For which intervals is the function positive?
1 answer:
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The average squared distance between the points is their variance
The average squared distance is 8/3
<h3>How to determine the average squared distance?</h3>The given parameters are:
R={(x,y): 0<=x<=2, 0<=y<=2}
The point = (2,2).
f(x,y) = (x-2)² + (y-2)²
The squared distance is calculated as:
D² = f(x,y) = (x-2)² + (y-2)²
Where the area (A) is:
A = xy
Substitute the maximum values of x and y in A = xy
A = 2 * 2
A = 4
The minimum values of x and y are 0.
So, the limits for the integrals are 0 to 2
The integral becomes
Expand
Evaluate the inner integral with respect to x from 0 to 2
Expand
Simplify
Expand
Evaluate the integral with respect to y from 0 to 2
Expand
D² = 32/3
The average squared distance (AD²) is calculated as:
AD² = D²/A
So, we have:
AD² = 32/3 4
Evaluate the quotient
AD² = 32/12
Simplify
AD² = 8/3
Hence, the average squared distance is 8/3
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