Please help (integrals)

  • Thread starter LinkMage
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  • #1
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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:

Answers and Replies

  • #2
Tom Mattson
Staff Emeritus
Science Advisor
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If you show how you got your answer then we will point out your mistake.
 
  • #3
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[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:
  • #4
VietDao29
Homework Helper
1,423
3
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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