Convergence of Improper Integral with Limit Evaluation

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Homework Help Overview

The discussion revolves around determining the convergence or divergence of an improper integral, specifically the integral from 0 to 1 of the function 1/√[3]{x}. Participants are exploring the nature of this integral and its evaluation using limits.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to classify the integral as a p-series based on its form but questions this classification due to the limits of integration. Other participants clarify that it is not a p-series but an improper integral, suggesting a need to evaluate it by taking a limit as the lower limit approaches zero.

Discussion Status

Participants are actively discussing the classification of the integral and the appropriate method for evaluation. There is a clear indication that some guidance has been offered regarding the evaluation process, although no consensus has been reached on the initial classification of the integral.

Contextual Notes

There is a focus on the limits of integration and the implications for convergence, with some participants questioning the assumptions made about the nature of the integral.

flyingpig
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Homework Statement



Determine whether the improper integral diverge or converge. If it is convergent, evaluate it with a limit

[tex]\int_{0}^{1}\frac{1}{\sqrt[3]{x}}dx[/tex]Just from inspection, I thought this was a p-series with 1/3 < 1, but then I noticed the limits of integration is from 0 to 1.

So my question is, is there a way, just from inspection, to notice this is convergent without resorting to any test?
 
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Just fixed question
 
It's not a p-series. It's not any kind of series at all. It's an improper integral. Integrate it and take the limit as the lower limit approaches 0.
 
Dick said:
It's not a p-series. It's not any kind of series at all. It's an improper integral. Integrate it and take the limit as the lower limit approaches 0.

I.E. Evaluate:

[tex]\int_{a}^{1}\frac{1}{\sqrt[3]{x}}dx[/tex]

Take the limit of the result as a → 0+.
 

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