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Math Help - Riemann Sums

  1. #1
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    Riemann Sums

    How do you find the area under the curve using Riemann sums with

    f(x) = 9 - x^2, a=0 b=4

    How are you supposed to set it up so that the constant doesn't become insignificant when you take the limit? Thank you.
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  2. #2
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    Let's use the right endpoint method:

    {\Delta}x=\frac{b-a}{n}=\frac{4-0}{n}=\frac{4}{n}

    c_{1}=a+k({\Delta}x)=0+\frac{4k}{n}

    (9-(\frac{4k}{n})^{2})(\frac{4}{n})=\frac{36}{n}-\frac{64k^{2}}{n^{3}}

    \frac{36}{n}\sum_{k=1}^{n}(1)-\frac{64}{n^{3}}\sum_{k=1}^{n}k^{2}

    But, k^{2}=\frac{n(n+1)(2n+1)}{6}

    36-\left(\frac{64}{n^{3}}(\frac{n(n+1)(2n+1)}{6})\rig  ht)

    =\frac{-32}{n}-\frac{32}{3n^{2}}+\frac{44}{3}

    Now, take the limit:

    -32\lim_{n\rightarrow{\infty}}\frac{1}{n}-\frac{32}{3}\lim_{n\rightarrow{\infty}}\frac{1}{n^  {2}}+\frac{44}{3}

    See what the limit is?. That's your solution.
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