1. **Problem statement:** A particle travels on the x-axis with velocity given by $v(t) = 6 \sin\left(\frac{t}{2}\right)$. The particle is at position $x=1$ when $t=0$. Find the position $x$ when $t=\pi$. 2. **Formula used:** Position is the integral of velocity. So, $$x(t) = x(0) + \int_0^t v(s) \, ds$$ where $x(0) = 1$. 3. **Calculate the integral:** $$x(\pi) = 1 + \int_0^{\pi} 6 \sin\left(\frac{s}{2}\right) ds$$ 4. **Integrate:** Let $u = \frac{s}{2}$, so $ds = 2 du$. $$\int_0^{\pi} 6 \sin\left(\frac{s}{2}\right) ds = 6 \int_0^{\pi} \sin\left(\frac{s}{2}\right) ds = 6 \int_0^{\frac{\pi}{2}} \sin(u) \cdot 2 du = 12 \int_0^{\frac{\pi}{2}} \sin(u) du$$ 5. **Evaluate the integral:** $$12 \int_0^{\frac{\pi}{2}} \sin(u) du = 12 \left[-\cos(u)\right]_0^{\frac{\pi}{2}} = 12 \left[-\cos\left(\frac{\pi}{2}\right) + \cos(0)\right] = 12 (0 + 1) = 12$$ 6. **Find the position:** $$x(\pi) = 1 + 12 = 13$$ **Final answer:** The position of the particle at $t=\pi$ is $\boxed{13}$, which corresponds to option (E).