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Math Help - Improper Integrals

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    Improper Integrals

    I'm having trouble with a couple of questions. I'm not sure about the last two, but I know that I need to use the limit to the lower bound for the first problem.

    1) Integral of (3)/(x^2) dx bounded by -1 (lower) to 1 (upper)

    2) Integral of (x^(-5/4)) dx bounded by 1 (lower) to infiniti (upper)

    3) Integral of (1)/((8-x)^1/3) dx bounded by 0 (lower) to 8 (upper)

    Thanks in advance for the help.
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    Quote Originally Posted by jake210 View Post
    I'm having trouble with a couple of questions. I'm not sure about the last two, but I know that I need to use the limit to the lower bound for the first problem.

    1) Integral of (3)/(x^2) dx bounded by -1 (lower) to 1 (upper)

    2) Integral of (x^(-5/4)) dx bounded by 1 (lower) to infiniti (upper)

    3) Integral of (1)/((8-x)^1/3) dx bounded by 0 (lower) to 8 (upper)

    Thanks in advance for the help.
    The first is improper because x \neq 0.

    So break it up into two integrals, namely

    \int_{-1}^0{\frac{3}{x^2}\,dx} + \int_0^1{\frac{3}{x^2}\,dx}

     = \left[-\frac{3}{x}\right]_{-1}^0 + \left[-\frac{3}{x}\right]_0^1

     = \left[\lim_{\varepsilon \to 0}\left(-\frac{3}{\varepsilon}\right) - 3\right] - \left[-3 - \lim_{\varepsilon \to 0}\left(-\frac{3}{\varepsilon}\right)\right].

    Can you go from here?
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    thanks for the help...i understand this problem now...how about the other two?
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    Quote Originally Posted by jake210 View Post
    I'm having trouble with a couple of questions. I'm not sure about the last two, but I know that I need to use the limit to the lower bound for the first problem.

    1) Integral of (3)/(x^2) dx bounded by -1 (lower) to 1 (upper)

    2) Integral of (x^(-5/4)) dx bounded by 1 (lower) to infiniti (upper)

    3) Integral of (1)/((8-x)^1/3) dx bounded by 0 (lower) to 8 (upper)

    Thanks in advance for the help.
    2. \int_1^\infty{x^{-\frac{5}{4}}\,dx} = \left[-4x^{-\frac{1}{4}}\right]_1^\infty

     = \lim_{\varepsilon \to \infty}\left(-4\varepsilon^{-\frac{1}{2}}\right) - (-4).

    Can you go from here?


    3. \frac{1}{(8 - x)^{\frac{1}{3}}} = (8 - x)^{-\frac{1}{3}}.


    So evaluate \int_0^8{(8 - x)^{-\frac{1}{3}}\,dx}.

    Can you go from here?
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