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Limit diverges or converges

  1. Sep 21, 2008 #1
    integral 2/(x-6)^2 respects to x from 0 to 8. Shouldn't the answer to this be converges and is =-4/3. The true answer to this is it diverges towards infinity... can someone please explain.
     
  2. jcsd
  3. Sep 21, 2008 #2

    statdad

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    Two things to consider:

    1. isn't there a number between [tex] 0 [/tex] and [tex] 8 [/tex] that could cause a problem?

    2. The integrand [tex] 2/{(x-6)^2}[/tex] is positive throughout the interval of integration,
    so how could the integral be negative?
     
  4. Sep 21, 2008 #3
    I think I've found the problem there should be an asymptote when x=6. My teacher never taught me to look for these.
     
  5. Sep 21, 2008 #4

    statdad

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    Whether you are graphing, integrating, differentiating, or simply contemplating the function and its domain for their inherent beauty :rofl: you should always look for the possible existence, and influence of, an asymptote for rational functions.
     
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