1. **Problem Statement:** Find the Laplace Transform of the function $$f(t) = \begin{cases} \frac{t}{k}, & 0 < t < k \\ 1, & t > k \end{cases}$$ where $k$ is a positive constant. 2. **Formula:** The Laplace Transform of a function $f(t)$ is defined as $$\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st} f(t) \, dt$$ where $s$ is a complex number parameter. 3. **Step 1: Break the integral according to the piecewise definition:** $$F(s) = \int_0^k e^{-st} \frac{t}{k} \, dt + \int_k^\infty e^{-st} \cdot 1 \, dt$$ 4. **Step 2: Evaluate the first integral:** $$I_1 = \frac{1}{k} \int_0^k t e^{-st} \, dt$$ Use integration by parts where: - Let $u = t \Rightarrow du = dt$ - Let $dv = e^{-st} dt \Rightarrow v = -\frac{1}{s} e^{-st}$ Then, $$I_1 = \frac{1}{k} \left[ -\frac{t}{s} e^{-st} \Big|_0^k + \frac{1}{s} \int_0^k e^{-st} dt \right]$$ 5. **Step 3: Evaluate the remaining integral:** $$\int_0^k e^{-st} dt = \left[-\frac{1}{s} e^{-st} \right]_0^k = \frac{1}{s} (1 - e^{-sk})$$ 6. **Step 4: Substitute back:** $$I_1 = \frac{1}{k} \left[ -\frac{k}{s} e^{-sk} + \frac{1}{s} \cdot \frac{1}{s} (1 - e^{-sk}) \right] = \frac{1}{k} \left( -\frac{k}{s} e^{-sk} + \frac{1 - e^{-sk}}{s^2} \right)$$ Simplify: $$I_1 = -\frac{e^{-sk}}{s} + \frac{1 - e^{-sk}}{k s^2}$$ 7. **Step 5: Evaluate the second integral:** $$I_2 = \int_k^\infty e^{-st} dt = \left[-\frac{1}{s} e^{-st} \right]_k^\infty = \frac{e^{-sk}}{s}$$ 8. **Step 6: Add both parts:** $$F(s) = I_1 + I_2 = \left(-\frac{e^{-sk}}{s} + \frac{1 - e^{-sk}}{k s^2} \right) + \frac{e^{-sk}}{s} = \frac{1 - e^{-sk}}{k s^2}$$ 9. **Final answer:** $$\boxed{\mathcal{L}\{f(t)\} = \frac{1 - e^{-sk}}{k s^2}}$$ This result shows the Laplace Transform of the piecewise function $f(t)$ defined above.