How do you find the average value of a function of two variables on a rectangle?

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Homework Help Overview

The discussion revolves around finding the average value of a function of two variables defined on a rectangular domain. Participants explore the integration process and its implications for the average value calculation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss integrating a function of two variables over a rectangle and the method of dividing by the area. Questions arise regarding the behavior of the integrand when integrating with respect to different variables.

Discussion Status

The discussion is active, with participants sharing their integration attempts and questioning the results. Some guidance is offered regarding the expected outcomes of the integration process, particularly concerning the integrand's behavior.

Contextual Notes

There is mention of specific limits for integration and the nature of the integrand, which leads to confusion about the results when integrating in different orders. The participants are navigating through these assumptions and interpretations.

Master J
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To find the average value of a function of one variable, you integrate it and multiply it by 1/(b - a) , where b and a are the limits of integration , uper and lower, respectively.

But how does one do this for a function of two variables, defined on a rectangle?
 
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The same way, basically. Integrate the function over the rectangle and divide by the area of the rectangle.
 
Thanks.

Well I am trying a question. The integrand is (x^2)y. The domain is [2,4] * [-6,6].

Now when I integrate it with respect to y first, it goes to zero. It does not go to zero when you start with x. How is this possible?
 
Master J said:
How is this possible?

Bad math. :wink:...It should be zero in both cases, try showing us what you are doing when you integrate it over x first...:smile:
 
Possibly!:biggrin:

Ah I see what I've been doing wrong! I don't even need to work out the average value, since it integrates to zero, as it should, since it is evenly distributed in the positve and negative y areas!
 
Last edited:

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