Final answer:
The average value of the function f(x, y) = y^3 over the given triangle is 22.5.
Step-by-step explanation:
To find the average value of the function f(x, y) = y^3 over the given triangle, we need to evaluate the integral of f(x, y) over the triangle and divide it by the area of the triangle.
First, let's find the area of the triangle using the formula:
A = 1/2 * base * height = 1/2 * |(0 - 3)(1 - 2) - (1 - 3)(0 - 2)| = 1/2 * |-3 - (-2)| = 1/2 * |-1| = 1/2.
Next, let's evaluate the integral of f(x, y) over the triangle:
I = ∫∫(T) y^3 dA.
Since the triangle is defined by (0, 2), (1, 1), and (3, 2), we have:
I = ∫∫(T) y^3 dA = ∫∫(T) y^3 dx dy,
where the limits of integration are: 0 ≤ x ≤ 3 and 1 ≤ y ≤ 2.
Using the properties of double integrals, we can write the integral as:
I = ∫(1 to 2) ∫(0 to 3) y^3 dx dy.
Now, let's evaluate the integral:
I = ∫(1 to 2) y^3x |(0 to 3) dy = ∫(1 to 2) y^3(3 - 0) dy = 3∫(1 to 2) y^3 dy = 3 * [1/4 * y^4] (1 to 2) = 3 * [1/4 * 2^4 - 1/4 * 1^4] = 3 * [1/4 * 16 - 1/4] = 3 * [4 - 1/4] = 3 * [16/4 - 1/4] = 3 * [15/4] = 45/4.
Finally, let's find the average value of f(x, y) over the triangle by dividing the integral by the area:
Average value = I / A = (45/4) / (1/2) = (45/4) * (2/1) = 90/4 = 45/2 = 22.5.
Therefore, the average value of the function f(x, y) = y^3 over the given triangle is 22.5.