The expression in the image appears to be:

$\int_0^2 \int_0^4 (6xy - 2y) \, dx \, dy$

This represents a double integral, where the region of integration is a rectangle with limits $0 \leq x \leq 4$ and $0 \leq y \leq 2$, and the function being integrated is $6xy - 2y$.

Let's solve this step by step.

Step 1: Integrate with respect to $x$

The inner integral is:

$\int_0^4 (6xy - 2y) \, dx$

Treating $y$ as a constant, we can integrate term by term:

[ \int_0^4 6xy , dx = 6y \int_0^4 x , dx = 6y \left[\frac{x^2}{2}\right]_0^4 = 6y \left[\frac{16}{2}\right] = 48y ]

$\int_0^4 -2y \, dx = -2y \int_0^4 1 \, dx = -2y [x]_0^4 = -2y(4 - 0) = -8y$

So the result of the inner integral is:

$48y - 8y = 40y$

Step 2: Integrate with respect to $y$

Now, we integrate the result from the first step with respect to $y$:

$\int_0^2 40y \, dy = 40 \int_0^2 y \, dy = 40 \left[\frac{y^2}{2}\right]_0^2 = 40 \times \frac{4}{2} = 80$

Final Result:

The value of the double integral is $80$.

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

Here are five related questions:

1. How do you determine the limits of integration in a double integral?
2. What is the significance of changing the order of integration in double integrals?
3. Can you evaluate a double integral over non-rectangular regions?
4. What happens if the function within the integral is not separable?
5. How would the result change if the integrand were $6xy^2 - 2y$?

Tip: Always check if the function inside the integral allows for simplifications or factorization, as it can make the computation easier.