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Integral Question

  1. Feb 5, 2008 #1
    Need to solve this using substitution:

    [tex]\int[/tex][tex]^{0}[/tex] y[tex]^{2}[/tex] (1- [tex]\frac{y^{3}}{a^{2}}[/tex])[tex]^{-2}[/tex] dy
    [tex]_{-a}[/tex]

    This is my first post so I'm not that good with the LaTeX code yet, but I hope you can read it correctly. I was thinking about substituting u for y^3. What do you guys think? Thanks
     
  2. jcsd
  3. Feb 5, 2008 #2

    rock.freak667

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    [tex]\int_{-a} ^{0} y^2(1-\frac{y^3}{a^2})^{-2} dy[/tex]


    instead of u=y^3 why not let u=1-[itex]\frac{y^3}{a^2}[/itex]?
     
  4. Feb 5, 2008 #3
    You set it up right.

    Hmm... If y^3 is used, can't that be used later to cancel out y^2?
     
  5. Feb 5, 2008 #4

    rock.freak667

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    If you let u=y^3 the y^2 will cancel out but you will be left with

    [tex]\int \frac{1}{3}(1-\frac{u}{a^2})^{-2} du[/tex]
     
  6. Feb 5, 2008 #5
    Yea that's what I got, so the y's are out. The 1/3 is constant so that comes out. The tricky part here is the ^-2. Ideas?
     
  7. Feb 6, 2008 #6
    but if you use [tex] 1-\frac{y^3}{a^2} [/tex] like suggested, you get

    [tex] du = - \frac{y^2}{3 a^2} dx[/tex]

    so

    [tex] \int y^2 (1-\frac{y^3}{a^2}) dx =- 3 a^2 \int \frac{1}{u^2} du[/tex]

    this is a bit easier
     
  8. Feb 6, 2008 #7
    That works too, but you guys are forgetting the -a to 0 part.
     
  9. Feb 6, 2008 #8

    rock.freak667

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    Putting back the 0 to -a with a u substitution is wrong, since 0 and -a are the limits for y not u.

    If the OP wanted to put back limits,this would have to be done

    [tex]u=1-\frac{y^3}{a^2}[/tex]

    y=0;u=1
    y=-a;u=1+a

    so that

    [tex]\int_{-a} ^{0} y^2(1-\frac{y^3}{a^2})^{-2} dy \equiv \frac{-a^2}{3} \int_{1+a} ^{1} u^{-2} du[/tex]
     
  10. Feb 6, 2008 #9
    yes, always remeber the new boundaries ;-)
     
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