I am trying to find the Laplace transform of
$$F(t) = \begin{cases} t & 0 \le t < 1 \\ 2 - t & 1 \le t < 2 \\ 0 & \text{otherwise} \end{cases}$$
My work is as follows:
$$\begin{align} \mathcal{L} \{ F(t) \} &= \int_0^1 t e^{-st} \ dt + \int_1^2 (2 - t)e^{-st} \ dt \\ &= \left\{ \left[ \dfrac{-1}{s} e^{-st} \times t \right]_0^1 + \dfrac{1}{s} \int_0^1 e^{-st} \ dt \right\} + \left\{ \left[ (2 - t) \times \dfrac{-1}{s} e^{-st} \right]_1^2 - \dfrac{1}{s} \int_1^2 e^{-st} \ dt \right\} \\ &= \dfrac{-1}{s}e^{-s} + \left[ \dfrac{-1}{s^2}e^{-st} \right]_0^1 - \dfrac{1}{s}e^{-s} + \left[ \dfrac{1}{s^2} e^{-st} \right]_1^2 \\ &= \dfrac{-1}{s}e^{-s} - \dfrac{1}{s^2}e^{-s} + \dfrac{1}{s^2} - \dfrac{1}{s}e^{-s} + \dfrac{1}{s^2}e^{-2s} - \dfrac{1}{s^2} e^{-s} \\ &= \dfrac{1}{s^2}e^{-2s} - \dfrac{2}{s^2}e^{-s} - \dfrac{2}{s}e^{-s} + \dfrac{1}{s^2} \end{align}$$
However, the solution is said to be $\dfrac{1}{s^2}(1 - e^{-s})^2$.
I would greatly appreciate it if people would please take the time to review my work and help me find my error.