Answer
$\frac{3}{s^2}-\frac{4}{s}$
Work Step by Step
Given: $f(t)=3t-4$
Obtain: $F(s)=\int^{\infty}_0 e^{-st}f(t)dt\\
=\int^{\infty}_0 e^{-st} (3t-4)dt\\
=3\int^{\infty}_0 te^{-st}dt-4\int^{\infty}_0e^{-st}dt\\
=3\lim [-\frac{t}{s}e^{-st}]^n_0-\frac{3}{s^2}\lim[e^{-st}]^n_0+\frac{4}{s}\lim [e^{-st}]^n_0\\
=3\lim [-\frac{n}{s}e^{-sn}]-\frac{3}{s^2}\lim[e^{-sn}-1]+\frac{4}{s}\lim [e^{-sn}-1]\\
=\frac{3}{s^2}-\frac{4}{s}$
