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Math Help - [SOLVED] Integral of abs(x)^(1/2)

  1. #1
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    [SOLVED] Integral of abs(x)^(1/2)

    Hello there,

    I was wondering how to integrate  |x|^{1/2} . I don't think that  \frac{2|x|^{3/2}}{3} + C is correct.

    Thank you!


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  2. #2
    Junior Member
    Joined
    Jul 2008
    Posts
    74
    Hello there,

    I figured out the answer.

    Since:

     |x|^{1/2} = \left\{\begin{array}{cc}\sqrt{x},&\mbox{ if }<br />
x\geq 0\\\sqrt{-x}, & \mbox{ if } x < 0\end{array}\right.

    Therefore:

     \\\int |x|^{1/2} = \left\{\begin{array}{cc}\int x^{\frac{1}{2}} \ dx,&\mbox{ if }<br />
x\geq 0\\\int (-x)^{\frac{1}{2}} \ dx, & \mbox{ if } x < 0\end{array}\right.

    For  x < 0 , let  u = -x \Rightarrow -du = dx and take the antiderivative.

     \\\int |x|^{1/2} = \left\{\begin{array}{cc}\frac{2x^{3/2}}{3} + C,&\mbox{ if }<br />
x\geq 0\\-\frac{2(-x)^{3/2}}{3} + C, & \mbox{ if } x < 0\end{array}\right.
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