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Math Help - Comparison Theorem Integrals

  1. #1
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    Comparison Theorem Integrals

    Use the comparison Theorem to determine whether the integral is convergent or divergent

    integral from 0,1

    [e^(-x)]/sqrt(x)

    thanks guys
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  2. #2
    Ted
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    Quote Originally Posted by Asuhuman18 View Post
    Use the comparison Theorem to determine whether the integral is convergent or divergent

    integral from 0,1

    [e^(-x)]/sqrt(x)

    thanks guys
    \frac{e^{-x}}{\sqrt{x}} \leq \frac{e^{\sqrt{x}}}{\sqrt{x}} \,\ \forall \,\ x \in (0,1].
    and it is easy to show that the improper integral \int_0^1 \frac{e^{\sqrt{x}}}{\sqrt{x}} \,\ dx converges.
    Hence, your improper integral converges.


    T
    Last edited by Ted; February 20th 2010 at 07:10 AM.
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  3. #3
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    or \frac{e^{-x}}{\sqrt{x}}\le \frac{1}{\sqrt{x}} for x>0 and in particular for x\in(0,1] holds and \int_0^1\frac{dx}{\sqrt x} verifies the convergence of the original integral.
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  4. #4
    Senior Member bkarpuz's Avatar
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    Similar to what Krizalid has shown, we can also find a lower bound.
    For all x\in[0,1], we have
    \frac{1}{\mathrm{e}}\leq\mathrm{e}^{-x}\leq1, which yields by integrating after dividing by \sqrt{x}
    \frac{2}{\mathrm{e}}\leq\int_{0}^{1}\frac{\mathrm{  e}^{-x}}{\sqrt{x}}\mathrm{d}x\leq2.
    Note that above, we apply improper integration and that
    \int_{a}^{b}\frac{1}{\sqrt{x}}\mathrm{d}x=2\sqrt{x  }\big|_{x=a}^{x=b} provided that b>a>0.
    Last edited by bkarpuz; February 20th 2010 at 07:24 AM. Reason: Ted pointed out a correction.
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  5. #5
    Ted
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    Quote Originally Posted by bkarpuz View Post
    Similar to what Krizalid has shown, we can also find a lower bound.
    For all x\in[0,1], we have
    \frac{1}{\mathrm{e}\sqrt{x}}\leq\frac{\mathrm{e}^{-x}}{\sqrt{x}}\leq\frac{1}{\sqrt{x}}, which yields by integrating
    \frac{2}{\mathrm{e}}\leq\int_{0}^{1}\frac{\mathrm{  e}^{-x}}{\sqrt{x}}\mathrm{d}x\leq2.
    Recall that
    \int_{a}^{b}\frac{1}{\sqrt{x}}\mathrm{d}x=2\sqrt{x  }\big|_{x=a}^{x=b}.
    It should be {\color{red} (}0,1].
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