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anonymous

  • 4 years ago

I really need help on improper integrals, please respond?

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  1. anonymous
    • 4 years ago
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    ok

  2. anonymous
    • 4 years ago
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    Okay, so I'm really confused as to what to do. Let's do an example problem, \[\int\limits_{0}^{1} dx/3x-2\]

  3. anonymous
    • 4 years ago
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    this isn't an improper integral

  4. anonymous
    • 4 years ago
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    It's dx over 3x-2, i'm looking at my book right now.

  5. anonymous
    • 4 years ago
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    ok

  6. anonymous
    • 4 years ago
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    okay. so I don't know what to do.

  7. anonymous
    • 4 years ago
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    well we can't have a zero denominator, which means we can't have x=2/3. This lies within our limits of integration which is why the integral is "improper".

  8. anonymous
    • 4 years ago
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    We need to split this integral

  9. anonymous
    • 4 years ago
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    what are the steps?

  10. anonymous
    • 4 years ago
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    \[\int\limits_{0}^{2/3}\frac{dx}{3x-2}+\int\limits_{2/3}^{1}\frac{dx}{3x-2}\]

  11. anonymous
    • 4 years ago
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    do you know what to do next?

  12. anonymous
    • 4 years ago
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    We evaluate the one-sided limits of each term separately:\[\int\limits_{0}^{2/3}\frac{dx}{3x-2}=\lim_{t \rightarrow 2/3^{-1}}\int\limits_{0}^{t}\frac{dx}{3x-2}\]

  13. anonymous
    • 4 years ago
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    I get:\[\lim_{t \rightarrow (2/3)^{-1}}\frac{1}{3}(\ln \left| 3t-2 \right|-\ln \left| 0-2 \right|)\]

  14. anonymous
    • 4 years ago
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    I'll let you take it from here

  15. anonymous
    • 4 years ago
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    Yes, it diverges so you wouldn't continue. Therefore the answer you just be that.

  16. anonymous
    • 4 years ago
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    yeppers

  17. anonymous
    • 4 years ago
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    dont forget to medal me to close this question

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