I am trying to express the following function as a unit step function so that I can use Laplace:
$ f(x) = \left\{ \begin{array}{lr} 0 & : t < 1\\ t^2-4t+5 & : 1\leq t \leq3 \\ 0 & : t>3 \\ \end{array} \right.$
Is the following correct, and if so, is it a "valid" way to solve it:
$g(t)=U_1(t)[t^2-4t+5]+U_3(t)[-(t^2-4t+5)]$
$\implies \mathcal{L}[g(t)]=\exp(-s)\mathcal{L}[(t+1)^2-4(t+1)+5]+\exp(-3s)\mathcal{L}[-((t+3)^2-4(t+3)+5)]$
$\implies \exp(-s)(\frac{2}{s^3}-\frac{2}{s^2}+\frac{2}{s})-\exp(-3s)(\frac{2}{s^3}+\frac{2}{s^2}+\frac{2}{s})$
I appreciate any hints if I have done something wrong here, or if I can improve something!
Your solution is correct, but instead of trying to recast something as a multiplies of step functions, we could go straight to the integration. \begin{align} \mathcal{L}\{f(t)\} &= \int_0^{\infty}f(t)e^{-st}dt\\ &= \int_0^10\cdot e^{-st}dt + \int_1^3(t^2-4t+5)e^{-st}dt + \int_3^{\infty}0\cdot e^{-st}dt\\ &= \int_1^3(t^2-4t+5)e^{-st}dt\\ &= \frac{\partial^2}{\partial s^2}\int_1^3 e^{-st}dt + 4\frac{\partial}{\partial s}\int_1^3e^{-st}dt + 5\int_1^3e^{-st}dt\tag{1} \end{align} Finally, the result simplifies to $$ \mathcal{L}\{f(t)\} = \frac{4e^{-2s}\bigl[(1+s^2)\sinh(s) - s\cosh(s)\bigr]}{s^3} $$ We have the derivatives in $(1)$ since $\frac{\partial^2}{\partial s^2}\bigl[e^{-st}\bigr] = t^2e^{-st}$ and $\frac{\partial}{\partial s}\bigl[e^{-st}\bigr] = -te^{-st}$.