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Improper Integration

  1. Apr 17, 2010 #1
    The problem statement, all variables and given/known data
    ixw7ie.jpg


    3The attempt at a solution

    5d3wbr.jpg

    Is this correct?
     
  2. jcsd
  3. Apr 17, 2010 #2

    Dick

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    Sure, it looks fine to me.
     
  4. Apr 17, 2010 #3
    Do I need to show that I'm approaching 4 from the right?
     
  5. Apr 17, 2010 #4

    Dick

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    I would say you are approaching 4 from the left. t<4, correct? Why would you want to approach from the right?
     
  6. Apr 17, 2010 #5
    Because if I were approaching ln|x-4| from the right, the graph goes to -∞. How could you approach it from the left?

    Or am I supposed to approach the original graph of 1/(x^2 -3x -4) from the left?
     
  7. Apr 17, 2010 #6

    Dick

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    The problem with your integral is at x=4. To resolve it as an improper integral you want to integrate from x=0 to x=4-epsilon where epsilon>0. That means you are approaching the upper limit from the left. You don't care what the limit is from the right.
     
  8. Apr 17, 2010 #7
    So even though the limit is approaching 4 from the left at ln|x-4|, we're infact evaluating the limit as it approaches 4 from the left of 1/(x^2-3x-4)?
     
  9. Apr 17, 2010 #8

    Dick

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    Now you are just confusing me. You are approaching x=4 from the left. Period. Approaching ln|x-4| from the left gives you the behavior of the integral of 1/(x^2-3x-4) on the interval [0,4].
     
  10. Apr 18, 2010 #9
    This is what I was confused about.

    66mc28.jpg

    In the graph above, you can approach 4 from the right but not from the left.
     
  11. Apr 18, 2010 #10

    Dick

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    It's an absolute value, temaire. |x-4|. Doesn't that mean anything to you? :)
     
  12. Apr 18, 2010 #11
    Oh, so this is the graph. (x is from 3 to 5)

    11maiqc.jpg

    I understand now.
     
  13. Apr 18, 2010 #12

    Dick

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    You've got it.
     
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