Results 1 to 4 of 4

Math Help - Integration by parts

  1. #1
    Junior Member
    Joined
    Sep 2009
    Posts
    65

    Integration by parts

    I am kinda of looking for some step by step of this problem below because I am having some trouble grasping parts.

    \int t^2 e^(3t) dt

    Thanks, Tyler

    PS: 3t is all in e^, I guess that didn't show up well.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,685
    Thanks
    616
    Hello, Tyler!

    If you're "having some trouble grasping parts",
    . . this may be over your head . . .


    I \;=\;\int t^2 e^{3t} \,dt

    By parts: . \begin{array}{ccccccc}<br />
u &=& t^2 && dv &=& e^t\,dt \\<br />
du &=& 2t\,dt && v &=& e^t \end{array}

    Then: . I \;=\;t^2e^t - 2\!\!\int te^t\,dt


    By parts again: . \begin{array}{ccccccc}u &=& t && dv &=& e^t\,dt \\<br />
du &=& dt && v &=& e^t \end{array}

    Then: . I \;=\;t^2e^t - 2\bigg[te^t - \int e^t\,dt\bigg] \;=\;t^2e^t - 2\bigg[te^t - e^t + C\bigg] \;=\;t^2e^t - 2te^t + 2e^t + C


    Therefore: . I \;=\;e^t(t^2 - 2t + 2) + C

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2009
    Posts
    65
    why would v be just e^t and not e^3t?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by Latszer View Post
    why would v be just e^t and not e^3t?
    a typo that was carried through the whole solution. it shouldn't be difficult for you to go through and put in the 3's and 1/3's everywhere they should go.
    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