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Math Help - Riemann Sums

  1. #1
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    Riemann Sums

    Calculate Rn for f(x)=-[(x^2)/3]+4 on the interval [0,4] and write the answer as a function of n without any summation signs. Also find lim as n-->infinity of Rn.

    I never have trouble calculating actual numbers (ie; R4 or L6, etc) but I always have trouble when I'm given a variable. If someone could work this out for me to put in my notes I would really appreciate it!
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    Quote Originally Posted by A5HLEY View Post
    Calculate Rn for f(x)=-[(x^2)/3]+4 on the interval [0,4] and write the answer as a function of n without any summation signs. Also find lim as n-->infinity of Rn.

    I never have trouble calculating actual numbers (ie; R4 or L6, etc) but I always have trouble when I'm given a variable. If someone could work this out for me to put in my notes I would really appreciate it!

    Here is the jist of it...

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

    and

     x_i=a+i \cdot \Delta x=0 +i \Delta x=i \Delta x =i\frac{4}{n}

    So we get the sum is

     R_n = \sum_{i=0}^{n}f(x_i) \Delta x = \sum_{i=0}^{n} \left( \frac{((i \frac{4}{n} )^2)}{3}+4 \right)  \frac{4}{n}

     \frac{64}{3n^3} \underbrace{ \sum_{i=0}^{n} i^2}_{=\frac{n(n+1)(2n+1)}{6}}+ \frac{16}{n} \underbrace{ \sum_{i=0}^{n} 1}_{=n}

     R_n=\frac{64}{3n^3}\frac{n(n+1)(2n+1)}{6}+\frac{16  }{n}n

    You should be able to simplify and take the limit from here.
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