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

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In summary, the student attempted to solve an integral by using the chain rule and the log properties, but then got confused. They finally solved it by treating (x+1) as a constant and using the following equation: \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.
  • #1
jonroberts74
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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
 
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  • #2
jonroberts74 said:

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
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:
  • #4
jonroberts74 said:
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]
 

1. What is the definition of an integral?

An integral is a mathematical concept that represents the area under a curve or the accumulation of a function over a certain interval. It is used to solve problems related to finding the total change, total distance, and total accumulated value of a function.

2. What does the notation "R=[0,1]x[0,1]" mean?

The notation "R=[0,1]x[0,1]" refers to a rectangular region in the Cartesian coordinate system. The region is bounded by the x-values 0 and 1, and the y-values 0 and 1. This notation is commonly used to define the limits of integration for a double integral.

3. How is a double integral evaluated?

A double integral is evaluated by first determining the limits of integration based on the given region. Then, the function to be integrated is multiplied by the infinitesimal area element, which is represented by "dA". The resulting expression is then solved using integration techniques, such as the Riemann sum or the substitution method.

4. What is the purpose of evaluating an integral over a given region?

Evaluating an integral over a given region allows us to find the total accumulation or area under a curve within that region. This can be useful in various fields of science, such as physics, engineering, and economics, to solve real-world problems related to rates of change, volumes, and probabilities.

5. Can the limits of integration for an integral change?

Yes, the limits of integration for an integral can change depending on the given problem. They can be defined by specific values or by variables, and they can also be different for each variable in a double integral. It is important to carefully define the limits of integration to accurately evaluate the integral.

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