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Homework Help: Integration by parts of a function

  1. Feb 2, 2010 #1
    the function is c = 15te-.2t

    the goal is to integrate it from t = 0 to t = 3

    so to set up the integral i took out the 15 first so i got:

    15 * integral from 0 to 3 of t*e-.2t

    i set u = t and so du = dt
    dv = e-.2tdt so v= -5e-.2t

    so following the integration by parts formula i got:

    15(-5*e-.2t + 5* integral of e-.2tdt

    then after integrating the second part i got

    15(-5e-.2t - 25e-.2t) evaluated from 0 to 3

    when evaluated i got

    (-123/4826181 - 205.8043635)-(-75-375) = 120.7130184

    but thats wrong,

    anyone have any hints or sees any mistakes that i made?

    thanks
     
  2. jcsd
  3. Feb 2, 2010 #2

    rock.freak667

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    this should be

    15(-5te-0.2t+5∫e-.2tdt
     
  4. Feb 2, 2010 #3
    your right, thats just a typo, i had the t there in my calculations
     
  5. Feb 2, 2010 #4

    rock.freak667

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    it looks like you did it correct, is your answer supposed to be in terms of e?
     
  6. Feb 2, 2010 #5
    i dont think so, it just asks for the exact answer
     
  7. Feb 2, 2010 #6

    rock.freak667

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    Then you need to write it in terms of the constant e. So recalculate the answer, just not using a calculator.
     
  8. Feb 2, 2010 #7
    so just leave it as with the base e but calculate the exponents of e?
     
  9. Feb 2, 2010 #8

    rock.freak667

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    Yes.
     
  10. Feb 2, 2010 #9
    ok, thanks alot
     
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