Step 1: Plan:
Area is the integral of the absolute value. The curve $y=x\sin(x/2)$ crosses the axis at $x=0, 2\pi, 4\pi$, so there are two loops.
Step 2: Antiderivative:
Find $\int x\sin(x/2)\,dx$ by parts with $u=x$ and $dv=\sin(x/2)\,dx$. The result is $-2x\cos(x/2)+4\sin(x/2)$.
Step 3: Loop 1:
From 0 to $2\pi$: $-2(2\pi)\cos\pi + 4\sin\pi - 0 = 4\pi$. This loop is above the axis.
Step 4: Loop 2:
From $2\pi$ to $4\pi$: $[-2(4\pi)\cos2\pi + 4\sin2\pi] - 4\pi = -8\pi - 4\pi = -12\pi$. This loop is below the axis, so its area is $12\pi$.
Step 5: Total:
$4\pi + 12\pi = 16\pi$ square units. Option (D).
Final Answer:
Add the two loop areas, $4\pi$ and $12\pi$, to get $16\pi$.
\[ \boxed{16\pi} \]