We need to evaluate the definite integral $I = \int_{-4}^{0} \frac{1}{\sqrt{4-x}} dx$.

1. Problem Description

We need to evaluate the definite integral $I = \int_{-4}^{0} \frac{1}{\sqrt{4-x}} dx$.

2. Solution Steps

We can solve this integral using a simple substitution. Let $u = 4 - x$.

Then, $du = -dx$, which implies $dx = -du$.

When $x = -4$, $u = 4 - (-4) = 8$.

When $x = 0$, $u = 4 - 0 = 4$.

So, the integral becomes:

$I = \int_{8}^{4} \frac{1}{\sqrt{u}} (-du) = - \int_{8}^{4} u^{-1/2} du$.

We can reverse the limits of integration and change the sign:

$I = \int_{4}^{8} u^{-1/2} du$.

Now, we can integrate $u^{-1/2}$ with respect to $u$:

$\int u^{-1/2} du = \frac{u^{(-1/2)+1}}{(-1/2)+1} + C = \frac{u^{1/2}}{1/2} + C = 2\sqrt{u} + C$.

Therefore,

$I = \int_{4}^{8} u^{-1/2} du = \left[2\sqrt{u}\right]_{4}^{8} = 2\sqrt{8} - 2\sqrt{4} = 2\sqrt{4 \cdot 2} - 2(2) = 2(2\sqrt{2}) - 4 = 4\sqrt{2} - 4$.

3. Final Answer

$4\sqrt{2} - 4$