The problem is to evaluate the double integral $\int_0^{\pi/2} [\frac{1}{2} \int_4^0 \sqrt{u} \, du ] \, dx$.

1. Problem Description

The problem is to evaluate the double integral $\int_0^{\pi/2} [\frac{1}{2} \int_4^0 \sqrt{u} \, du ] \, dx$.

2. Solution Steps

First, we evaluate the inner integral with respect to $u$.

$$\int_4^0 \sqrt{u} \, du = \int_4^0 u^{1/2} \, du = \left[ \frac{u^{3/2}}{3/2} \right]_4^0 = \left[ \frac{2}{3} u^{3/2} \right]_4^0 = \frac{2}{3}(0^{3/2} - 4^{3/2}) = \frac{2}{3}(0 - (\sqrt{4})^3) = \frac{2}{3}(0 - 2^3) = \frac{2}{3}(0-8) = -\frac{16}{3}.$$

Next, we multiply this result by $\frac{1}{2}$:

$$\frac{1}{2} \int_4^0 \sqrt{u} \, du = \frac{1}{2} \left( -\frac{16}{3} \right) = -\frac{8}{3}.$$

Finally, we evaluate the outer integral with respect to $x$:

$$\int_0^{\pi/2} \left( -\frac{8}{3} \right) dx = -\frac{8}{3} \int_0^{\pi/2} dx = -\frac{8}{3} [x]_0^{\pi/2} = -\frac{8}{3} \left(\frac{\pi}{2} - 0\right) = -\frac{8}{3} \cdot \frac{\pi}{2} = -\frac{4\pi}{3}.$$

3. Final Answer

The final answer is $-\frac{4\pi}{3}$.