Results 1 to 2 of 2

Math Help - intergration by parts

  1. #1
    Member
    Joined
    Feb 2009
    Posts
    91

    intergration by parts

    Hi

    So I am solving integral sign (S) (x-1)e^-2xdx by using integration by parts

    u = x-1 dv = e^-2x dx
    du = dx v= (-1/2)e^-2x

    so S udv = uv - S vdu

    so = (x-1)((-1/2)e^-2x) - S (-1/2)e^-2x dx

    Now my question is when I solve the S dx part does it become x-1 or just x?

    Thanks

    Calculus beginner
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Bruno J.'s Avatar
    Joined
    Jun 2009
    From
    Canada
    Posts
    1,266
    Thanks
    1
    Awards
    1
    I'm not sure I understand your question. Your calculations are correct. Just solve the remaining integral \frac{-1}{2}\int e^{-2x}dx=\frac{1}{4}e^{-2x}.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: December 8th 2009, 01:30 AM
  2. intergration by parts
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 4th 2009, 02:18 PM
  3. intergration by parts
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2009, 01:19 PM
  4. Intergration by parts
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 12th 2009, 04:46 AM
  5. Intergration by Parts
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 22nd 2007, 01:13 PM

Search Tags


/mathhelpforum @mathhelpforum