Results 1 to 3 of 3

Math Help - Calculus I Help?

  1. #1
    Junior Member
    Joined
    Jan 2006
    From
    Oakland
    Posts
    55

    Calculus I Help?

    Find the Riemman Sum of the integration of x^3 from 1 to 2.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Nimmy
    Find the Riemman Sum of the integration of x^3 from 1 to 2.
    It depends how you want to partion the region.
    I will partition the region with right-endpoint.

    Take interval [1,2] divide it in n equal regions. Thus, each one has length \Delta x=\frac{2-1}{n}=\frac{1}{n}.
    Thus, the Riemann sum is,
    \sum_{k=1}^n f(a+\Delta x\cdot k)\Delta x
    Rewrite is as,
    \sum_{k=1}^n f\left(1+\frac{k}{n} \right) \frac{1}{n}
    Thus,
    \sum_{k=1}^n \left( \frac{n+k}{n} \right)^3\cdot \frac{1}{n}
    Which simplifies to,
    \sum_{k=1}^n \frac{(n+k)^3}{n^4}

    Do you also want to know its value?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    Do you mean the tedious technical approach?.

    Using right endpoint.

    y=x^{3}\;\text{over the interval}\;\  [1,2]

    {\Delta}x=\frac{2-1}{n}=\frac{1}{n}

    x_{k}=a+k{\Delta}x=1+k{\Delta}x=1+\frac{k}{n}

    \left(1+\frac{k}{n}\right)^{3}{\Delta}x=(1+\frac{k  }{n})^{3}(\frac{1}{n})

    Expanding and summing:

    \sum_{k=1}^{n}\left(\frac{k^{3}}{n^{4}}+\frac{3k^{  2}}{n^{3}}+\frac{3k}{n^{2}}+\frac{1}{n}\right)

    Now, remember the 'identities':

    k^{3}=\frac{n^{2}(n+1)^{2}}{4}

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

    k=\frac{n(n+1)}{2}

    Now, use those and it all whittles down to:

    \lim_{n\to\infty}\frac{15n^2+14n+3}{4n^{2}}=\frac{  15}{2}

    Perform the 'easy' integration and see if you get the same thing.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: December 13th 2011, 09:11 PM
  2. Replies: 2
    Last Post: June 25th 2010, 10:41 PM
  3. Replies: 1
    Last Post: February 11th 2010, 07:09 AM
  4. Calculus III But doesn't require Calculus :)
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 19th 2009, 04:23 PM
  5. Replies: 1
    Last Post: June 23rd 2008, 09:17 AM

Search Tags


/mathhelpforum @mathhelpforum