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Improper integrals

  1. Apr 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine whether the integral is divergent or convergent. If it is convergent, evaluate it. If not, give the answer -1. :
    Integral from 6 to infinity x(e^(-3x))dx

    2. Relevant equations



    3. The attempt at a solution
    In this case, i first made the limit as t goes to infinity Integral from 6 to t x(e^(-3x))dx and then i did a u sub (u=-3x and du/-3=dx) but when i tried to replace x in the original equation which i thought would be x=u/-3. From there, i replaced the values and solved it but it's wrong...and i know for sure that's its convergent but i don't know where i'm going wrong... thanks
     
  2. jcsd
  3. Apr 4, 2009 #2

    Dick

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    It is convergent. To get the antiderivative you need to integrate by parts.
     
  4. Apr 4, 2009 #3
    okay so i did the ibp and set u=x du=dx and dv=e^(-3x) dx V=e^(-3x)/-3 ..i'm not sure but do i evaluate the entire equation from 6 to t or do i have to change the limits of integration?? I tried leaving it as it and evaluated it as t goes to infinity but i got the wrong answer... thanks
     
  5. Apr 4, 2009 #4

    Dick

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    It looks like you are on the right track. What did you finally get for the antiderivative? What did you get for an answer? What do you think the answer should be?
     
  6. Apr 4, 2009 #5
    I finally got the answer! You were very helpful! But i have one small question about the question when i was evaluating the limit as t goes to infinity for [(te^(-3t)/-3)-(e^(-3t)/9)]...i was just wondering about the e^(-3t) and as it goes to infinity, does it go towards negative infinity? and if i divide infinity by a number does it head to zero?
     
  7. Apr 4, 2009 #6

    Dick

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    The limit as t goes to infinity of e^(-3t) is zero isn't it? The t*e^(-3t) part is the one you might worry about. But you can show that goes too by zero to using l'Hopital's theorem.
     
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