Comparison Test on Interval 0 to 1

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SUMMARY

The discussion focuses on evaluating the convergence of the improper integral from 0 to 1 of the function (5ln(x)) / (x^(3/2)). Participants clarify that the integral diverges as x approaches 0 due to the integrand tending towards negative infinity. To determine convergence, limits must be applied, specifically using the limit as b approaches 0 from the positive side. The conclusion is that if the limit exists, the integral converges; if it is infinite, the integral diverges.

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theRukus
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Homework Statement


Determine whether or not the integral from 0 to 1 of (5ln(x)) / ( x^(3/2) ) converges or not.


Homework Equations





The Attempt at a Solution


I just need to know which end of the integral they are talking about. As x=>0, y=>-infinity. As x=>1, y=>1. I'm assuming they want to know whether the function converges or diverges as x=>0, correct?
 
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theRukus said:

Homework Statement


Determine whether or not the integral from 0 to 1 of (5ln(x)) / ( x^(3/2) ) converges or not.


Homework Equations





The Attempt at a Solution


I just need to know which end of the integral they are talking about. As x=>0, y=>-infinity. As x=>1, y=>1. I'm assuming they want to know whether the function converges or diverges as x=>0, correct?
Since this is an improper integral (the integrand is undefined at one of the interval endpoints), you need to use limits to evaluate the integral. If the limit below exists, the integral converges. If the limit doesn't exist or is infinite, the integral diverges.

\lim_{b \to 0^+}\int_b^1 \frac{5ln(x)dx}{x^{3/2}}
 
look for lnx ,x>1 but u should attention that its ln definition but when we solve that integral from 0 to 1 we want to account a surface from 0 to 1 so u should draw ln x it definition from 0 to 1 and when u draw that u know that. i hope that u understand my reason
 

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