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Math Help - Integral Application

  1. #1
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    Integral Application

    Consider the function .


    In this problem you will calculate by using the definition


    The summation inside the brackets is which is the Riemann sum where the sample points are chosen to be the right-hand endpoints of each sub-interval.

    Calculate for on the interval and write your answer as a function of without any summation signs.



    I need to find and Please help!
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  2. #2
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    right endpoint method: x_{i}=a+k{\Delta}x

    {\Delta}x=\frac{3-0}{n}=\frac{3}{n}

    x_{i}=\frac{3k}{n}

    f(x_{i}){\Delta}x=\left[\frac{(\frac{3k}{n})^{2}}{4}-9\right]\cdot\frac{3}{n}

    =\frac{27k^{2}}{4n^{3}}-\frac{27}{n}

    \sum_{i=1}^{n}f(x_{i}){\Delta}x=\frac{27}{4n^{3}}\  sum_{i=1}^{n}k^{2}-\frac{27}{n}\sum_{i=1}^{n}1

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

    So, we have:

    \frac{27}{8n}+\frac{9}{8n^{2}}-\frac{99}{4}

    Now, take the limit:

    \lim_{n\rightarrow{\infty}}\left[\frac{27}{8n}+\frac{9}{8n^{2}}-\frac{99}{4}\right]

    See, the limit?. You should get the same as if you integrate the 'easy' way.
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