1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
  2. jcsd
  3. Nov 4, 2005 #2

    Tom Mattson

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

    User Avatar
    Homework Helper

    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]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Please help (integrals)
Loading...