1. Limited time only! Sign up for a free 30min personal 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!

Homework Help: Evaluate integral if R=[0,1]x[0,1]

  1. Jul 24, 2014 #1
    1. The problem statement, all variables and given/known data
    evaluate integral if R=[0,1]x[0,1]

    [tex]\iint_R ln[(x+1)(y+1)]dA[/tex]





    3. The attempt at a solution

    [tex]\int_{0}^{1}\int_{0}^{1} ln[(x+1)(y+1)] dydx[/tex]

    by parts

    [tex]u=ln[(x+1)(y+1)]; du = \frac{1}{(x+1)(y+1)}dy;dv=dy;v=y[/tex]

    [tex]\int_{0}^{1} \Bigg[\frac{y}{(x+1)(y+1)} - \frac{1}{x+1}\int_{0}^{1} 1 - \frac{1}{y+1}dy\Bigg]dx[/tex]

    [tex]\int_{0}^{1} \Bigg[\frac{1}{2(x+1)} - \frac{1}{x+1} - \frac{ln(2)}{(x+1)}\Bigg]dx[/tex]


    [tex] \frac{1}{2}ln(2) - ln(2) - ln^2(2)[/tex]

    not sure what I did wrong but I know thats the incorrect answer
     
  2. jcsd
  3. Jul 24, 2014 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    An error was made when you stated [tex]u=ln[(x+1)(y+1)];\ \ du = \frac{1}{(x+1)(y+1)}dy\ .[/tex]

    Treat (x+1) as a constant (with respect to u). Then: Either use the chain rule, or write ##\ \ln[(x+1)(y+1)]\ ## as ##\ \ln(x+1)+\ln(y+1)\ ##
     
  4. Jul 24, 2014 #3
    so I could do

    [tex]\int_{0}^{1}ln(x+1)dx+\int_{0}^{1}ln(y+1)dy[/tex]

    forgot about the log properties for a second

    or even

    [tex]2\int_{0}^{1}ln(x+1)dx[/tex] or the y counterpart

    [tex]2[(x+1)ln(x+1)-(x+1)]\Bigg|_{0}^{1} = 4ln(2)-2 = ln(16)-2[/tex]
     
    Last edited: Jul 24, 2014
  5. Jul 24, 2014 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You can't generally do that sort of split for a double integral (or any iterated integral for that matter).

    However, the following is true.

    [itex]\displaystyle \int_{0}^{1}\int_{0}^{1} ln[(x+1)(y+1)]\, dydx=\int_{0}^{1}\int_{0}^{1} ln(x+1)\, dydx+\int_{0}^{1}\int_{0}^{1} ln(y+1)\, dydx[/itex]
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted