Does the integral of 2/(x-6)^2 from 0 to 8 converge or diverge?

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Homework Help Overview

The discussion revolves around the convergence or divergence of the integral of the function 2/(x-6)^2 from 0 to 8. Participants are exploring the implications of the integrand's behavior within the specified interval.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning the presence of potential issues within the interval, particularly concerning the behavior of the integrand at x=6. There is also discussion about the nature of the integral and its expected value.

Discussion Status

The conversation is actively exploring the implications of the asymptote at x=6 and its effect on the integral. Some participants are reflecting on the importance of identifying such features in rational functions, indicating a productive direction in the discussion.

Contextual Notes

There is an indication that the original poster may have misconceptions about the behavior of the integral and the role of asymptotes, which has not been explicitly resolved in the discussion.

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integral 2/(x-6)^2 respects to x from 0 to 8. Shouldn't the answer to this be converges and is =-4/3. The true answer to this is it diverges towards infinity... can someone please explain.
 
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Two things to consider:

1. isn't there a number between 0 and 8 that could cause a problem?

2. The integrand 2/{(x-6)^2} is positive throughout the interval of integration,
so how could the integral be negative?
 
I think I've found the problem there should be an asymptote when x=6. My teacher never taught me to look for these.
 
Whether you are graphing, integrating, differentiating, or simply contemplating the function and its domain for their inherent beauty :smile: you should always look for the possible existence, and influence of, an asymptote for rational functions.
 

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