Question

Find the average squared distance between the points of r{(x,y): 0x, 0y} and the point (,)

Answer 1

The average squared distance between the points is their variance

The average squared distance is 8/3

How to determine the average squared distance?

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

$D\² = \int \int (x-2)\² + (y-2)\² dx dy$

Expand

$D\² = \int \int x\² -4x + 4 + (y-2)\² dx dy$

Evaluate the inner integral with respect to x from 0 to 2

$D\² = \int \frac 13x^3 -2x^2 + 4x + x(y-2)\² |\limits^2_0 dy$

Expand

$D\² = \int [\frac 13(2)^3 -2(2)^2 + 4(2) + (2)(y-2)\²] dy$

Simplify

$D\² = \int \frac 83 + (2)(y-2)\² dy$

Expand

$D\² = \int \frac 83 + 2y^2-8y + 8 \ dy$

Evaluate the integral with respect to y from 0 to 2

$D\² = [\frac 83y + \frac 23y^3- 4y^2 + 8y ]|\limits^2_0$

Expand

$D\² = \frac 83*2 + \frac 23*2^3- 4*2^2 + 8*2$

D² = 32/3

The average squared distance (AD²) is calculated as:

AD² = D²/A

So, we have:

AD² = 32/3 $\div$ 4

Evaluate the quotient

AD² = 32/12

Simplify

AD² = 8/3

Hence, the average squared distance is 8/3

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