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!

Antidifferentiation for a larger problem

  1. Feb 21, 2007 #1

    nealh149

    User Avatar

    1. The problem statement, all variables and given/known data
    I'm trying to antidifferentiate a function in order to find the volume of a solid. The function comes out to be (2x+1)/(x^2+8x+16)


    2. Relevant equations

    N/A

    3. The attempt at a solution
    I've tried a multitude of things. I figured that partial fractions would work, but it turns out that since the denominator is a square that doesn't work. I'm not asking you to solve, just a nudge in the right direction
     
    nealh149, Feb 21, 2007
  2. jcsd
    Phys.org - latest science and technology news stories on Phys.org
  3. Feb 21, 2007 #2

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    What do you mean by this? Try the partial fraction decomposition [tex]\frac{A}{(x+4)}+\frac{B}{(x+4)^2}[/tex]
     
    cristo, Feb 21, 2007
  4. Feb 21, 2007 #3

    nealh149

    User Avatar

    ALright. I don't understand though. Where did the third (x+4) come from. The denominator only factors to (x+4)(x+4)
     
    nealh149, Feb 21, 2007
  5. Feb 21, 2007 #4

    nealh149

    User Avatar

    Okay even when i try that I get

    A(x+4)^2+B(x+4) = 2x+1

    When I plug in to eliminate one of the variables (x=-4) then I get 0+0 = -7. I'm probably doing something wrong.
     
    nealh149, Feb 21, 2007
  6. Feb 21, 2007 #5

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    We know that the number of unknown constants must be equal to the degree of the denominator, and so the easiest way to set it up is to take the two fractions as A/(x+4) and B/(x+4)2

    This is not correct: You should have the following expression: [tex]\frac{2x+1}{(x+4)^2}\equiv \frac{A}{(x+4)}+\frac{B}{(x+4)^2}[/tex] Multiplying this expression by (x+4)2 does not give what you have above.
     
    cristo, Feb 21, 2007
  7. Feb 21, 2007 #6

    nealh149

    User Avatar

    Ok. I'll try that. Why did you use those denominators though, for the partials. Intuitively I would use x+4 and x+4 because they multiply together to get x^2 +8x+16

    Thanks though
     
    nealh149, Feb 21, 2007
  8. Feb 21, 2007 #7

    nealh149

    User Avatar

    Alright. So I solve that equation and get 2x+1 = A(x+4) + B

    A = 1 and B = -7

    WHen I antidifferentiate I get ln(x+4) + 7/(x+4) Seems that I'm missing a factor of 2 in front of the first term. Sorry for the problems.
     
    nealh149, Feb 21, 2007
  9. Feb 21, 2007 #8

    nealh149

    User Avatar

    nevermind on that last part, i found a mistake. but the previous question is still troubling me
     
    nealh149, Feb 21, 2007
  10. Feb 21, 2007 #9

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Well, if we use (x+4) as the only denominator we obtain something like [tex]\frac{2x+1}{(x+4)^2}\equiv \frac{A}{x+4}+\frac{B}{x+4}=\frac{A+B}{x+4}[/tex]; notice that this can never be in the required form (since the denominator will always be linear). Thus, we take the two fractions as (x+4) and (x+4)2, so that the fraction looks like[tex]\frac{A}{x+4}+\frac{B}{(x+4)^2}=\frac{A(x+4)+B}{(x+4)^2}[/tex] which is of the required form.
     
    cristo, Feb 21, 2007
  11. Feb 21, 2007 #10

    nealh149

    User Avatar

    Alright that makes sense. I'm trying to solve that equation now. I can solve for B=-7, but I can't solve for A because I have two variables: x and A.
     
    nealh149, Feb 21, 2007
  12. Feb 21, 2007 #11

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    Just put, say, x=0.
     
    cristo, Feb 21, 2007
  13. Feb 21, 2007 #12

    nealh149

    User Avatar

    Oh nvm. I got it. I have the problem solved now. I really appreciate the help cristo, you've enhanced my understanding of partial fractions.
     
    nealh149, Feb 21, 2007
  14. Feb 21, 2007 #13

    cristo

    User Avatar
    Staff Emeritus
    Science Advisor

    You're very welcome; glad to be of help!
     
    cristo, Feb 21, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Antidifferentiation for a larger problem

  1. Algebra Question(Part of a Larger Problem) (Replies: 4)

  2. Antidifferentiation integral (Replies: 11)

  3. A simple antidifferentiation question (Replies: 7)

  4. Does the denominator become larger faster (Replies: 2)

  5. Constructing a Larger Collection of Functions (Replies: 4)

Loading...