Small confusion about an improper integral example.

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Discussion Overview

The discussion revolves around the evaluation of the improper integral \(\int^{1}_{0}\frac{1}{\sqrt{1-x^{2}}}\) and the reasoning behind using limits at different points of discontinuity. Participants explore the implications of continuity at the endpoints of the integral.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant asserts that the integral should be evaluated using the limit as \(t\) approaches \(0^{+}\) because the function is continuous at that endpoint, leading to a result of \(\frac{\pi}{2}\).
  • Another participant argues that the integral should instead use the limit as \(t\) approaches \(1^{-}\) due to the function's discontinuity at \(1\), suggesting that the issue lies at that endpoint.
  • There is a request for clarification on why the first limit approach was chosen over the second, indicating confusion about the reasoning behind the selection of limits.
  • A later reply questions the rationale of the original choice, suggesting it may have been a mistake or an attempt to illustrate a point.
  • One participant expresses confusion about discrepancies between their answer and the provided solution, hinting at potential errors in the exercise.

Areas of Agreement / Disagreement

Participants express differing views on which limit should be used for evaluating the integral, indicating that there is no consensus on the correct approach.

Contextual Notes

Participants highlight the dependence on the continuity of the function at the endpoints, which remains unresolved in the discussion.

funcalys
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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:
 
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funcalys said:
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:


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

DonAntonio
 
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]
 
funcalys said:
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]


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
 
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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