Presumably,
[tex]f(t)=\begin{cases}0&\text{for }0\le t<4\\t-4&\text{for }t\ge4\end{cases}[/tex]
By definition of the Laplace transform,
[tex]\mathscr L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_4^\infty(t-4)e^{-st}\,\mathrm dt[/tex]
Integrate by parts, taking
[tex]u=t-4\implies\mathrm du=\mathrm dt[/tex]
[tex]\mathrm dv=e^{-st}\,\mathrm dt\implies v=-\dfrac1se^{-st}[/tex]
so we get
[tex]\displaystyle\mathscr L_s\{f(t)\}=\left(-\frac1se^{-st}(t-4)\right)\bigg|_{t=4}^{t\to\infty}+\frac1s\int_4^\infty e^{-st}\,\mathrm dt[/tex]
[tex]=-\dfrac1{s^2}e^{-st}\bigg|_{t=4}^{t\to\infty}[/tex]
[tex]=\dfrac{e^{-4s}}{s^2}[/tex]
