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Stupid Tabular Method for Integration by parts

  1. Oct 28, 2007 #1
    Is this the correct way to use the Tabular Method for
    [tex]\int x^2e^{-5x}dx[/tex]


    repeated diff:

    [tex]x^2[/tex]

    [tex]2x[/tex]

    [tex]2[/tex]


    repeated Integration:

    [tex]e^{-5x}[/tex]

    [tex]-\frac{1}{5}e^{-5x}[/tex]

    [tex]\frac{1}{25}e^{-5x}[/tex]

    [tex]-\frac{1}{125}e^{-5x}[/tex]


    =[tex]-\frac{x^2}{5}e^{-5x}+\frac{2x}{25}e^{-5x}+\frac{2}{125}e^{-5x}+C[/tex]

    I hate this.

    Casey
     
  2. jcsd
  3. Oct 28, 2007 #2
    yes but the signs of the 2nd/3rd terms are wrong.

    You should write out the table and include the alternating +/-

    writing it out this way might be easier to do since all you do is read across to find the terms
    Code (Text):

              [tex]e^{-5x}[/tex]
    +   [tex]x^2[/tex]   [tex]\frac{-e^{-5x}}{5}[/tex]

    -   [tex]2x[/tex]   [tex]\frac{e^{-5x}}{25}[/tex]
     
     
  4. Oct 28, 2007 #3
    Then I clearly have no idea what I am doing. I am looking at the tabular method in my book, and from what i have read I take the diagonal product of x^2*(-1/5)e^{-5x} and then ADD it to the next product of 2x*(1/25)e^{-5x} and now subtract the product of 2*(-1/125)e^{-5x}

    Am I off on all of them by a step?
     
  5. Oct 28, 2007 #4
    OHHHH...I get it ...form the product then multiply by +1 or -1 THEN add all of the terms...that makes sense. so it is

    [tex]-\frac{x^2}{5}e^{-5x}-\frac{2x}{25}e^{-5x}-\frac{2}{125}e^{-5x}+C[/tex]

    Casey
     
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