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Homework Help: Please help (integrals)

  1. Nov 4, 2005 #1
    How do I do this?
    [tex] \int {\frac {1} {(\sqrt {-x})}} dx [/tex]
    I got [tex] 2 \sqrt {-x} [/tex] but my teacher got [tex] -2 \sqrt {-x} [/tex] and I don't know how he got there.
     
    Last edited: Nov 4, 2005
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  3. Nov 4, 2005 #2

    Tom Mattson

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    If you show how you got your answer then we will point out your mistake.
     
  4. Nov 4, 2005 #3
    [tex] \int {\frac {1} {\sqrt{-x}}} dx = \int {\frac {1} {(-x)^{1/2}} dx = \int {(-x)^{-1/2}} dx = \frac {(-x)^{-1/2+1}} {-1/2+1} = \frac {(-x)^{1/2}} {1/2} = 2 \sqrt {-x} [/tex]
     
    Last edited: Nov 4, 2005
  5. Nov 4, 2005 #4

    VietDao29

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    Nope,
    You are wrong when going from:
    [tex]\int (-x) ^ {-\frac{1}{2}} dx[/tex] to [tex]\frac{(-x) ^ {-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C[/tex].
    The reason is that, you only have:
    [tex]\int x ^ \alpha dx = \frac{x ^ {\alpha + 1}}{\alpha + 1} + C[/tex].
    You do not have:
    [tex]\int (-x) ^ \alpha dx = \frac{(-x) ^ {\alpha + 1}}{\alpha + 1} + C[/tex]
    So, from [tex]\int (-x) ^ {-\frac{1}{2}} dx[/tex], you can use u-substitution:
    Let u = -x, so du = -dx (or you can say dx = -du). So the integral becomes:
    [tex]\int u ^ {-\frac{1}{2}}(-du) = - \int u ^ {-\frac{1}{2}}du = -2 u ^ {\frac{1}{2}} + C[/tex]. Since u = -x, so change u back to x, you'll have:
    [tex]-2\sqrt{-x} + C[/tex]
     
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