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Small confusion about an improper integral example.

  1. Oct 28, 2012 #1
    We have that
    [itex]\int^{1}_{0}\frac{1}{\sqrt{1-x^{2}}}=lim_{\stackrel{}{t \rightarrow 0^{+}}}\int^{1}_{t}\frac{1}{\sqrt{1-x^{2}}}=lim_{\stackrel{}{t \rightarrow 0^{+}}}[arcsin(x)]^{1}_{t}=\frac{\pi}{2}[/itex]
    However, I think [itex]\int^{1}_{0}\frac{1}{\sqrt{1-x^{2}}}[/itex] should equal to [itex]lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}[/itex]
    since f is not continuous at 1, not 0.:confused:
     
  2. jcsd
  3. Oct 28, 2012 #2

    Yes, of course: the problem is at 1, not at zero.

    DonAntonio
     
  4. Oct 29, 2012 #3
    Thanks but I'm wondering why they used the first one instead of the second.
    [itex]lim_{\stackrel{}{t \rightarrow 0^{+}}}\int^{1}_{t}\frac{1}{\sqrt{1-x^{2}}}[/itex]
    [itex]lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}[/itex]
     
  5. Oct 29, 2012 #4

    I don't know who is "they" but it most probably is a mistake or, perhaps, "they" wanted to make some point. Anyway, it is reduntant.

    DonAntonio
     
  6. Oct 29, 2012 #5
    Thanks, I was just confused when my answer doesn't match the exercise's solution, maybe some mistakes were made . Thank again :biggrin:.
     
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