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Thread: Improper Integrals

  1. #1
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    Improper Integrals

    I'm having trouble with the following problem:

    Show that $\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by larson View Post
    I'm having trouble with the following problem:

    Show that $\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$
    Show that this is what?

    $\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$

    $\displaystyle = \lim_{b \to \infty}
    \int_{1}^b \frac {du}{u^\frac{3}{2}}$

    $\displaystyle = \lim_{b \to \infty}-2 \left ( \frac{1}{b^{1/2}} - \frac{1}{1^{1/2}} \right )$

    $\displaystyle = 2$

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Show that this is what?

    $\displaystyle \int_{1}^{\infty} \frac {du}{u^\frac{3}{2}}$

    $\displaystyle = \lim_{b \to \infty}
    \int_{1}^b \frac {du}{u^\frac{3}{2}}$

    $\displaystyle = \lim_{b \to \infty}-2 \left ( \frac{1}{b^{1/2}} - \frac{1}{1^{1/2}} \right )$

    $\displaystyle = 2$

    -Dan
    How do you know to take the 2 out? Thats the only thing thats confusing me, and by the way the answer is correct, although I'm sure you knew .
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Hello
    $\displaystyle
    \int x^{\alpha} \,\mathrm{d}x= \frac{t^{\alpha+1}}{\alpha+1}$ with $\displaystyle \alpha \neq -1$
    Applying it with $\displaystyle \alpha=-\frac{3}{2}$ gives us :$\displaystyle
    \int \frac{1}{x^{\frac{3}{2}}} \,\mathrm{d}x= \int x^{-\frac{3}{2}} \,\mathrm{d}x = \frac{t^{-\frac{3}{2}+1}}{\frac{-3}{2}+1}=\frac{t^{-\frac{1}{2}}}{-\frac{1}{2}}=-\frac{2}{\sqrt{t}} $
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