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!

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 here

    I dont see how they got that answer...
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?