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Improper integrals with infinite discontinuities

  1. Mar 25, 2009 #1
    1. The problem statement, all variables and given/known data

    [tex]\int[/tex][tex]\frac{13}{(x-8)^2}[/tex]dx

    2. Relevant equations

    it is integrated from 7 to 9 and i am aware that there is an infinite discontinuity at x=8 so we have to take the limit from both sides individually.


    3. The attempt at a solution

    The only thing I can think that I might be doing wrong is just integrating it wrong but it seems like such an easy integration. The integration that I come up with:

    -[tex]\frac{13}{x-8}[/tex]


    as a goes to 8 [[tex]\frac{-13}{a-8}[/tex]+[tex]\frac{13}{7-8}[/tex]]+[[tex]\frac{-13}{9-8}[/tex]+[tex]\frac{13}{a-8}[/tex]]

    and this comes out to be -26 which has to be wrong because the graph is above the positive x-axis and also it is not one of my answer choices.
     
  2. jcsd
  3. Mar 25, 2009 #2

    Hurkyl

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    I thought you said you were going to take the limit of both sides individually? This looks more like you threw both sides together first, and then took a limit.
     
  4. Mar 25, 2009 #3
    I'm not really sure how to write it in there with this program but I did take the limits individually when I did the work on my paper. The first bracket is the limit from 7 to a as a approaches 8 from the left and the second bracket is the limit from a to 9 as a approaches 8 from the right. Did i do that right?
     
  5. Mar 25, 2009 #4
    then don't you just add them in the end?
     
  6. Mar 25, 2009 #5

    Hurkyl

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    What were the limits on the two sides then? What two numbers did you add together to get -26?
     
  7. Mar 25, 2009 #6
    both of them came out to -13
     
  8. Mar 25, 2009 #7

    Hurkyl

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    Interesting. Could you show how you calculated that?
     
  9. Mar 25, 2009 #8
    lim as a --> 8- [ [tex]\frac{-13}{x-8}[/tex] ]

    from 7 to a =

    [[tex]\frac{-13}{(a-8)}[/tex]+[tex]\frac{13}{(7-8)}[/tex]]

    [ 0 + [tex]\frac{13}{-1}[/tex]]

    = -13
     
  10. Mar 25, 2009 #9
    its the same basically for the other side
     
  11. Mar 25, 2009 #10

    Hurkyl

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    The 13/-1 makes sense. How did you get that 0?
     
  12. Mar 25, 2009 #11
    well doesn't the 8- with the negative sign to the right of the number mean +8 but that it is approaching from the left side?
    or are you saying that it means I should have plugged in a negative 8 for a?

    or does that mean that that factor goes to infinity?
     
    Last edited: Mar 25, 2009
  13. Mar 25, 2009 #12

    Hurkyl

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    Bingo. ([itex]+\infty[/itex], I believe? You should check)
     
  14. Mar 25, 2009 #13
    thanks for your help It's just been a while since calc I and i kinda forgot how to do limits

    and yes one of the answer choices is that the integral diverges
     
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