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

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The integral of 2/(x-6)^2 from 0 to 8 diverges due to the presence of an asymptote at x=6, which lies within the interval of integration. The integrand is positive throughout the interval, leading to confusion about the integral's value being negative. The discussion highlights the importance of identifying asymptotes when working with rational functions, as they can significantly affect the convergence of integrals. Participants emphasize that overlooking such critical points can lead to incorrect conclusions about the integral's behavior. Understanding the function's domain and its asymptotic characteristics is essential for accurate analysis.
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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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