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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.

Evaluate the iterated double integral: ∫_1^6 ∫_0^4 (1/4)x^3 y dx dy

Learn how to evaluate the double integral ∫_1^6 ∫_0^4 (1/4)x^3 y dx dy, including detailed steps for solving the inner and outer integrals, and applying Fubini's theorem.

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