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peggiepenguin

  • 3 years ago

does the integral of 1/x+ srqtx from 1 to infinity converge?

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  1. anonymous
    • 3 years ago
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    i doubt it

  2. anonymous
    • 3 years ago
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    the degree of the denominator has to be larger than the degree of the numerator by more than 1

  3. electrokid
    • 3 years ago
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    is it \[\int_1^\infty\left({1\over x}+\sqrt{x}\right)dx\] OR \[\int_1^\infty{dx\over x+\sqrt{x}} \]

  4. peggiepenguin
    • 3 years ago
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    the second one

  5. electrokid
    • 3 years ago
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    @satellite73 aha... @peggiepenguin next time please use the "equation editor on this website" or "TEX" to enter your equations.. so we know what you are talking about!!

  6. electrokid
    • 3 years ago
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    so, it DOES converge coz, the denominator increases continuously and hence, the function decreases continuously

  7. anonymous
    • 3 years ago
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    oh no!!

  8. electrokid
    • 3 years ago
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    oh yes! :D

  9. anonymous
    • 3 years ago
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    just because the function decreases does not mean the integral converges that is a necessary but not a sufficient condition it must decrease fast enough

  10. anonymous
    • 3 years ago
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    simple example \[\int_1^{\infty}\frac{dx}{x}\] does not converge

  11. anonymous
    • 3 years ago
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    neither does the one above

  12. electrokid
    • 3 years ago
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    aah.. jumping to conclusions too soon. \[ u=\sqrt{x}\implies u^2=x\implies 2udu=dx\\ \int_1^\infty\frac{2u}{u^2+u}du\\ =2\int_1^\infty {du\over u+1}=2[\ln (u+1)]_1^\infty=\boxed{\infty} \]

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