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Math Help - Evaluate the function by using the Riemann sum

  1. #1
    Member dokrbb's Avatar
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    Evaluate the function by using the Riemann sum

    Consider the function  f(x) = -\frac{x^2}{2} + 6}

    I need to calculate \displaystyle \int_0^3 ({-\frac{x^2}{2} + 6})dx by using the definition of Riemann sum for R_n - right end-points,

     \sum_{n=1}^{n} (-\frac{x^2}{2} + 6)

    so, I started by finding \delta x = \frac{3-0}{n} = \frac{3}{n}

    after that

     a) R_n = \sum_{n=1}^{n} (-\frac{(\frac{3i}{n})^2}{2} + 6) (\frac{3}{n}) =
     b) R_n = \sum_{n=1}^{n} (-\frac{(\frac{9i^2}{n^2})}{2} + 6) (\frac{3}{n}) =
     c) R_n = \sum_{n=1}^{n} (-\frac{9i^2}{2n^2} + 6) (\frac{3}{n}) =
     d) R_n = \sum_{n=1}^{n} (-\frac{9i^2 + 12n^2}{2n^2}) (\frac{3}{n}) =
     e) R_n = \sum_{n=1}^{n} (-\frac{3(3i^2 + 4n^2)}{2n^2}) (\frac{3}{n}) =
     f) R_n =(\frac{3}{n}) (-\frac{3}{2n^2})\sum_{n=1}^{n} [(3i^2 + 4n^2)]  =
     g) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{3(n(n+1)(2n+1)}{6} + 4n^2)]  =
     h) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1)}{2} + 4n^2)]  =
     i) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1) + 8n^2}{2}]  =

    so, I'm basically asked to provide

    R_n = as a function of n, without summation symbols, and for this I entered step i) , but it was considered as wrong, so obviously, my mistake is somewhere above,

    \lim_{n\to infty} R_n, here I obtained  -\frac{9}{2} , but it is actually wrong since all my work-up before is wrong,

    can you, please help me with this,

    thanks,
    dokrbb
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  2. #2
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    Re: Evaluate the function by using the Riemann sum

    Quote Originally Posted by dokrbb View Post
    Consider the function  f(x) = -\frac{x^2}{2} + 6}

    I need to calculate \displaystyle \int_0^3 ({-\frac{x^2}{2} + 6})dx by using the definition of Riemann sum for R_n - right end-points,

     \sum_{n=1}^{n} (-\frac{x^2}{2} + 6)

    so, I started by finding \delta x = \frac{3-0}{n} = \frac{3}{n}

    after that

     a) R_n = \sum_{n=1}^{n} (-\frac{(\frac{3i}{n})^2}{2} + 6) (\frac{3}{n}) =
    I'll stop you here, first of all, your counter in your sum is "i", not "n", so they should all look like \displaystyle \begin{align*} \sum_{i = 1}^n \end{align*}

     b) R_n = \sum_{n=1}^{n} (-\frac{(\frac{9i^2}{n^2})}{2} + 6) (\frac{3}{n}) =
     c) R_n = \sum_{n=1}^{n} (-\frac{9i^2}{2n^2} + 6) (\frac{3}{n}) =
     d) R_n = \sum_{n=1}^{n} (-\frac{9i^2 + 12n^2}{2n^2}) (\frac{3}{n}) =
     e) R_n = \sum_{n=1}^{n} (-\frac{3(3i^2 + 4n^2)}{2n^2}) (\frac{3}{n}) =
     f) R_n =(\frac{3}{n}) (-\frac{3}{2n^2})\sum_{n=1}^{n} [(3i^2 + 4n^2)]  =
    I'll stop you here too. After this point, you are actually evaluating the sum and getting a closed expression for it. You should NOT continue to put in the summation signs as you aren't going to be summing anything after this point...

     g) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{3(n(n+1)(2n+1)}{6} + 4n^2)]  =
     h) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1)}{2} + 4n^2)]  =
     i) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1) + 8n^2}{2}]  =

    so, I'm basically asked to provide

    R_n = as a function of n, without summation symbols, and for this I entered step i) , but it was considered as wrong, so obviously, my mistake is somewhere above,

    \lim_{n\to infty} R_n, here I obtained  -\frac{9}{2} , but it is actually wrong since all my work-up before is wrong,

    can you, please help me with this,

    thanks,
    dokrbb
    So once you have fixed everything, you should have

    \displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

    which you can now simplify as much as you like
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  3. #3
    Member dokrbb's Avatar
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    Re: Evaluate the function by using the Riemann sum

    Quote Originally Posted by Prove It View Post
    I'll stop you here, first of all, your counter in your sum is "i", not "n", so they should all look like \displaystyle \begin{align*} \sum_{i = 1}^n \end{align*}



    I'll stop you here too. After this point, you are actually evaluating the sum and getting a closed expression for it. You should NOT continue to put in the summation signs as you aren't going to be summing anything after this point...



