Question

Find the laplace transform of f(t)={0,(t−4),0≤t<4t≥4

Answer 1

Presumably,

$f(t)=\begin{cases}0&\text{for }0\le t$

By definition of the Laplace transform,

$\mathscr L_s\{f(t)\}=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_4^\infty(t-4)e^{-st}\,\mathrm dt$

Integrate by parts, taking

$u=t-4\implies\mathrm du=\mathrm dt$
$\mathrm dv=e^{-st}\,\mathrm dt\implies v=-\dfrac1se^{-st}$

so we get

$\displaystyle\mathscr L_s\{f(t)\}=\left(-\frac1se^{-st}(t-4)\right)\bigg|_{t=4}^{t\to\infty}+\frac1s\int_4^\infty e^{-st}\,\mathrm dt$
$=-\dfrac1{s^2}e^{-st}\bigg|_{t=4}^{t\to\infty}$
$=\dfrac{e^{-4s}}{s^2}$