The given problem asks to evaluate the double integral:

$\iint_{R} (4y - 1) \, dA$

where $R = [0, 5] \times [0, 4]$, meaning that $x$ ranges from 0 to 5 and $y$ ranges from 0 to 4.

We can solve this step-by-step.

Step 1: Set up the integral

The integral can be written as:

$\int_0^5 \int_0^4 (4y - 1) \, dy \, dx$

Step 2: Evaluate the inner integral

We first integrate with respect to $y$:

$\int_0^4 (4y - 1) \, dy = \left[ 2y^2 - y \right]_0^4$

Now, evaluate this at the bounds $y = 4$ and $y = 0$:

$\left( 2(4)^2 - (4) \right) - \left( 2(0)^2 - (0) \right) = (2 \cdot 16 - 4) = 32 - 4 = 28$

So, the result of the inner integral is 28.

Step 3: Evaluate the outer integral

Now we integrate with respect to $x$:

$\int_0^5 28 \, dx = 28x \big|_0^5 = 28(5) - 28(0) = 140$

Final Answer:

The value of the double integral is:

$\boxed{140}$

Would you like further details or have any questions?

Related Questions:

1. What is the geometrical interpretation of a double integral?
2. How would the answer change if the function was different, such as $(2y - 3)$?
3. Can you evaluate the same integral over a different region, such as $R = [1, 4] \times [0, 3]$?
4. How does changing the limits of integration affect the solution?
5. Can you perform the integration using polar coordinates for a circular region?

Tip:

When solving double integrals, always make sure to carefully choose the correct order of integration and verify the bounds for each variable.