1. **Problem Statement:** We need to find the Laplace Transform of the CPU load function: $$f(t) = \begin{cases} t^3 & m+1 < t < m+2 \\ 4 - t^2 & m+2 < t < m+4 \\ 1 & \text{otherwise} \end{cases}$$ where $m$ is the last digit of the student ID. Given student ID ends with 6, so $m=6$. 2. **Rewrite the function with $m=6$:** $$f(t) = \begin{cases} t^3 & 7 < t < 8 \\ 4 - t^2 & 8 < t < 10 \\ 1 & \text{otherwise} \end{cases}$$ 3. **Laplace Transform definition:** $$\mathcal{L}\{f(t)\} = \int_0^\infty e^{-st} f(t) dt$$ 4. **Break the integral into intervals:** $$\mathcal{L}\{f(t)\} = \int_0^7 e^{-st} \cdot 1 \, dt + \int_7^8 e^{-st} t^3 \, dt + \int_8^{10} e^{-st} (4 - t^2) \, dt + \int_{10}^\infty e^{-st} \cdot 1 \, dt$$ 5. **Calculate each integral separately:** **Integral 1:** $$I_1 = \int_0^7 e^{-st} dt = \left[-\frac{e^{-st}}{s}\right]_0^7 = \frac{1 - e^{-7s}}{s}$$ **Integral 2:** $$I_2 = \int_7^8 e^{-st} t^3 dt$$ Use integration by parts repeatedly or known formula for $\int t^n e^{-st} dt$. Set $I_n = \int t^n e^{-st} dt$. Using reduction formula: $$I_n = -\frac{t^n e^{-st}}{s} + \frac{n}{s} I_{n-1}$$ Calculate stepwise: - For $n=3$: $$I_3 = -\frac{t^3 e^{-st}}{s} + \frac{3}{s} I_2$$ - For $n=2$: $$I_2 = -\frac{t^2 e^{-st}}{s} + \frac{2}{s} I_1$$ - For $n=1$: $$I_1 = -\frac{t e^{-st}}{s} + \frac{1}{s} I_0$$ - For $n=0$: $$I_0 = \int e^{-st} dt = -\frac{e^{-st}}{s}$$ Substitute back: $$I_1 = -\frac{t e^{-st}}{s} - \frac{1}{s^2} e^{-st}$$ $$I_2 = -\frac{t^2 e^{-st}}{s} + \frac{2}{s} \left(-\frac{t e^{-st}}{s} - \frac{1}{s^2} e^{-st}\right) = -\frac{t^2 e^{-st}}{s} - \frac{2 t e^{-st}}{s^2} - \frac{2 e^{-st}}{s^3}$$ $$I_3 = -\frac{t^3 e^{-st}}{s} + \frac{3}{s} \left(-\frac{t^2 e^{-st}}{s} - \frac{2 t e^{-st}}{s^2} - \frac{2 e^{-st}}{s^3}\right) = -\frac{t^3 e^{-st}}{s} - \frac{3 t^2 e^{-st}}{s^2} - \frac{6 t e^{-st}}{s^3} - \frac{6 e^{-st}}{s^4}$$ Evaluate $I_3$ from 7 to 8: $$I_2 = \left[ I_3 \right]_7^8 = \left(-\frac{t^3 e^{-st}}{s} - \frac{3 t^2 e^{-st}}{s^2} - \frac{6 t e^{-st}}{s^3} - \frac{6 e^{-st}}{s^4}\right) \Bigg|_7^8$$ 6. **Integral 3:** $$I_3 = \int_8^{10} e^{-st} (4 - t^2) dt = 4 \int_8^{10} e^{-st} dt - \int_8^{10} t^2 e^{-st} dt$$ Calculate each part: - $$\int_8^{10} e^{-st} dt = \left[-\frac{e^{-st}}{s}\right]_8^{10} = \frac{e^{-8s} - e^{-10s}}{s}$$ - For $$\int t^2 e^{-st} dt$$ use formula from above: $$I_2 = -\frac{t^2 e^{-st}}{s} - \frac{2 t e^{-st}}{s^2} - \frac{2 e^{-st}}{s^3}$$ Evaluate from 8 to 10: $$\left[ I_2 \right]_8^{10} = \left(-\frac{t^2 e^{-st}}{s} - \frac{2 t e^{-st}}{s^2} - \frac{2 e^{-st}}{s^3}\right) \Bigg|_8^{10}$$ So, $$I_3 = 4 \cdot \frac{e^{-8s} - e^{-10s}}{s} - \left[ I_2 \right]_8^{10}$$ 7. **Integral 4:** $$I_4 = \int_{10}^\infty e^{-st} dt = \left[-\frac{e^{-st}}{s}\right]_{10}^\infty = \frac{e^{-10s}}{s}$$ 8. **Combine all integrals:** $$\mathcal{L}\{f(t)\} = I_1 + I_2 + I_3 + I_4$$ 9. **Final compact expression:** $$\boxed{\begin{aligned} \mathcal{L}\{f(t)\} &= \frac{1 - e^{-7s}}{s} + \left(-\frac{t^3 e^{-st}}{s} - \frac{3 t^2 e^{-st}}{s^2} - \frac{6 t e^{-st}}{s^3} - \frac{6 e^{-st}}{s^4}\right) \Bigg|_7^8 \\ &+ 4 \cdot \frac{e^{-8s} - e^{-10s}}{s} - \left(-\frac{t^2 e^{-st}}{s} - \frac{2 t e^{-st}}{s^2} - \frac{2 e^{-st}}{s^3}\right) \Bigg|_8^{10} + \frac{e^{-10s}}{s} \end{aligned}}$$ This expression fully represents the Laplace Transform of the given piecewise function with $m=6$. **Note:** For handwritten submission, write each step carefully showing the integration by parts and evaluation at limits.