Results 1 to 6 of 6

Math Help - integration by parts

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    91

    integration by parts

    HI

    calculating integral sign ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

    So if u = x then du = dx

    then I need a dv .... if dv = e^-1 then v = e^-1 doesn't it? and how does this help me if I now introduce V?

    how can I get all of these "e"s into "u"s?

    Thanks

    Calculus Beginner
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454
    Quote Originally Posted by calcbeg View Post
    HI

    calculating integral sign ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

    So if u = x then du = dx

    then I need a dv .... if dv = e^-1 then v = e^-1 doesn't it? and how does this help me if I now introduce V?

    how can I get all of these "e"s into "u"s?

    Thanks

    Calculus Beginner
    did you mean ...

    \int_0^2 \frac{e^{-x}}{1-e^{-2x}} \, dx

    or what you posted?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2009
    Posts
    91
    what I posted - there is an x then the e stuff
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,698
    Thanks
    454
    Quote Originally Posted by calcbeg View Post
    what I posted - there is an x then the e stuff
    your notation is a bit confusing ...

    ("S") b=2; a=0 (e^-x/(1-e^-2)) dx

    is the integrand

    \frac{e^{-x}}{1-e^{-2}}

    or is it

    e^{\frac{-x}{1-e^{-2}}}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,701
    Thanks
    1470
    If it really is, as you said, ("S") b=2; a=0 (e^-x/(1-e^-2)) dx
    which is \int\frac{e^{-x}}{1- e^{-2}}dx, then that is the same as \frac{1}{1- e^{-1}}\int e^{-x}dx and that should be easy to integrate.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Oct 2009
    From
    Currently: Maryland, USA
    Posts
    1
    Quote Originally Posted by HallsofIvy View Post
    If it really is, as you said, ("S") b=2; a=0 (e^-x/(1-e^-2)) dx
    which is \int\frac{e^{-x}}{1- e^{-2}}dx, then that is the same as \frac{1}{1- e^{-1}}\int e^{-x}dx and that should be easy to integrate.
    Why did you take out \frac{1}{1- e^{-1}}? Wouldn't the e be to the negative 2 power-- \frac{1}{1- e^{-2}} since all of it is a constant?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: January 11th 2012, 02:30 PM
  2. Replies: 8
    Last Post: September 2nd 2010, 12:27 PM
  3. Replies: 0
    Last Post: April 23rd 2010, 03:01 PM
  4. Integration by Parts!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: January 22nd 2010, 03:19 AM
  5. Replies: 1
    Last Post: February 17th 2009, 06:55 AM

Search Tags


/mathhelpforum @mathhelpforum