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anonymous
 4 years ago
Testing for convergence or divergence of an integral: Use the direct comparison test or limit comparison test to test the integral for convergence.
integral (0 to 1) of dt/(tsint)
How do I know which test to use, and how should I proceed from there?
anonymous
 4 years ago
Testing for convergence or divergence of an integral: Use the direct comparison test or limit comparison test to test the integral for convergence. integral (0 to 1) of dt/(tsint) How do I know which test to use, and how should I proceed from there?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I wish I had my calculator on me :( Try comparing it to 1/t.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0does it matter whether I compare it to 1/t versus 1/t^2 or 1/t^3, etc?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep, this diverges ;) 1/(t  sin t) > 1/t from t:[0,1] \[\int\limits_{0}^{1} \frac {dt}{t} = \left[ \ln t \right]_{0}^{1}\] That is divergent. Since 1/t (the smaller function) diverges, then 1/(t sin t) (the bigger function) must diverge as well.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It doesn't matter for this case, but for other integrals, you might wanna choose carefully.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I see. thanks. Could I use LCT for this integral as well?
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