Answer
$$L[f(t)]=L[2t]= \frac{2}{s^2}$$
Work Step by Step
Given $$f(t)=2t$$
So, we get
\begin{aligned} L[f(t)] & =\int_{0}^{\infty} e^{-s t}f(t) \ d t\\
& =\int_{0}^{\infty} e^{-s t}(2t) d t \\
&=2\int_{0}^{\infty} t e^{-s t} d t\\
\text{let}\ I_1=\int_{0}^{\infty} t e^{-s t} d t
\ \ \ \ \ \ \ \ \ \ \\
\text{integrate with partition}
\end{aligned}
let
$$u=t, \ \ \ \ \ du=e^{-st}dt$$
$$du=dt, \ \ \ \ \ v=\ -\frac{1}{s}e^{-st}$$
so, we get
\begin{aligned}
I_1&= uv |_0^{\infty}- \int_{0}^{\infty} v \ du \\
&= -\frac{t}{s}e^{-st} |_0^{\infty}+\int_{0}^{\infty} \frac{1}{s}e^{-st} \ dt \\
&=0- \frac{1}{s^2}e^{-st}|_{0}^{\infty} \\
&= \frac{1}{s^2}
\end{aligned}
So, we get
$$L[f(t)]=L[2t]= \frac{2}{s^2}$$
