Evaluate integral if R=[0,1]x[0,1]

  • #1
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Homework Statement


evaluate integral if R=[0,1]x[0,1]

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





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
 

Answers and Replies

  • #2
SammyS
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Homework Statement


evaluate integral if R=[0,1]x[0,1][tex]\iint_R ln[(x+1)(y+1)]dA[/tex]

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 that's the incorrect answer
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)\ ##
 
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  • #3
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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]
 
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  • #4
SammyS
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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]
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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