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Math Help - Improper Integral with one discontinuous integrand

  1. #1
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    Improper Integral with one discontinuous integrand

    \int_0^2 \ \frac{1}{\sqrt[3]{x-1}}\, dx.

    I know that it is undefined at x=1 because the denominator can't be 0, so I did this:
    \lim_{x \rightarrow 1-} \int_0^c \ \frac{1}{\sqrt[3]{x-1}}\, + \lim_{x \rightarrow 1+} \int_c^2 \ \frac{1}{\sqrt[3]{x-1}}\,

    Then I integrated it...
    \lim_{x \rightarrow 1-} \frac{3}{2}\ (x-1)^\frac{2}{3} |_0^C+ \lim_{x \rightarrow 1+} \frac{3}{2}\ (x-1)^\frac{2}{3} |_C^2

    Plugging everything in-

    \lim_{x \rightarrow 1-} \frac{3}{2}\ (c-1)^\frac{2}{3} - \frac{3}{2}\ (0-1)^\frac{2}{3} + \lim_{x \rightarrow 1+} \frac{3}{2}\ (2-1)^\frac{2}{3} - \frac{3}{2}\ (c-1)^\frac{2}{3}

    Then I started to find the limit but somehow I got the wrong answer. I got 1.5-1.5+1.5-1.5 for some reason, because I got that the one sided limits go to 1.5. The answer is supposed to be 3.

    Thank you for your help!!
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  2. #2
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    Quote Originally Posted by juicysharpie View Post
    \int_0^2 \ \frac{1}{\sqrt[3]{x-1}}\, dx.

    I know that it is undefined at x=1 because the denominator can't be 0, so I did this:
    \lim_{x \rightarrow 1-} \int_0^c \ \frac{1}{\sqrt[3]{x-1}}\, + \lim_{x \rightarrow 1+} \int_c^2 \ \frac{1}{\sqrt[3]{x-1}}\,

    Then I integrated it...
    \lim_{x \rightarrow 1-} \frac{3}{2}\ (x-1)^\frac{2}{3} |_0^C+ \lim_{x \rightarrow 1+} \frac{3}{2}\ (x-1)^\frac{2}{3} |_C^2

    Plugging everything in-

    \lim_{x \rightarrow 1-} \frac{3}{2}\ (c-1)^\frac{2}{3} - \frac{3}{2}\ (0-1)^\frac{2}{3} + \lim_{x \rightarrow 1+} \frac{3}{2}\ (2-1)^\frac{2}{3} - \frac{3}{2}\ (c-1)^\frac{2}{3}

    Then I started to find the limit but somehow I got the wrong answer. I got 1.5-1.5+1.5-1.5 for some reason, because I got that the one sided limits go to 1.5. The answer is supposed to be 3.

    Thank you for your help!!
    \lim_{c \rightarrow 1^{-}} \frac{3}{2}\ (c-1)^\frac{2}{3} = 0. Similarly for the second one.
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