to

    \frac {3}{2} + 3 \sqrt{3}
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  5. #5
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    I'm not sure where I'm going wrong.. I've tried re-working this quite a few times.
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  6. #6
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    Anyone?
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  7. #7
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    Let me ask my professor today, I can't figure it out either.
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