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

[jonroberts74]

Homework Statement

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

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

The Attempt at a Solution

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

by parts

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

\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

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


\frac{1}{2}ln(2) - ln(2) - ln^2(2)

not sure what I did wrong but I know that's the incorrect answer

 

[SammyS]

Homework Statement

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

The Attempt at a Solution

\int_{0}^{1}\int_{0}^{1} ln[(x+1)(y+1)] dydx
by partsu=ln[(x+1)(y+1)]; du = \frac{1}{(x+1)(y+1)}dy;dv=dy;v=y
\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
\int_{0}^{1} \Bigg[\frac{1}{2(x+1)} - \frac{1}{x+1} - \frac{ln(2)}{(x+1)}\Bigg]dx
\frac{1}{2}ln(2) - ln(2) - ln^2(2)
not sure what I did wrong but I know that's the incorrect answer

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

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)\$

 

[jonroberts74]

so I could do

\int_{0}^{1}ln(x+1)dx+\int_{0}^{1}ln(y+1)dy

forgot about the log properties for a second

or even

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

2[(x+1)ln(x+1)-(x+1)]\Bigg|_{0}^{1} = 4ln(2)-2 = ln(16)-2

 

[SammyS]

so I could do

\int_{0}^{1}ln(x+1)dx+\int_{0}^{1}ln(y+1)dy

forgot about the log properties for a second

or even

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

2[(x+1)ln(x+1)-(x+1)]\Bigg|_{0}^{1} = 4ln(2)-2 = ln(16)-2

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.

\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