The problem is asking to evaluate a double integral:

$\int_{-2}^{2} \int_{x^2}^{2 - x} y \, dx \, dy$

Steps to solve:

1. Set up the double integral: The integral is written as: $\int_{-2}^{2} \left( \int_{x^2}^{2 - x} y \, dy \right) dx$

2. Solve the inner integral (involving $y$): The inner integral is with respect to $y$, so we treat $x$ as constant: $\int_{x^2}^{2 - x} y \, dy = \left[ \frac{y^2}{2} \right]_{x^2}^{2 - x}$ Plug in the limits: $\frac{(2 - x)^2}{2} - \frac{x^4}{2}$

3. Simplify: Expand and simplify the expression: $\frac{(2 - x)^2}{2} = \frac{4 - 4x + x^2}{2}$ Therefore, the result of the inner integral is: $\frac{4 - 4x + x^2 - x^4}{2}$

4. Set up the outer integral: Now integrate this expression with respect to $x$: $\int_{-2}^{2} \frac{4 - 4x + x^2 - x^4}{2} \, dx$

5. Split the integral: Split the integral into parts: $\frac{1}{2} \left( \int_{-2}^{2} 4 \, dx - \int_{-2}^{2} 4x \, dx + \int_{-2}^{2} x^2 \, dx - \int_{-2}^{2} x^4 \, dx \right)$

   Now calculate each part.

6. Evaluate the integrals:

   • $\int_{-2}^{2} 4 \, dx = 4 \times (2 - (-2)) = 16$
   • $\int_{-2}^{2} 4x \, dx = 4 \times \left[ \frac{x^2}{2} \right]_{-2}^{2} = 0$ (since it's an odd function over a symmetric interval)
   • $\int_{-2}^{2} x^2 \, dx = 2 \times \left[ \frac{x^3}{3} \right]_{0}^{2} = 2 \times \frac{8}{3} = \frac{16}{3}$
   • $\int_{-2}^{2} x^4 \, dx = 2 \times \left[ \frac{x^5}{5} \right]_{0}^{2} = 2 \times \frac{32}{5} = \frac{64}{5}$

7. Combine the results: The total integral is: $\frac{1}{2} \left( 16 - 0 + \frac{16}{3} - \frac{64}{5} \right)$

   Simplify this expression to get the final result.

I'll calculate this now.The result of the evaluated double integral is approximately $4.2667$.

Converting this into a fraction:

$\frac{64}{15}$

None of the provided options matches exactly with this result, so it seems there may be a mistake in the setup or an issue with the problem constraints. Would you like to go over any part of the solution in more detail or discuss any adjustments?

Here are five related questions you might explore:

1. How do you determine the correct bounds for a double integral in a different scenario?
2. What happens if you reverse the order of integration?
3. How do you compute double integrals over non-rectangular regions?
4. What are some techniques for approximating double integrals if exact calculation is difficult?
5. Can you solve this problem using polar coordinates?

Tip: When solving double integrals, always double-check the limits of integration to ensure they match the problem's region.