The given problem is to evaluate the double integral:

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

Let's solve this step by step.

Step 1: Inner integral with respect to $y$

The inner integral is:

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

Here, $x^2$ is treated as a constant since we are integrating with respect to $y$. We integrate term by term:

1. For $x^2$: $\int x^2 \, dy = x^2 y$

2. For $2y$: $\int 2y \, dy = y^2$

So, the result of the inner integral is: $x^2 y + y^2 \bigg|_{0}^{1}$

Substitute the limits $y = 1$ and $y = 0$: $\left( x^2 (1) + (1)^2 \right) - \left( x^2 (0) + (0)^2 \right) = x^2 + 1$

Step 2: Outer integral with respect to $x$

The outer integral is: $\int_{0}^{2} \left( x^2 + 1 \right) \, dx$

Again, integrate term by term:

1. For $x^2$: $\int x^2 \, dx = \frac{x^3}{3}$

2. For $1$: $\int 1 \, dx = x$

So, the result of the outer integral is: $\frac{x^3}{3} + x \bigg|_{0}^{2}$

Substitute the limits $x = 2$ and $x = 0$: $\left( \frac{2^3}{3} + 2 \right) - \left( \frac{0^3}{3} + 0 \right) = \frac{8}{3} + 2 = \frac{8}{3} + \frac{6}{3} = \frac{14}{3}$

Final Answer:

The value of the double integral is: $\frac{14}{3} \quad \text{or} \quad 4\frac{2}{3}$

The correct choice is: $4\frac{2}{3}$.

Would you like a more detailed explanation of any step?

5 Related Questions:

1. How do you set up double integrals for more complex regions of integration?
2. What are the general rules for switching the order of integration in double integrals?
3. Can double integrals always be solved by separating into two single integrals?
4. How would the integral change if the region of integration were reversed (e.g., $y$ from 0 to 2)?
5. What are some practical applications of double integrals in physics and engineering?

Tip:

Always verify the limits of integration and carefully handle constants when integrating with respect to one variable.