The given problem is a double integral of the function $\frac{1}{4} x^3 y$ over the region defined by the limits $x \in [0, 4]$ and $y \in [1, 6]$. The integral is written as:

$\int_1^6 \left( \int_0^4 \frac{1}{4} x^3 y \, dx \right) dy$

Step 1: Solve the inner integral with respect to $x$.

We treat $y$ as a constant in the inner integral.

$\int_0^4 \frac{1}{4} x^3 y \, dx$

First, factor out constants:

$\frac{y}{4} \int_0^4 x^3 \, dx$

Now solve the integral $\int_0^4 x^3 \, dx$:

$\int_0^4 x^3 \, dx = \left[ \frac{x^4}{4} \right]_0^4 = \frac{4^4}{4} - 0 = \frac{256}{4} = 64$

Thus, the inner integral becomes:

$\frac{y}{4} \times 64 = 16y$

Step 2: Solve the outer integral with respect to $y$.

Now, substitute $16y$ into the outer integral:

$\int_1^6 16y \, dy$

Solve this:

$16 \int_1^6 y \, dy = 16 \left[ \frac{y^2}{2} \right]_1^6 = 16 \left( \frac{6^2}{2} - \frac{1^2}{2} \right)$

$= 16 \left( \frac{36}{2} - \frac{1}{2} \right) = 16 \times \frac{35}{2} = 16 \times 17.5 = 280$

Final Answer:

The value of the double integral is $280$.

Let me know if you would like more details or further clarification.

Here are 5 related questions to expand your understanding:

1. How would you compute the integral if the limits of $x$ were different?
2. What changes if the function inside the integral is $x^2y$ instead of $x^3y$?
3. How does the result change if the limits of $y$ are from 0 to 4 instead of 1 to 6?
4. What are the properties of double integrals over rectangular regions?
5. Can you explain how Fubini's theorem is applied to this double integral?

Tip: When solving iterated integrals, always handle the inner integral first and treat all outer variables as constants.