To solve this integral by interchanging the order of integration, we begin by looking at the given integral:

$\int_1^4 \int_2^5 x y^2 \, dy \, dx$

Step 1: Identify the bounds and function

• Outer integral (with respect to $x$): $x$ ranges from 1 to 4.
• Inner integral (with respect to $y$): $y$ ranges from 2 to 5.

The function to integrate is $x y^2$.

Step 2: Interchange the order of integration

When interchanging the order of integration, we switch the roles of $x$ and $y$. Therefore:

• $y$ will now range from 2 to 5.
• $x$ will now range from 1 to 4.

The new integral is: $\int_2^5 \int_1^4 x y^2 \, dx \, dy$

Step 3: Perform the inner integral (with respect to $x$)

The inner integral becomes: $\int_1^4 x \, dx = \left[ \frac{x^2}{2} \right]_1^4 = \frac{4^2}{2} - \frac{1^2}{2} = \frac{16}{2} - \frac{1}{2} = \frac{15}{2}$

Step 4: Perform the outer integral (with respect to $y$)

Now, we have: $\int_2^5 \frac{15}{2} y^2 \, dy = \frac{15}{2} \int_2^5 y^2 \, dy$

The integral of $y^2$ is: $\int_2^5 y^2 \, dy = \left[ \frac{y^3}{3} \right]_2^5 = \frac{5^3}{3} - \frac{2^3}{3} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39$

Step 5: Multiply the results

Now, multiplying the result by $\frac{15}{2}$: $\frac{15}{2} \times 39 = \frac{15 \times 39}{2} = \frac{585}{2} = 292.5$

Final Answer:

The value of the integral is: $292.5$

Would you like more details on any part of this calculation?

Here are 5 related questions for further exploration:

1. How does changing the order of integration affect the outcome of double integrals?
2. What are the conditions for interchanging the order of integration in a double integral?
3. How would you compute the same integral using polar coordinates?
4. Can you explain the geometric interpretation of double integrals?
5. What happens if the bounds of integration are not constants but functions of each other?

Tip: When interchanging the order of integration, always ensure that the bounds correspond correctly to the new order.