Small confusion about an improper integral example.

[funcalys]

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.

 

[DonAntonio]

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.

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

DonAntonio

 

[funcalys]

Thanks but I'm wondering why they used the first one instead of the second.
$lim_{\stackrel{}{t \rightarrow 0^{+}}}\int^{1}_{t}\frac{1}{\sqrt{1-x^{2}}}$
$lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}$

 

[DonAntonio]

Thanks but I'm wondering why they used the first one instead of the second.
$lim_{\stackrel{}{t \rightarrow 0^{+}}}\int^{1}_{t}\frac{1}{\sqrt{1-x^{2}}}$
$lim_{\stackrel{}{t \rightarrow 1^{-}}}\int^{t}_{0}\frac{1}{\sqrt{1-x^{2}}}$

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

 

[funcalys]

Thanks, I was just confused when my answer doesn't match the exercise's solution, maybe some mistakes were made . Thank again .