To evaluate the double integral:

$\int_{1}^{2} \int_{0}^{3} \left( x + 3x^{\frac{1}{2}} y \cdot \frac{1}{3} \right) \, dx \, dy$

Step 1: Simplify the integrand

First, let's simplify the integrand:

$x + 3x^{\frac{1}{2}} y \cdot \frac{1}{3} = x + x^{\frac{1}{2}} y$

So, the double integral becomes:

$\int_{1}^{2} \int_{0}^{3} \left( x + x^{\frac{1}{2}} y \right) \, dx \, dy$

Step 2: Integrate with respect to $x$

We now integrate the inner integral with respect to $x$, while treating $y$ as a constant.

$\int_{0}^{3} \left( x + x^{\frac{1}{2}} y \right) \, dx$

This splits into two integrals:

$\int_{0}^{3} x \, dx + \int_{0}^{3} x^{\frac{1}{2}} y \, dx$

1. For the first integral, $\int_{0}^{3} x \, dx$:

$\int x \, dx = \frac{x^2}{2} \quad \text{so} \quad \left[\frac{x^2}{2}\right]_{0}^{3} = \frac{9}{2}$

2. For the second integral, $\int_{0}^{3} x^{\frac{1}{2}} y \, dx$:

Since $y$ is treated as a constant, we can factor it out:

$y \int_{0}^{3} x^{\frac{1}{2}} \, dx = y \cdot \left[\frac{2}{3} x^{\frac{3}{2}}\right]_{0}^{3} = y \cdot \frac{2}{3} \cdot \left(3^{\frac{3}{2}} - 0\right)$

Since $3^{\frac{3}{2}} = 3 \sqrt{3}$, this becomes:

$y \cdot \frac{2}{3} \cdot 3\sqrt{3} = 2y \sqrt{3}$

Step 3: Combine results of the inner integral

Now, combining the results of both integrals:

$\int_{0}^{3} \left( x + x^{\frac{1}{2}} y \right) \, dx = \frac{9}{2} + 2y \sqrt{3}$

Step 4: Integrate with respect to $y$

Now we integrate this with respect to $y$ from 1 to 2:

$\int_{1}^{2} \left( \frac{9}{2} + 2y \sqrt{3} \right) \, dy$

This splits into two integrals:

$\int_{1}^{2} \frac{9}{2} \, dy + \int_{1}^{2} 2y \sqrt{3} \, dy$

1. For the first integral, $\int_{1}^{2} \frac{9}{2} \, dy$:

$\frac{9}{2} \cdot (2 - 1) = \frac{9}{2}$

2. For the second integral, $\int_{1}^{2} 2y \sqrt{3} \, dy$:

Since $\sqrt{3}$ is constant, we have:

$2\sqrt{3} \int_{1}^{2} y \, dy = 2\sqrt{3} \cdot \left[\frac{y^2}{2}\right]_{1}^{2} = 2\sqrt{3} \cdot \left( \frac{4}{2} - \frac{1}{2} \right) = 2\sqrt{3} \cdot \frac{3}{2} = 3\sqrt{3}$

Step 5: Final result

Now, adding the two results together:

$\frac{9}{2} + 3\sqrt{3}$

Thus, the value of the double integral is:

$\boxed{\frac{9}{2} + 3\sqrt{3}}$

Would you like further details or clarification? Here are some related questions you might find useful:

1. What is the method of changing the order of integration in double integrals?
2. How do you approach integrals involving square roots like $x^{1/2}$?
3. Can you solve double integrals in polar coordinates?
4. How do you handle integrals where one variable is treated as constant?
5. How can you check the correctness of your integral results?

Tip: When evaluating double integrals, always simplify the integrand before attempting the integration step by step to avoid unnecessary complexity!