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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

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    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

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    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]
     
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