By definition of the transform, and the definition of [tex]f[/tex], we have
[tex]\displaystyle\int_0^\infty f(t)e^{-st}\,\mathrm dt=9\int_0^5e^{-st}\,\mathrm dt[/tex]
Substitute [tex]u=-st[/tex], so that [tex]\mathrm du=-s\,\mathrm dt[/tex].
[tex]\displaystyle-\frac9s\int_0^{-5s}e^u\,\mathrm du=-\frac9s(e^{-5s}-1)=\boxed{\frac9s-\frac{9e^{-5s}}s}[/tex]
Alternatively, if you're familiar with the unit step function, you can write
[tex]f(t)=9(u(t)-u(t-5))[/tex]
where the unit step function is defined as
[tex]u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t<c\end{cases}[/tex]
Either way, the transform is the same.
