Small confusion about an improper integral example.

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SUMMARY

The discussion centers on the evaluation of the improper integral \(\int^{1}_{0}\frac{1}{\sqrt{1-x^{2}}}dx\). The correct limit approach is established as \(\lim_{t \rightarrow 0^{+}}\int^{1}_{t}\frac{1}{\sqrt{1-x^{2}}}dx = \frac{\pi}{2}\), indicating that the discontinuity occurs at \(x = 1\) rather than \(x = 0\). The confusion arises from the alternative limit \(\lim_{t \rightarrow 1^{-}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}dx\), which is deemed incorrect for this integral evaluation.

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funcalys
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We have that
\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}
However, I think \int^{1}_{0}\frac{1}{\sqrt{1-x^{2}}} should equal to lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}
since f is not continuous at 1, not 0.:confused:
 
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funcalys said:
We have that
\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}
However, I think \int^{1}_{0}\frac{1}{\sqrt{1-x^{2}}} should equal to lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}
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.
lim_{\stackrel{}{t \rightarrow 0^{+}}}\int^{1}_{t}\frac{1}{\sqrt{1-x^{2}}}
lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}
 
funcalys said:
Thanks but I'm wondering why they used the first one instead of the second.
lim_{\stackrel{}{t \rightarrow 0^{+}}}\int^{1}_{t}\frac{1}{\sqrt{1-x^{2}}}
lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}


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:.
 
I've encountered a few different definitions of "indefinite integral," denoted ##\int f(x) \, dx##. any particular antiderivative ##F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)## the set of all antiderivatives ##\{F:\mathbb{R} \to \mathbb{R}, F'(x) = f(x)\}## a "canonical" antiderivative any expression of the form ##\int_a^x f(x) \, dx##, where ##a## is in the domain of ##f## and ##f## is continuous Sometimes, it becomes a little unclear which definition an author really has in mind, though...

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