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Math Help - Calculus - Riemann sum

  1. #1
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    Calculus - Riemann sum

    Consider the function f(x) = -x^2/2 -8

    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.


    Okay, I can't seem to get the right answer for this one. Here's what I have done so far:

    The distance in play here is from 0 to 4, so with n number of rectangles, each width should be 4/n. I replaced x with 4/n(i) and replaced dx with 4/n.

    Since in this equation we have x^2, I end up with
    -(4/n(i))^2 / 2 - 8 (4/n)

    I can't use "i" in my answer so I substituted it with the special sum formula for i^2 which is n(n+1)(2n+1)/6

    All together then, what I came to is

    ((-[(4^2/n^2)(n(n+1)(2n+1)/6)]/2) - 8 )(4/n)

    sorry for all the parentheses.

    I know that this answer is wrong but I have no idea where I went wrong.
    Thank you for any help!!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi,
    Quote Originally Posted by littlejodo View Post
    All together then, what I came to is

    ((-[(4^2/n^2)(n(n+1)(2n+1)/6)]/2) - 8 )(4/n)
    You just forgot one letter : -\left(\frac{4^2}{n^2}\cdot \frac{n(n+1)(2n+1)}{6\times 2}+8{\color{red}n}\right)\cdot \frac{4}{n}. The red n comes from \sum_{i=1}^n8=8n\ldots

    Now you have to compute

    \begin{aligned}-\int_0^4\frac{x^2}{2}+8\,\mathrm{d}x&=\lim_{n\to\i  nfty}-\left(\frac{4^2}{n^2}\cdot \frac{n(n+1)(2n+1)}{6\times 2}+8{\color{red}n}\right)\cdot \frac{4}{n}\\<br />
&=\lim_{n\to\infty}  -\frac{32(n+1)(2n+1)}{6n^2}-32\\<br />
&= \ldots\end{aligned}
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  3. #3
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    THANK YOU!!!

    I'm happy to see I wasn't terribly off base.
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