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Math Help - Riemann Sums not summing

  1. #1
    Junior Member beebe's Avatar
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    Riemann Sums not summing

    I'm working on using Riemann sums to find definite integrals. The answers aren't coming out as what they should be. Here's what I did. Sorry for all the stuff, but I'm trying to show as many steps as possible.
    f(x)=x^2+x=x(x+1)
    \int_{-2}^{0}(x^2+x)dx, \triangle x=\frac{2}{n}, x_i=(\frac{2i}{n}-2)
    \lim_{n\to\infty}\sum_{i=1}^{n}f((\frac{2i}{n}-2))*\frac{2}{n}=\lim_{n\to\infty}\frac{2}{n}*\sum_  {i=1}^{n}f((\frac{2i}{n}-2))
    =\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}(\frac{  2i}{n}-2)(\frac{2i}{n}-1)=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n}(\fra  c{4i^2}{n^2}-\frac{6i}{n}+2)
    next I split the sums and distribute
    =\lim_{n\to\infty}[\frac{8}{n^3}\sum_{i=1}^{n}i^2-\frac{12}{n^2}\sum_{i=1}^{n}i+\frac{4}{n}]=\lim_{n\to\infty}[\frac{8}{n^3}\sum_{i=1}^{n}i^2-\frac{12}{n^2}\sum_{i=1}^{n}i]+\lim_{n\to\infty}\frac{4}{n}
    4/n goes to zero, then I substitute in for each of the sums using formulas:
    =\lim_{n\to\infty}\frac{8}{n^3}*\frac{n(n+1)(2n+1)  }{6}-\frac{12}{n^2}*\frac{n(n+1)}{2}
    }=\lim_{n\to\infty}\frac{4}{3}*\frac{2n^2+3n+1}{n^  2}-6*(1+\frac{1}{n})
    =\lim_{n\to\infty}\frac{4}{3}*(2+\frac{3}{n}+\frac  {1}{n^2})-6*(1+\frac{1}{n})
    =\frac{8}{3}-6=\frac{-10}{3}

    Integrating by using the antiderivative gives me:
    \int_{-2}^{0}(x^2+x)dx=(\frac{x^3}{3}+\frac{x^2}{2})]_{-2}^0=0-(\frac{-8}{3}+2)=\frac{2}{3}

    Where did I go wrong?
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  2. #2
    MHF Contributor
    Jester's Avatar
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    Re: Riemann Sums not summing

    You have \frac{4}{n} but it should be \sum_{i=1}^n \frac{4}{n}
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  3. #3
    Junior Member beebe's Avatar
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    Re: Riemann Sums not summing

    Thanks, I see that I was skipping a step. I'm getting the right answer now.
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