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Math Help - Help Imporper integral by comparison??

  1. #1
    Member zangestu888's Avatar
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    Help Imporper integral by comparison??

    Integral of:
    dx / sqrt (x^4 + 4x^3) from upper limit b=infinte and a=1

    Integral of:

    1 + 2t(cos(t))^2(dt)/ (t)^3 from upper limit infinte an a=lowerlimit=pi

    Please I need help using the comparison method to determine if integral converges or diverges.!
    THANKS
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  2. #2
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    Quote Originally Posted by zangestu888 View Post
    Integral of:
    dx / sqrt (x^4 + 4x^3) from upper limit b=infinte and a=1

    Integral of:

    1 + 2t(cos(t))^2(dt)/ (t)^3 from upper limit infinte an a=lowerlimit=pi

    Please I need help using the comparison method to determine if integral converges or diverges.!
    THANKS
    Since x^4 < x^4 + 4x^3 for x \ge 1 then \frac{1}{x^4 + 4x^3} < \frac{1}{x^4}\,dx then

    \int_1^{\infty}\frac{1}{x^4 + 4x^3}\,dx < \int_1^{\infty}\frac{1}{x^4}\,dx < \infty

    so your integral converges.

    For the second problem try \cos^2 t \le 1 and see if you can get an inequality like above.
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  3. #3
    Member zangestu888's Avatar
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    OKay thank i will work on it
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  4. #4
    Member zangestu888's Avatar
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    ??

    so (cost)^2 <=1

    am still not quite understanding it, can you give me a hint on how to begin i know i have to find a function that I know is either convergent or divergent 1/t^3 would be my function? am so lost? i wana know how to start if u may.
    And for the first question I would like to knnow what hapened to the sqrt( of the function ) the radical u showed me disspeard?
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  5. #5
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    Quote Originally Posted by zangestu888 View Post
    so (cost)^2 <=1

    am still not quite understanding it, can you give me a hint on how to begin i know i have to find a function that I know is either convergent or divergent 1/t^3 would be my function? am so lost? i wana know how to start if u may.
    And for the first question I would like to knnow what hapened to the sqrt( of the function ) the radical u showed me disspeard?
    I totally missed the square root. Hopefully this fixes it.

    Since x^4 < x^4 + 4x^3 for x \ge 1 then \sqrt{x^4} < \sqrt{x^4 + 4x^3} then \frac{1}{\sqrt{x^4 + 4x^3}} < \frac{1}{x^2} then

    \int_1^{\infty}\frac{1}{\sqrt{x^4 + 4x^3}}\,dx < \int_1^{\infty}\frac{1}{x^2}\,dx < \infty

    so your integral converges.

    As for a hint on the second one

    \cos^2t < 1\;\;\; \Rightarrow \;\;\;1 + 2 t \cos^2 t < 1 + 2t \;\;\; \Rightarrow\;\;\; \frac{1 + 2 t \cos^2 t }{t^3} < \frac{1 + 2t}{t^3}
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  6. #6
    Member zangestu888's Avatar
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    I think i am gettin this, wats the trick t o usually doing this is to know a bunch of functions that are divergent or convergent so you can compare thas about it really write? thanks for the help much appreciatedd
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