Results 1 to 4 of 4

Math Help - improper integrals

  1. #1
    Newbie
    Joined
    May 2009
    Posts
    6

    improper integrals

    Hello i have a question,

    Find the area of region R that lies between y = 1/x and y = 1/(x+1) to the right of x = 1

    I assume it has to do with bounding divergents can someone lend a hand?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    You need to calculate R=\int_1^\infty\left(\frac{1}{x}-\frac{1}{1+x}\right)dx=\ln2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    May 2009
    Posts
    6
    okay the first part is what I wrote out too, where does the ln2 take place in the question?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    R=\int_1^\infty\left(\frac{1}{x}-\frac{1}{1+x}\right)dx=\left[\ln x-\ln (1+x)\right]_1^\infty=\left[\ln{\frac{x}{1+x}}\right]_1^\infty

    The limit \lim_{x \to \infty}\ln{\frac{x}{1+x}}=0, so all we end up with is -\ln\frac{1}{2}=\ln 2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Improper Integrals
    Posted in the Differential Geometry Forum
    Replies: 9
    Last Post: February 15th 2011, 01:09 AM
  2. improper integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 7th 2009, 06:06 AM
  3. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 17th 2009, 09:29 PM
  4. Improper Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 5th 2008, 12:57 PM
  5. Improper Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 24th 2007, 06:05 PM

Search Tags


/mathhelpforum @mathhelpforum