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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.

Solve the double integral: ∫∫ x(x + y) dy dx over the region [0, 2] x [0, 2]

Learn how to solve the double integral ∫∫ x(x + y) dy dx over the region [0, 2] x [0, 2] with a step-by-step solution. This calculus problem explores concepts of multivariable integration, iterated integrals, and the use of algebraic simplifications.

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