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Math Help - Converging integrals

  1. #1
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    Converging integrals

    Determine whether or not the following intervals converge,showing your reasoning

    a)


    b)

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  2. #2
    Moo
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    Hello,

    The first one can be compared to a Riemann integral :

    \int x^{-\frac{1}{2}} dx = \int \frac{1}{x^{\frac{1}{2}}} dx

    Which diverges.

    You can also remember that an antiderivative of x^n is \frac{x^{n+1}}{n+1}.

    Here, n=\frac{-1}{2}

    So \int_0^\infty x^{-\frac{1}{2}} dx=\frac{1}{1-\frac{1}{2}} [x^{1/2}]_0^\infty

    Which also diverges.



    An antiderivative for \frac{1}{x^2+1} is arctan function.

    \int_1^\infty \frac{1}{x^2+1} dx=[Arctan(x)]_1^\infty=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}




    Edit : 5th post ^^
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  3. #3
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    For the second one observe that \frac{1}<br />
{{1 + x^2 }} \le \frac{1}<br />
{{x^2 }}, hence the integral converges by direct comparison.
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  4. #4
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    Quote Originally Posted by Krizalid View Post
    For the second one observe that \frac{1}<br />
{{1 + x^2 }} \le \frac{1}<br />
{{x^2 }}, hence the integral converges by direct comparison.
    You need to be careful here. You should actually say,
    0\leq \frac1{1+x^2}\leq \frac1{x^2}.

    You need the \geq 0 part, I am sure you can see why.

    Or a stronger way to say this is,
    \left| \frac1{1+x^2}\right| \leq \frac1{x^2}.

    In general if |f(x)|\leq g(x) and the integral of g(x) converges then f(x) integral will converge also.
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