Results 1 to 6 of 6

Math Help - Checking over limit process

  1. #1
    Member
    Joined
    Oct 2011
    Posts
    153

    Checking over limit process

    Evaluate the definite integral by the limit process:
    12(x2+1)dx
    Answer from book: 10/3


    This is what I did order:
    1. I used the right justified Riemann Sum (by the way, on the top of ∑ is n, and the bottom is i=1, so like i=1n, except the n is on the top and the i=1 is on the bottom, but I don't know how to type it like that, so just letting you know that I'll do ∑ alone)
    (2 - 1)/n ∑ f(1 + ((2-1)/n)i)

    2. I simplified it
    1/n ∑ f(1 + (1/n)i)

    3. I replaced f with the original equation
    1/n ∑ (1 + (1/n)i)2 + 1

    4. I foiled it
    1/n ∑ (1 + (2/n)i + (1/n2)i2) + 1

    5. I used the sigma notation and multiplied the 'n' into the function
    1/n [(1n + 2n(n(n+1))/2) + 1/n2 ((n(n+1)(2n+1))/6) + 1n]

    6. I multiplied the 1/n outside into everything inside
    1 + 2/n2 (n2/2 + n/2) + 1/n3 (2n3/6 + 3n3/6 + n/6) + 1

    7. I got rid of everything that had an n inside it, since the limit was n -> infinity, so the closer to 0
    That meant I was left with:
    1 + 2 + 2/6 + 1, which equals 6/6 + 12/6 + 2/6 + 6/6, which equals 26/6, which equals 13/3

    I think I might have done something wrong, can someone explain to me if I was doing this process correctly, but there might have been an error as well, so if you can help me point that out where I went wrong.

    Limit process has so many multiplications and a long process, so I may have gotten lost in some ways.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Checking over limit process

    Your method in general is good; let's pick up at step 4:

    S_n=\frac{1}{n}\sum_{k=1}^n\left(2+\frac{2k}{n}+ \frac{k^2}{n^2} \right)

    Using the summation formulas, we have:

    S_n=\frac{1}{n}\left(2n+\frac{2}{n}\cdot\frac{n(n+  1)}{2}+\frac{1}{n^2}\cdot\frac{n(n+1)(2n+1)}{6} \right)

    S_n=2+\frac{(n+1)}{n}+\frac{(n+1)(2n+1)}{6n^2}

    and so:

    \lim_{n\to\infty}S_n=2+1+\frac{1}{3}=\frac{10}{3}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Oct 2011
    Posts
    153

    Re: Checking over limit process

    Quote Originally Posted by MarkFL2 View Post
    Your method in general is good; let's pick up at step 4:

    S_n=\frac{1}{n}\sum_{k=1}^n\left(2+\frac{2k}{n}+ \frac{k^2}{n^2} \right)

    Using the summation formulas, we have:

    S_n=\frac{1}{n}\left(2n+\frac{2}{n}\cdot\frac{n(n+  1)}{2}+\frac{1}{n^2}\cdot\frac{n(n+1)(2n+1)}{6} \right)

    S_n=2+\frac{(n+1)}{n}+\frac{(n+1)(2n+1)}{6n^2}

    and so:

    \lim_{n\to\infty}S_n=2+1+\frac{1}{3}=\frac{10}{3}
    Ooo, I see thanks!
    I might have messed up with those n(n+1)/2 thing, since I turned it into n^2/2 + n/2, so I might have messed up the multiplications.

    But after seeing it in a shorter form, I can see how you got to 10/3
    Thanks!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Oct 2011
    Posts
    153

    Re: Checking over limit process

    Quote Originally Posted by MarkFL2 View Post
    Your method in general is good; let's pick up at step 4:

    S_n=\frac{1}{n}\sum_{k=1}^n\left(2+\frac{2k}{n}+ \frac{k^2}{n^2} \right)

    Using the summation formulas, we have:

    S_n=\frac{1}{n}\left(2n+\frac{2}{n}\cdot\frac{n(n+  1)}{2}+\frac{1}{n^2}\cdot\frac{n(n+1)(2n+1)}{6} \right)

    S_n=2+\frac{(n+1)}{n}+\frac{(n+1)(2n+1)}{6n^2}

    and so:

    \lim_{n\to\infty}S_n=2+1+\frac{1}{3}=\frac{10}{3}
    Just wondering real quick, how did you get 2n on the left side of the equation.
    I'm a bit confused, like after you foiled it, the left ended up a 1? o.o
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor MarkFL's Avatar
    Joined
    Dec 2011
    From
    St. Augustine, FL.
    Posts
    1,988
    Thanks
    734

    Re: Checking over limit process

    \left(1+\frac{1}{n}k \right)^2+1=1+\frac{2}{n}k+\frac{k^2}{n^2}+1=2+ \frac{2}{n}k+ \frac{k^2}{n^2}
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Oct 2011
    Posts
    153

    Re: Checking over limit process

    Quote Originally Posted by MarkFL2 View Post
    \left(1+\frac{1}{n}k \right)^2+1=1+\frac{2}{n}k+\frac{k^2}{n^2}+1=2+ \frac{2}{n}k+ \frac{k^2}{n^2}
    Oh right!
    I forgot about the 1 at the end!
    Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. deritive by limit process
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 8th 2010, 10:57 AM
  2. Finding the Derivative by Limit Process
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 2nd 2010, 12:57 PM
  3. Area of a region using the limit process
    Posted in the Calculus Forum
    Replies: 2
    Last Post: January 12th 2010, 06:09 PM
  4. Slope by limit process
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: May 29th 2009, 09:12 PM
  5. Replies: 1
    Last Post: December 16th 2008, 12:27 AM

Search Tags


/mathhelpforum @mathhelpforum