    So once you have fixed everything, you should have

    \displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

    which you can now simplify as much as you like
    and what about my lim at infinity?
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    Re: Evaluate the function by using the Riemann sum

    Quote Originally Posted by dokrbb View Post
    and what about my lim at infinity?
    Well, have you tried simplifying? You should easily be able to see what the limit is after simplifying...
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    Re: Evaluate the function by using the Riemann sum

    Quote Originally Posted by Prove It View Post
    Well, have you tried simplifying? You should easily be able to see what the limit is after simplifying...
    actually that's my problem - I got the same result as you showed me, well just it needs to be written without summation symbol, but after simplifying it, I have got -(9/2), which is not correct, I checked my steps, and got another example (-36/2) = -18, but it's again not what it's expected,

    can you just show me these few steps, I have got a dozen of pages with corrections and simplifications, and have no idea which one to put here in order to show you that I really tried this exercise,
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  6. #6
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    Re: Evaluate the function by using the Riemann sum

    Start by expanding everything and then dividing top and bottom by the highest power of n.
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  7. #7
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    Re: Evaluate the function by using the Riemann sum

    Quote Originally Posted by Prove It View Post
    Start by expanding everything and then dividing top and bottom by the highest power of n.
    ok, I have:

    \displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

    \displaystyle \begin{align*} \lim_{n \to \infty}R_n =  \lim_{n \to \infty}\left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

    \displaystyle \begin{align*} \left( -\frac{9}{2n^3} \right) \lim_{n \to \infty}\left[ \frac{n(n + 1)(2n + 1)}{2} + \frac{8n^2}{2} \right] \end{align*}

    \displaystyle \begin{align*} \left( -\frac{9}{2n^3} \right) \lim_{n \to \infty}\left[ (n^3( 1 + \frac{1}{n})(1 + \frac{1}{2n}) + (4n^2) \right] \end{align*}

    \displaystyle \begin{align*} \left( -\frac{9}{2n^3} \right)* \left( n^3 \right) \lim_{n \to \infty}\left[ ( 1 + \frac{1}{n})(1 + \frac{1}{2n}) + ( \frac{4}{n}) \right] \end{align*}

    and I'm left with  -\frac{9}{2}, which is not correct, or I really miss something really at the basic arithmetic... and, something else,

    this \displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*} as I entered it in the webwork, is also considered as wrong - maybe they want me to contract the sum, to multiply \left( -\frac{9}{2n^3} \right) with the rest of the sum???

    thanks, Prove It, for your patience
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  8. #8
    Junior Member FelixFelicis28's Avatar
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    Re: Evaluate the function by using the Riemann sum

    Quote Originally Posted by dokrbb View Post
    ok, I have:

    \displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}
    The issue is your R_n, you've evaluated the limit correctly.

    I haven't looked through your working to pinpoint an error, but I make R_n = -\frac{9n(n+1)(2n+1)}{4n^3} + 18.

    Now let n\to \infty} and it should give \tfrac{27}{2} which I believe is correct.
    Last edited by FelixFelicis28; September 22nd 2013 at 06:59 PM.
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    Re: Evaluate the function by using the Riemann sum

    Quote Originally Posted by FelixFelicis28 View Post
    The issue is your R_n, you've evaluated the limit correctly.

    I haven't looked through your working to pinpoint an error, but I make R_n = -\frac{9n(n+1)(2n+1)}{4n^3} + 18.

    Now let n\to \infty} and it should give \tfrac{27}{2} which I believe is correct.
    it is, indeed, correct,

    so I was just loosing this +18 at the end of my evaluation, because of keeping separately  \left( -\frac{9}{2n^3} \right) through all the steps?

    thanks a lot,
    dokrbb
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