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Math Help - Comaprison tests for convergence

  1. #1
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    Comaprison tests for convergence

    Hi, I'm having some difficulty in recognising what new improper integral to create when doing a comparison test.
    IS there a step-by-step logical approach that could be used?
    My understanding is, we do comparison tests because the improper integral being dealt with is too difficult to integrate. I understand the rules in that if f(x)>g(x)
    then f(x) is divergent if g(x) is divergent and g(x) is convergent if f(x) is convergent.

    But, I don't understand a few problems:

    So, for example,

    http://i.imgur.com/OF9Ip.jpg

    If you could look at 6b, and the solutions to 6b.

    I don't understand what the solutions mean when they say "Use x< x^1/4"

    The same with 6c. Why have they said 'use 1/3y^4' ?

    I managed to work out 6a myself. are there flaws in the solutions? What am I missing here..

    Additionally, in regards to the matrices question at the top, what method is being referred to there? Is it asking for me to use row operations?

    Thanks!
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  2. #2
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    Re: Comaprison tests for convergence

    They use x \le x^{\frac{1}{4}} so that 1 - x^{\frac{1}{4}} \le 1 - x

    \frac{1}{1 - x^{\frac{1}{4}}} \ge \frac{1}{1-x} (along the domain)

    Therefore

    \int_{0}^{1} \frac{1}{1 - x^{\frac{1}{4}}}\, dx \ge \int_{0}^{1} \frac{1}{1-x}\, dx
    Last edited by richard1234; November 4th 2012 at 07:28 PM.
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  3. #3
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    Re: Comaprison tests for convergence

    ^Thanks!

    makes sense

    although, I'm still not sure about why they use 3y^4 ?
    Would it not make more sense to use y^4?
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