We are asked to find the average of $x^2 y$ over the rectangular region $[0, 2] \times [0, 5]$, where $(X, Y)$ is uniformly distributed. The average or expected value of $x^2 y$ for a uniform distribution is calculated as:

$\mathbb{E}[X^2 Y] = \frac{1}{\text{Area}} \int_0^2 \int_0^5 x^2 y \, dy \, dx$

Step 1: Determine the area of the rectangle

The area of the rectangle is:

$\text{Area} = 2 \times 5 = 10$

Step 2: Set up the double integral

We need to compute the double integral:

$\mathbb{E}[X^2 Y] = \frac{1}{10} \int_0^2 \int_0^5 x^2 y \, dy \, dx$

Step 3: Evaluate the inner integral with respect to $y$

First, we integrate $x^2 y$ with respect to $y$ from 0 to 5:

$\int_0^5 x^2 y \, dy = x^2 \left[ \frac{y^2}{2} \right]_0^5 = x^2 \left( \frac{5^2}{2} - \frac{0^2}{2} \right) = x^2 \cdot \frac{25}{2} = \frac{25}{2} x^2$

Step 4: Evaluate the outer integral with respect to $x$

Now, we integrate the result with respect to $x$ from 0 to 2:

$\int_0^2 \frac{25}{2} x^2 \, dx = \frac{25}{2} \int_0^2 x^2 \, dx = \frac{25}{2} \left[ \frac{x^3}{3} \right]_0^2 = \frac{25}{2} \cdot \frac{8}{3} = \frac{25 \times 8}{6} = \frac{200}{6} = \frac{100}{3}$

Step 5: Divide by the area to find the average

Finally, divide by the area of the rectangle (which is 10):

$\mathbb{E}[X^2 Y] = \frac{1}{10} \times \frac{100}{3} = \frac{100}{30} = \frac{10}{3}$

Final Answer:

The average of $x^2 y$ over the interval $[0, 2] \times [0, 5]$ is $\frac{10}{3}$.

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Would you like more details or clarifications on any part of the solution?

Here are 5 related questions for further practice:

1. What is the expected value of $X^2$ over the interval [0, 2]?
2. What is the expected value of $Y$ over the interval [0, 5]?
3. How would the result change if the region were [0, 3] × [0, 4]?
4. Can you compute the average of $X^3 Y^2$ over the same region?
5. What is the expected value of $XY$ over the rectangle [0, 2] × [0, 5]?

Tip: For double integrals, it's useful to evaluate the inner integral first before working on the outer one to simplify the overall computation.