The given problem is a double integral, which appears to be of the form:

$\int_0^2 \int_0^2 x(x + y) \, dy \, dx$

Let's solve this step-by-step.

Step 1: Integrate with respect to $y$

The inner integral is: $\int_0^2 x(x + y) \, dy$

Since $x$ is a constant with respect to $y$, we can distribute $x$ inside: $\int_0^2 (x^2 + xy) \, dy$

Now, integrate term by term: $\int_0^2 x^2 \, dy + \int_0^2 xy \, dy$

Step 2: Evaluate the inner integrals

• For the first term: $\int_0^2 x^2 \, dy = x^2 \int_0^2 \, dy = x^2 [y]_0^2 = x^2 (2 - 0) = 2x^2$

• For the second term: $\int_0^2 xy \, dy = x \int_0^2 y \, dy = x \left[ \frac{y^2}{2} \right]_0^2 = x \cdot \frac{2^2}{2} = 2x$

So, the result of the inner integral is: $\int_0^2 x(x + y) \, dy = 2x^2 + 2x$

Step 3: Integrate with respect to $x$

Now, evaluate the outer integral: $\int_0^2 (2x^2 + 2x) \, dx$

Integrate term by term: $\int_0^2 2x^2 \, dx + \int_0^2 2x \, dx$

Step 4: Evaluate the outer integrals

• For the first term: $\int_0^2 2x^2 \, dx = 2 \int_0^2 x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_0^2 = 2 \cdot \frac{2^3}{3} = \frac{16}{3}$

• For the second term: $\int_0^2 2x \, dx = 2 \left[ \frac{x^2}{2} \right]_0^2 = 2 \cdot \frac{2^2}{2} = 4$

Step 5: Combine the results

The final result of the double integral is: $\int_0^2 \int_0^2 x(x + y) \, dy \, dx = \frac{16}{3} + 4 = \frac{28}{3}$

So, the solution to the integral is $\frac{28}{3}$.

Would you like more details or have any questions?

Here are 5 related questions you might find interesting:

1. How does the order of integration affect the result of a double integral?
2. What is the geometric interpretation of a double integral over a region?
3. How do you evaluate double integrals over non-rectangular regions?
4. Can double integrals be used to find the volume under a surface?
5. How does changing the limits of integration affect the integral result?

Tip: When solving double integrals, simplifying the inner integral first can make the problem much more manageable.