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Integration question.

  1. Jan 15, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    I'd appreciate some help in computing the integral in the attachment.


    2. Relevant equations



    3. The attempt at a solution
    I presume it should be handled using L'hopital, but I am not sure how to demonstrate that that is indeed the case.
     

    Attached Files:

  2. jcsd
  3. Jan 15, 2013 #2

    CAF123

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    Yes, I'Hopital will be applied. Consider using the fundamental theorem of Calculus as well.
     
  4. Jan 15, 2013 #3
    But L'hopital applies for 0/0, for instance. How is that the case here?
     
  5. Jan 15, 2013 #4

    CAF123

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    As x → 0, the upper limit tends to 0 and since the lower limit is already zero, we know ##\int_{a}^{a} f dt = 0##. ( and the function f in your example is continous near x = 0)
     
  6. Jan 15, 2013 #5
    Is it rigorous/formal enough, to simply state that int [a,a] f dt = 0?
     
  7. Jan 15, 2013 #6

    CAF123

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    I would say: $$\int_0^0 f\,dt = F(0) - F(0) = 0,$$ since a function cannot have one to many as an option.

    If I recall correctly, and perhaps someone could confirm, it is not possible to evaluate that integral (analytically) if the limits are finite (i.e something like 0 to ##a##, ##a## a real number).
     
  8. Jan 15, 2013 #7
    Is it true then that the limit diverges?
     
  9. Jan 15, 2013 #8

    CAF123

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    I wouldn't figure so quickly. So we have justified using i'Hopital. Now apply i'Hopital.
    (the bit about your integrand not being integrable with finite limits was just for interest. It doesn't really relate to this problem since the region of int. is [0,0] and provided an integral exists this is always zero if those are your bounds).
     
  10. Jan 15, 2013 #9

    SammyS

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    I don't think so, but I hurried through it, so I could have made a mistake.
     
  11. Jan 15, 2013 #10
    No, it is equal to zero. Thanks!!
     
  12. Jan 15, 2013 #11

    SammyS

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    That's what I got too ... zero.
     
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