Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: F(3)=1/4 integration

  1. Apr 27, 2005 #1
    f(3)=1/4

    [tex]\int\frac{dy}{y}=\int (6-2x)dx[/tex]
    [tex]-\frac{1}{y}=6x-x^2+C[/tex]
    [tex]-4=18-9+C[/tex]
    [tex]C=-13[/tex]
    [tex]-\frac{1}{y}=6x-x^2-13[/tex]
    [tex]y=-\frac{1}{6x-x^2-13}[/tex]
    the answer is...
    [tex]y=\frac{1}{6x-x^2+13}[/tex]

    why?
     
  2. jcsd
  3. Apr 27, 2005 #2

    James R

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Looks like the answer is wrong.
     
  4. Apr 27, 2005 #3
    I think that the negative was distributed before the Constant was added.. I just dont know why?
     
  5. Apr 27, 2005 #4
    [tex] \int \frac{dy}{y} [/tex] is not [tex] \frac{-1}{y} [/tex]
     
  6. Apr 27, 2005 #5
    sorry, it's y^2

    f(3)=1/4

    [tex]\int\frac{dy}{y^2}=\int (6-2x)dx[/tex]
    [tex]-\frac{1}{y}=6x-x^2+C[/tex]
    [tex]-4=18-9+C[/tex]
    [tex]C=-13[/tex]
    [tex]-\frac{1}{y}=6x-x^2-13[/tex]
    [tex]y=-\frac{1}{6x-x^2-13}[/tex]
    the answer is...
    [tex]y=\frac{1}{6x-x^2+13}[/tex]

    why?
     
  7. Apr 27, 2005 #6
    [tex]-\frac{1}{y}=6x-x^2+C[/tex]

    [tex] y = -\frac{1}{6x-x^2+c} [/tex]

    [tex] -\frac{1}{4} = \frac{1}{9+C} [/tex]

    C = 13
     
  8. Apr 27, 2005 #7
    sub that back in... you get the same equation as I do...
     
  9. Apr 27, 2005 #8
    I'm sorry, C is -13, and yeah, the one you got is correct.
     
  10. Apr 27, 2005 #9
    the book's answer is http://home.earthlink.net/~urban-xrisis/phy001.jpg [Broken]

    I dont see how they got that answer...
     
    Last edited by a moderator: May 2, 2017
  11. Apr 27, 2005 #10
    You wrote the answer in correct in your first post, and unless theres a new math system where

    C + 9 = 4, and C = 18 theyre wrong. C is -13
     
  12. Apr 27, 2005 #11
    I took the answers right off of the college-board website! wow...
     
  13. Apr 27, 2005 #12
    Nobody's infall-yable.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook