Improper Integral Domain Question

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SUMMARY

The discussion centers on evaluating the improper integral \(\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}\) for an AP Calculus BC class. The correct evaluation involves recognizing that the integrand is undefined at \(x = 1\), thus classifying it as an improper integral. The limit approach, \(\lim_{a \rightarrow 1^-} \int_0^a \frac{dx}{\sqrt{1 - x^2}}\), is necessary to properly evaluate the integral, leading to the conclusion that the result is \(\frac{\pi}{2}\).

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Homework Statement


I have to integrate this equation for my AP Calc BC class:
[tex]\int_{0}^{1}\frac{dx}{\sqrt{1-x^2}}[/tex]

Homework Equations


I have to use the definition of an improper integral.

The Attempt at a Solution


I know that it is [tex]\arcsin(x)[/tex] from 0 to 1, and both are in the domain of the arcsin function, so i don't need to use the improper integral equation, and the answer is [tex]\frac{\pi}{2}[/tex]. But I find it weird that I'd be graded on a question that i don't need to use the topic for. Am I missing anything??
 
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Yes. The integrand isn't defined at x = 1, which makes it an improper integral. For this reason you need to use a limit to evaluate it. I.e.,
[tex] \lim_{a \rightarrow 1^-} \int_0^a \frac{dx}{\sqrt{1 - x^2}}[/tex]
 
oh. so it's the domain of the integrand. thanks :D
 

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