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The integral you've provided is a double integral of the function \( 2xy + 2y^2 \) over the region defined by the limits:

\[
\int_1^2 \left( \int_x^2 (2xy + 2y^2) \, dy \right) dx
\]

### Step 1: Evaluate the inner integral
First, we'll evaluate the inner integral with respect to \( y \):

\[
\int_x^2 (2xy + 2y^2) \, dy
\]

We'll integrate term by term:

- For the term \( 2xy \):
  \[
  \int 2xy \, dy = 2x \int y \, dy = 2x \cdot \frac{y^2}{2} = x y^2
  \]

- For the term \( 2y^2 \):
  \[
  \int 2y^2 \, dy = 2 \cdot \frac{y^3}{3} = \frac{2y^3}{3}
  \]

So, the result of the inner integral is:

\[
\left. x y^2 + \frac{2y^3}{3} \right|_{y=x}^{y=2}
\]

Substituting the limits \( y = 2 \) and \( y = x \):

- For \( y = 2 \):
  \[
  x(2^2) + \frac{2(2^3)}{3} = 4x + \frac{16}{3}
  \]

- For \( y = x \):
  \[
  x(x^2) + \frac{2x^3}{3} = x^3 + \frac{2x^3}{3} = \frac{3x^3}{3} + \frac{2x^3}{3} = \frac{5x^3}{3}
  \]

Thus, the inner integral result is:

\[
\left( 4x + \frac{16}{3} \right) - \frac{5x^3}{3} = 4x - \frac{5x^3}{3} + \frac{16}{3}
\]

### Step 2: Evaluate the outer integral
Now, we'll evaluate the outer integral with respect to \( x \):

\[
\int_1^2 \left( 4x - \frac{5x^3}{3} + \frac{16}{3} \right) dx
\]

We integrate term by term:

- For \( 4x \):
  \[
  \int 4x \, dx = 4 \cdot \frac{x^2}{2} = 2x^2
  \]

- For \( -\frac{5x^3}{3} \):
  \[
  \int -\frac{5x^3}{3} \, dx = -\frac{5}{3} \cdot \frac{x^4}{4} = -\frac{5x^4}{12}
  \]

- For \( \frac{16}{3} \):
  \[
  \int \frac{16}{3} \, dx = \frac{16}{3}x
  \]

Now, substituting the limits \( x = 1 \) and \( x = 2 \) into the final expression:

\[
\left( 2(2^2) - \frac{5(2^4)}{12} + \frac{32}{3} \right) - \left( 2(1^2) - \frac{5(1^4)}{12} + \frac{16}{3} \right)
\]

Let's calculate each part:

- For \( x = 2 \):
  \[
  2(4) - \frac{5(16)}{12} + \frac{32}{3} = 8 - \frac{80}{12} + \frac{32}{3} = 8 - \frac{20}{3} + \frac{32}{3} = 8 + \frac{12}{3} = 12
  \]

- For \( x = 1 \):
  \[
  2(1) - \frac{5}{12} + \frac{16}{3} = 2 - \frac{5}{12} + \frac{16}{3} = 2 + \frac{59}{12} = 6.9167
  \]

So, the final value of the double integral is:

\[
12 - 6.9167 = 5.0833
\]

This is the final result.

Would you like any further details or explanations, or have any questions?

### Related Questions:
1. How do you handle double integrals with more complex functions?
2. What are the applications of double integrals in physics and engineering?
3. How would the solution differ if the limits of integration were reversed?
4. What techniques can simplify solving double integrals?
5. Can double integrals always be computed analytically?

### Tip:
When solving double integrals, especially with variable limits, always ensure you properly handle the bounds and keep the expressions organized to minimize errors.

Learn how to solve the double integral of 2xy + 2y^2 with detailed step-by-step calculations and evaluations. This problem covers key concepts in calculus, including double integrals and the fundamental theorem of calculus. Suitable for college-level mathematics students.

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