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Math Help - Upper bounds

  1. #1
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    Upper bounds

    the question is does this converge or diverge and if it converges give its upper bound..

    \int\limits_1^\infty \frac{\sin(3x)+3}{x^3+x}dx

    what i have is by comparison
    x^3+x>x^3

    \frac{1}{x^3+x}<\frac{1}{x^3}

    and
    -1\leq\sin(3x)\leq1

    \frac{2}{x^3+x}\leq\frac{\sin(3x)}{x^3+x}\leq\frac  {4}{x^3+x}

    so
    \frac{\sin(3x)}{x^3+x}<\frac{4}{x^3}

    and i know that \frac{4}{x^3} converges

    but not sure how to get the upper bound, can someone explain to me how to get the upper bound please! thank you
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    What you actually found, it's the upper bound.

    What's this? An improper integral? A series? Also, you didn't say for what x you're working on.
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  3. #3
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    Sorry it is an improper integral, but were currently in the series section.
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  4. #4
    Math Engineering Student
    Krizalid's Avatar
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    Okay, then you used \frac{\sin (3x)+3}{x^{3}+x}\le \frac{4}{x^{3}+x}<\frac{4}{x^{3}}, and that was enough to show its convergence.
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