How do I calculate a 2d integration with the result being an average?

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SUMMARY

This discussion focuses on calculating a 2D integration to determine the average value of the function f(x, y) = x² + y² over the rectangular domain defined by [-1, 1] for x and [-3, 3] for y. The correct approach involves integrating the function over the specified domain and then dividing the result by the area of the rectangle, which is 6. The average value is thus calculated as the integral result divided by the area, confirming that the average can be represented as the integral divided by the area of the rectangle.

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hexa
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Hi,

I'm trying to calculate a 2d integration with the result being an average.

Let the sum be x^2 + y^2 and the domain [-1 x 1; -3 y 3]
(The actual sum is more complicated but that's too difficult to write here)

so I thought:

1/(3+3) integrate 1/(1+1) integrate x^2 + y^2 dxdy

but my problem is that I'm not quiet sure as to where I have to add the values for the average to the calculation. The 1/6 to the first integration and 1/2 to the second or can I simply say 1/12 times the result after both integrations? I tried both and got different results and I don't have sollutions so I'm not sure which way to go.

Hexa
 
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I have no idea what you mean by "calculate a 2d integration with the result being an average". An average of what? If you mean the average of the function f(x,y) (say, f(x,y)= x2+ y2), then that is the same as the integral of f(x,y) over that rectangle, divided by the area of the rectangle. The point of an average is that it "can be used in place of the values". Here, the integral of a constant over a rectangular region is just the constant times the area- so the constant is the integral divided by the area.
 

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