[tex]f(t)=(t-5)u_2(t)-(t-2)u_5(t)[/tex]
The Laplace transform is
[tex]F(s)=\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=\int_2^\infty(t-5)e^{-st}\,\mathrm dt-\int_5^\infty(t-2)e^{-st}\,\mathrm dt[/tex]
Integrate by parts; in the first integral, take
[tex]u=t-5\implies\mathrm du=\mathrm dt[/tex]
[tex]\mathrm dv=e^{-st}\,\mathrm dt\implies v=-\dfrac{e^{-st}}s[/tex]
[tex]\implies\displaystyle\int_2^\infty(t-5)e^{-st}\,\mathrm dt=-\frac{e^{-st}}s(t-5)\bigg|_2^\infty+\frac1s\int_2^\infty e^{-st}\,\mathrm dt[/tex]
[tex]=-\dfrac{3e^{-2s}}s-\dfrac{e^{-st}}{s^2}\bigg|_2^\infty[/tex]
[tex]=-\dfrac{3e^{-2s}}s+\dfrac{e^{-2s}}{s^2}=-\dfrac{(3s-1)e^{-2s}}{s^2}[/tex]
For the second integral, take
[tex]u=t-2\implies\mathrm du=\mathrm dt[/tex]
[tex]\mathrm dv=e^{-st}\,\mathrm dt\implies v=-\dfrac{e^{-st}}s[/tex]
[tex]\implies\displaystyle\int_5^\infty(t-2)e^{-st}\,\mathrm dt=-\dfrac{(t-2)e^{-st}}s\bigg|_5^\infty+\frac1s\int_5^\infty e^{-st}\,\mathrm dt[/tex]
[tex]=\dfrac{3e^{-5s}}s-\dfrac{e^{-st}}{s^2}\bigg|_5^\infty[/tex]
[tex]=\dfrac{3e^{-5s}}s+\dfrac{e^{-5s}}{s^2}=\dfrac{(3s+1)e^{-5s}}{s^2}[/tex]
So we have
[tex]F(s)=\dfrac{(3s+1)e^{-5s}-(3s-1)e^{-2s}}{s^2}[/tex]
