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Homework Help: Improper Integral

  1. Nov 27, 2009 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    Let f be continuous on the half-open interval (a, b] and suppose that
    [tex] \lim_{x \to a^{+}} |f(x)| = \infty[/tex]. Then

    [tex]\int_{a}^{b}f(x) dx = \lim_{ t \to a^{+}}\int_{t}^{b}f(x)

    3. The attempt at a solution


    = \lim_{ t \to -3^{+}}\int_{t}^{1}\frac{x}{\sqrt{9-x^{2}}}


    [tex] u = 9 - x^{2} [/tex]
    [tex] du = -2x dx [/tex]

    [tex] \lim_{ t \to -3^{+}}\int_{t}^{1}\frac{x}{\sqrt{9-x^{2}}}

    = \lim_{ t \to 0^{+}}-\frac{1}{2}\int_{t}^{8} u^{-1/2} du

    =\lim_{ t \to 0^{+}}-u^{1/2}|_{t}^{8}

    =-\sqrt{3sin(1)} + \lim_{ t \to -3^{+}}\sqrt{3sin(t)}

    =-1.588840129 + ?[/tex]

    I get 0 for the limit, but according to Maple and my graphing calculator, that does not give the correct value for this integral. The correct value is approximately -2.8. May I please have some guidance as to what may have went wrong?
  2. jcsd
  3. Nov 27, 2009 #2
    I don't understand that step where you introduce sine. You should just get:

    \lim_{ t \to 0^{+}}-u^{1/2}|_{t}^{8}

    =-\lim_{t \to 0^+}(\sqrt{8} - \sqrt{t})

    = 0 - 2\sqrt{2}

    = -2\sqrt{2}

    \approx -2.8
  4. Nov 27, 2009 #3
    Oh, god, I made an incredibly stupid mistake....Thank you.
  5. Nov 27, 2009 #4
    haha, it happens. no worries.
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