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Evaluating an integral when theta>0

  1. Mar 23, 2013 #1
    1. The problem statement, all variables and given/known data


    ∫ e^(-y/θ) dy
    0

    2. Relevant equations

    The directions say evaluate when theta>0.


    3. The attempt at a solution

    I have been integrating by using u of substitution and been getting -θ*e^(-y/θ) and evaluating from 0 to ∞ and getting the answer as undefined. I need help with evaluating the integral. Please help. Thanks.
     
    Last edited by a moderator: Mar 24, 2013
  2. jcsd
  3. Mar 23, 2013 #2

    Zondrina

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    Are you familiar with the anti derivative of ##e^{- \frac{x}{a}}##?
     
  4. Mar 23, 2013 #3

    SammyS

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    Your result should not be "undefined".

    You have the correct anti-derivative.

    Write [itex]\displaystyle \ \int_0^{\infty}e^{-y/\theta}\,dy\ [/itex] as [itex]\displaystyle \ \lim_{a\to\infty} \int_0^{a}e^{-y/\theta}\,dy\ .[/itex]

    Now use your anti-derivatve with the limits of integration, then take the limit.
     
  5. Mar 23, 2013 #4
    No, I'm not. How is it used in this situation?
     
    Last edited by a moderator: Mar 24, 2013
  6. Mar 23, 2013 #5
    You want to evaluate the integral [itex] \int e^{-y/\theta}, dy [/itex]. What happens when you make the substituion [itex] u = {\frac{-y}{\theta}} [/itex] ? What is [itex] du [/itex] ?

    Can you generalize to [itex] \int f(cx)\, dx [/itex] when you know what [itex] \int f(x)\, dx [/itex] is? This is a useful result to know when evaluating integrals so you don't always have to make these substitutions.
     
    Last edited: Mar 23, 2013
  7. Mar 23, 2013 #6
    How would i take the anti derivative if says to evaluate for all theta> 0 not quite sure how to do this.
     
    Last edited: Mar 23, 2013
  8. Mar 23, 2013 #7
    What happens to -θ*e^(-b/θ) as b goes to ∞? to 0? Remember that θ is just a constant.
     
    Last edited: Mar 23, 2013
  9. Mar 23, 2013 #8
    is theta a constant even though in the directions it says evaluate from theta>0.
     
  10. Mar 23, 2013 #9

    SammyS

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    No. It says evaluate when θ > 0 .

    If θ ≤ 0, then [itex]\displaystyle \ \int_0^{\infty}e^{-y/\theta}\,dy\ [/itex] does not converge.


    (Did you not understand my previous post?)
     
  11. Mar 23, 2013 #10
    For some reason I'm not understanding what you are saying, SammyS. HS scientist just said it was a constant, and I was disagreeing. Is there any way you could give me the answer and answer how you got it?

    Thanks
     
    Last edited by a moderator: Mar 24, 2013
  12. Mar 23, 2013 #11
    also since it says evaluate when theta > 0 this means that all values to infinity so doesnt this mean when evaluating ill get infinity/infinity is the exponent.
     
  13. Mar 23, 2013 #12

    SammyS

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    This just means θ is a constant, and it's positive.


    In your Original Post, you said that using u substitution, you got -θ*e^(-y/θ) .

    That's the correct result for the indefinite integral, [itex]\displaystyle \ \int e^{-y/\theta}\,dy\ .[/itex] Another way to say that is: The anti-derivative of [itex]\displaystyle \ e^{-y/\theta}\ [/itex] is -θ*e^(-y/θ) .

    Do you know how evaluate a definite integral (with limits of integration), if you know the indefinite integral (without limits of integration)?
     
  14. Mar 23, 2013 #13
    yes i know how to evaluate the definite but when the values of theta are : theta>0 how do you evaluate this when the limits of integration are also from 0 to infinity?
     
  15. Mar 23, 2013 #14

    SammyS

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    Again: θ is simply some constant -- it happens to be positive. The variable of integration is "y" . It's y that goes from 0 to ∞ as specified by the limits of integration.
     
  16. Mar 24, 2013 #15
    Let's try to do this step by step:
    [itex] \displaystyle\lim_{y\rightarrow \infty} {-\theta e^{-y/\theta}} =

    -\theta \displaystyle\lim_{y\rightarrow \infty} {e^{-y/\theta}} [/itex] =

    [itex] -\theta (\displaystyle\lim_{y\rightarrow \infty} {e^{-y}})^{1/\theta} [/itex].

    Since [itex] \theta [/itex] is greater than zero, what can you say about this limit?
    Similarly, what about [itex] -\theta (\displaystyle\lim_{y\rightarrow 0} {e^{-y}})^{1/\theta} [/itex]?
     
    Last edited: Mar 24, 2013
  17. Mar 24, 2013 #16
    Yes, thank you. I should be more careful when writing out solutions.
     
  18. Mar 24, 2013 #17

    SammyS

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    You may still be able to go back and Edit that post.

    Added in Edit:

    Now that you've made the change, I'll delete my post regarding the typo.

    (I think PF gives you something like a 700 minute window for Editing.)
     
    Last edited: Mar 24, 2013
  19. Mar 24, 2013 #18
    Wow, I did not realize that there is such a long allowed time to edit posts. Thanks!
     
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