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Math Help - help with Riemann sum (RHS) of Definite Inegral

  1. #1
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    help with Riemann sum (RHS) of Definite Inegral

    Could somone show me how the integral X^2-7X+10 on [2,5] would be equal to -4.5. I know that given RHS= F(Xi)delta x the delta is 3/n and Xi is 1+3i/n so I know how to start like this.

    (1+(3i/n)^2-7(1+(3i/n))+10)(3/n)

    I just don't see how I would break this down.

    I know that after it is simplifed I need to take the limit as n approaches postive infintiy.
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  2. #2
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    The algebra is the booger in these things more than anything.

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

    Using the right end method.

    x_{k}=a+k{\Delta}x

    x_{k}=2+\frac{3k}{n}

    \left[\left(2+\frac{3k}{n}\right)^{2}-7\left(2+\frac{3k}{n}\right)+10\right]\cdot\frac{3}{n}

    Now, expand out:

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

    Use the formulae for the sum of the squares and integers.

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

    Now, take the limit:

    \lim_{n\to \infty}\frac{9}{2n^{2}}-\lim_{n\to \infty}\frac{9}{2}

    Now, you can see the 4.5...right?.
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