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Integration by Parts

  1. May 26, 2005 #1
    hi guys
    just doing some revision and im stuck on this question

    *integral sign* x^2 . exponential ^ -3x . dx

    I know i have to use integration by parts, but i just cant seem to get it out
    any ideas?
  2. jcsd
  3. May 26, 2005 #2


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    HINT:Take [tex]dv=e^{-3x} \ dx [/tex] and [tex] u=x^2 [/tex].

    U'll figure out what to do next.

  4. May 26, 2005 #3
    but, what is the integral of exponential ^ -3x?
  5. May 26, 2005 #4
    is it -1/3 *exponential* ^ -3x?
  6. May 26, 2005 #5


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    Yes,as you can check by differentiation.

  7. May 26, 2005 #6
    The tabular method would work great on this question. Are you familiar with this?
  8. May 26, 2005 #7
    ok, once i sub u, du, v and dv into the integral by parts formula, i have to assign, u and dv again to:
    -1/3 exponential ^ -3x and 2x

    so is dv assigned to the exponential again, like in the first case?
  9. May 26, 2005 #8
    no, i am not aware of the tabular method, sorry
  10. May 26, 2005 #9
    ok, ive got my answer, unfortunately i dont kow if its right, we dont get solutions for this exercise

    -1/3 *exponential* ^ -3x . x^2 - 1/9*exponential*^-3x . 2x - 2/27*exponential*^-3x
  11. May 26, 2005 #10
    Maple agrees, good job.
  12. May 26, 2005 #11


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    Surely,you must add an arbitrary constant wrt "x" to your solution.

  13. May 27, 2005 #12
    arbitary constant would be + c
  14. May 27, 2005 #13
    In case you were interested in this method, http://marauder.millersville.edu/~bikenaga/calculus/parts/partspf.html [Broken] is a link to the explanation. If one of the two terms will eventually differentiate to zero, this method is much less time consuming.

    Last edited by a moderator: May 2, 2017
  15. May 28, 2005 #14
    i see, thanx for that link, i will probably apply this method from now on
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