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Homework Help: Computing an improper integral

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

    [tex]\int^{1}_{0}\frac{dx}{\sqrt{1-x^{2}}}[/tex]

    2. Relevant equations

    None

    3. The attempt at a solution

    [tex]\int^{1}_{0}\frac{dx}{\sqrt{1-x^{2}}} = sin^{-1}x\right|^{1}_{0}[/tex]

    [tex]sin^{-1}x\right|^{1}_{0} = \frac{\pi}{2} - 0[/tex]

    so the final answer is just pi/2. I have no problem computing the answer, but it's in the improper integrals section of the textbook....but I don't see this as being an improper integral. There's no need to deal with infinity at all, no asymptotes, no discontinuity on the closed interval [0,1].

    Am I missing something?
     
    Last edited: Jun 17, 2010
  2. jcsd
  3. Jun 17, 2010 #2

    D H

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    What happens to [tex]1/\sqrt{1-x^2}[/tex] as [itex]x\to1\quad[/itex]?
     
  4. Jun 17, 2010 #3
    Good call. Fixed it and showed a few extra steps, namely:

    [tex]\int^{1}_{0}\frac{dx}{\sqrt{1-x^{2}}} = lim_{t\rightarrow1^{-}}\int^{t}_{0}\frac{dx}{\sqrt{1-x^{2}}}[/tex]

    then just computed the integral which is arcsin x, and the limit as t approaches 1 from the left for arcsin t is just pi/2.

    Yay! Thanks!
     
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