1. **Problem Statement:** Show that the Laplace transform of the piecewise function $$f(t) = \begin{cases} 10 \sin 2t & 0 \leq t < \pi \\ 0 & t \geq \pi \end{cases}$$ is $$F(s) = \frac{20(1 - e^{-\pi s})}{s^2 + 4}.$$ 2. **Recall the Laplace transform definition:** $$\mathcal{L}\{f(t)\} = \int_0^{\infty} e^{-st} f(t) dt.$$ Since $f(t)$ is zero for $t \geq \pi$, the integral reduces to $$F(s) = \int_0^{\pi} e^{-st} 10 \sin 2t \, dt.$$ 3. **Use the formula for Laplace transform of $\sin(at)$:** $$\mathcal{L}\{\sin at\} = \frac{a}{s^2 + a^2}.$$ But here the integral is from 0 to $\pi$, not to infinity, so we compute directly: 4. **Compute the integral:** $$F(s) = 10 \int_0^{\pi} e^{-st} \sin 2t \, dt.$$ Use integration by parts or the known formula: $$\int e^{pt} \sin qt \, dt = \frac{e^{pt}}{p^2 + q^2} (p \sin qt - q \cos qt) + C.$$ Here, $p = -s$, $q = 2$. 5. **Evaluate definite integral:** $$F(s) = 10 \left[ \frac{e^{-st}}{(-s)^2 + 2^2} (-s \sin 2t - 2 \cos 2t) \right]_0^{\pi} = 10 \left[ \frac{e^{-st}}{s^2 + 4} (-s \sin 2t - 2 \cos 2t) \right]_0^{\pi}.$$ 6. **Calculate the boundary terms:** At $t=\pi$: $$\sin 2\pi = 0, \quad \cos 2\pi = 1,$$ so $$-s \sin 2\pi - 2 \cos 2\pi = -2.$$ At $t=0$: $$\sin 0 = 0, \quad \cos 0 = 1,$$ so $$-s \sin 0 - 2 \cos 0 = -2.$$ 7. **Substitute back:** $$F(s) = 10 \frac{1}{s^2 + 4} \left( e^{-s\pi} (-2) - 1 \cdot (-2) \right) = 10 \frac{1}{s^2 + 4} (-2 e^{-s\pi} + 2) = \frac{20 (1 - e^{-s\pi})}{s^2 + 4}.$$ **Final answer:** $$F(s) = \frac{20 (1 - e^{-\pi s})}{s^2 + 4}.$$