Step 1: Complete the square inside the square root.
$4x - x^2 = -(x^2 - 4x) = -(x^2 - 4x + 4 - 4) = 4 - (x-2)^2$. So $\sqrt{4x-x^2} = \sqrt{4-(x-2)^2}$.
Step 2: Substitute $x - 2 = 2\sin\theta$.
Then $dx = 2\cos\theta\,d\theta$ and $\sqrt{4-(x-2)^2} = 2\cos\theta$. When $x=0$: $\sin\theta = -1$, $\theta = -\pi/2$. When $x=4$: $\sin\theta = 1$, $\theta = \pi/2$. Also $x + 2 = 2\sin\theta + 4$.
Step 3: Rewrite the integral.
\[ I = \int_{-\pi/2}^{\pi/2} \frac{(2\sin\theta + 4)}{2\cos\theta} \cdot 2\cos\theta\,d\theta = \int_{-\pi/2}^{\pi/2} (2\sin\theta + 4)\,d\theta. \]
Step 4: Split into odd and even parts.
$2\sin\theta$ is odd on $[-\pi/2, \pi/2]$, so $\int_{-\pi/2}^{\pi/2} 2\sin\theta\,d\theta = 0$. And $4$ is even: $\int_{-\pi/2}^{\pi/2} 4\,d\theta = 4\pi$.
Step 5: Combine.
\[ I = 0 + 4\pi = 4\pi. \]
Step 6: Note on the answer choices.
The correct value is $4\pi$, which does not match any printed option. The answer blank in the original paper is empty, confirming no option gives $4\pi$.
\[ \boxed{4\pi} \]