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Finding the integral of t e^(-lambda t) from t=0 to t=infinity

  1. Oct 3, 2006 #1
    [tex] \int_{0}^{infty} t e^{-\lambda t} \lambda dt [/tex]


    [tex] \frac{1}{\lambda e^{\lambda t}} (1 + t \lambda) \right]_{0}^{\infty} = \frac{1}{\lambda} [/tex]

    is this corect??

    for hte infinity part we dont have to worry since the 1/exponential term goes to zero as does the t/exponential term. For t = 0 however t/exponentail is zero and what is left is 1/lambda
     
    Last edited: Oct 3, 2006
  2. jcsd
  3. Oct 3, 2006 #2

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    You're missing a factor of lambda. It's helpful if you think of t as being a time, so that lambda is a frequency and the integral needs to have dimensions of time^2.
     
  4. Oct 3, 2006 #3
    how am i missing it??

    did i not do everything right??

    see the whole work

    [tex] \frac{-1}{\lambda e^{\lambda t}} (1 + t \lambda) \right]_{0}^{\infty} = \lim_{t \rightarrow \infty} \left(\frac{-1}{\lambda e^{\lambda t}} \right) - \left( \frac{-1}{\lambda}\right) - 0 \right) = \frac{1}{\lambda} [/tex]

    where does the extra fact oro lambda come from??
     
    Last edited: Oct 3, 2006
  5. Oct 3, 2006 #4

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    The problem is before the first step you've shown, since the dimensions are wrong there as well. How did you take the antiderivative?
     
  6. Oct 3, 2006 #5
    are u talkin about hte negative sign i was missing??

    i did integration by parts..
     
  7. Oct 3, 2006 #6

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    No, you're missing a factor of lambda in the antiderivative. Go back through it step through step.
     
  8. Oct 3, 2006 #7
    since we are on this what about
    [tex] \int_{T}^{\infty} e^{-\lambda t} \lambda dt = -e^{-\lambda t} |_{T}^{\infty} = 0 + e^{-\lambda T} [/tex]

    am i missing a fact or of lambda here too.. WHY??
     
    Last edited: Oct 3, 2006
  9. Oct 3, 2006 #8
    i typed out the question wrong...

    it should be e^\lambda t * lambda
     
  10. Oct 3, 2006 #9

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    I'm confused. Your antiderivative in the first step is right except for a factor of lambda. If you meant for there not to be a t multiplying the exponential, then your original answer is completely wrong. So which is it?
     
  11. Oct 3, 2006 #10
    Ok let's restart

    [tex] \int_{0}^{\infty} t e^{-\lambda t} \lambda dt [/tex]

    and the integral is
    [tex] \frac{-1}{\lambda e^{\lambda t}} (1 + t \lambda) \right]_{0}^{\infty} = \frac{1}{\lambda} [/tex]

    good?
     
  12. Oct 3, 2006 #11

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    Yea, that's right.
     
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