The integral in the image is a double integral of the form:

$\int_0^2 \int_0^{\sqrt{x}} x^3 y \, dy \, dx$

Let's evaluate this step by step.

Step 1: Inner Integral with respect to $y$

The inner integral is:

$\int_0^{\sqrt{x}} x^3 y \, dy$

Since $x^3$ is treated as a constant with respect to $y$, we integrate $y$ alone:

$x^3 \int_0^{\sqrt{x}} y \, dy$

The integral of $y$ is:

$\frac{y^2}{2}$

So, applying the limits $y = 0$ to $y = \sqrt{x}$:

$x^3 \left[ \frac{y^2}{2} \right]_0^{\sqrt{x}} = x^3 \left( \frac{(\sqrt{x})^2}{2} - \frac{0^2}{2} \right) = x^3 \left( \frac{x}{2} \right)$

Thus, the result of the inner integral is:

$\frac{x^4}{2}$

Step 2: Outer Integral with respect to $x$

Now, we need to evaluate the outer integral:

$\int_0^2 \frac{x^4}{2} \, dx$

This simplifies to:

$\frac{1}{2} \int_0^2 x^4 \, dx$

The integral of $x^4$ is:

$\frac{x^5}{5}$

Applying the limits $x = 0$ to $x = 2$:

$\frac{1}{2} \left[ \frac{x^5}{5} \right]_0^2 = \frac{1}{2} \left( \frac{2^5}{5} - \frac{0^5}{5} \right) = \frac{1}{2} \left( \frac{32}{5} \right) = \frac{16}{5}$

Final Answer:

The value of the double integral is:

$\boxed{\frac{16}{5}}$

Would you like more details on any step or further explanation? Here are some questions to expand your understanding:

1. How does the order of integration affect the result of a double integral?
2. Can you change the limits of integration for this problem?
3. What happens if you reverse the order of integration in this particular case?
4. How would the problem change if the limits of $y$ were different?
5. How is this type of problem used in physical applications like finding volumes?

Tip: When integrating over a region, always check if reversing the order of integration might simplify the process.