Let's solve the iterated integral step by step:

$I = \int_0^4 \int_1^4 \sqrt{s \cdot x} \, ds \, dx$

Step 1: Evaluate the inner integral (with respect to $s$)

We have the inner integral:

$\int_1^4 \sqrt{s \cdot x} \, ds$

Since $x$ is treated as a constant with respect to $s$, we can factor out $\sqrt{x}$:

$\sqrt{x} \int_1^4 \sqrt{s} \, ds$

Now, integrate $\sqrt{s} = s^{1/2}$ with respect to $s$:

$\int s^{1/2} \, ds = \frac{2}{3} s^{3/2}$

Evaluating the limits from $s = 1$ to $s = 4$:

[ \frac{2}{3} \left[ 4^{3/2} - 1^{3/2} \right] = \frac{2}{3} \left[ 8 - 1 \right] = \frac{2}{3} \times 7 = \frac{14}{3} ]

Thus, the inner integral becomes:

$\sqrt{x} \times \frac{14}{3} = \frac{14}{3} \sqrt{x}$

Step 2: Evaluate the outer integral (with respect to $x$)

Now, we have:

$I = \int_0^4 \frac{14}{3} \sqrt{x} \, dx$

Factor out the constant $\frac{14}{3}$:

$I = \frac{14}{3} \int_0^4 \sqrt{x} \, dx$

Next, integrate $\sqrt{x} = x^{1/2}$ with respect to $x$:

$\int x^{1/2} \, dx = \frac{2}{3} x^{3/2}$

Evaluating the limits from $x = 0$ to $x = 4$:

$\frac{2}{3} \left[ 4^{3/2} - 0^{3/2} \right] = \frac{2}{3} \times 8 = \frac{16}{3}$

Thus, the outer integral becomes:

$I = \frac{14}{3} \times \frac{16}{3} = \frac{224}{9}$

Final Answer:

The value of the iterated integral is:

$\boxed{\frac{224}{9}}$

Would you like more details or have any questions? Here are some related questions to explore further:

1. How can we modify the integral if the limits were reversed?
2. What would happen if the integrand was $\sqrt{s \cdot x^2}$?
3. How would this change if $x$ were the inner integral and $s$ the outer?
4. What does the method of Fubini’s Theorem tell us about switching the order of integration?
5. Can you solve the same type of integral where the square root is replaced by a trigonometric function?

Tip: When evaluating iterated integrals, always check if you can simplify the integrand by factoring out constants, especially when dealing with powers or square roots.