Step 1: Set up the convolution integral directly in time.
\[ z(t)=\int_{-\infty}^{\infty} x(\tau)\,y(t-\tau)\,d\tau \]
Since $x(\tau)$ is $1$ only for $2\le \tau\le 3$ and $0$ elsewhere, this collapses to
\[ z(t)=\int_{2}^{3} y(t-\tau)\,d\tau=\int_2^3 e^{-4(t-\tau)}u(t-\tau)\,d\tau \]
Step 2: Work out the integral in the three time ranges.
The unit step $u(t-\tau)$ needs $\tau\le t$, so we split by where $t$ falls.
For $t<2$: no $\tau$ in $[2,3]$ satisfies $\tau\le t$, so $z(t)=0$.
For $2\le t<3$: $\tau$ runs from $2$ up to $t$,
\[ z(t)=\int_2^{t} e^{-4(t-\tau)}\,d\tau=\frac{1-e^{-4(t-2)}}{4} \]
For $t\ge 3$: $\tau$ runs over the full pulse, $2$ to $3$,
\[ z(t)=\int_2^{3} e^{-4(t-\tau)}\,d\tau=\frac{e^{-4(t-3)}-e^{-4(t-2)}}{4} \]
Step 3: Take the Laplace transform of this piecewise $z(t)$.
\[ Z(s)=\int_2^{3} \frac{1-e^{-4(t-2)}}{4}\,e^{-st}\,dt+\int_3^{\infty} \frac{e^{-4(t-3)}-e^{-4(t-2)}}{4}\,e^{-st}\,dt \]
Carrying out both integrals and collecting terms gives
\[ Z(s)=\frac{e^{-3s}\left(e^{s}-1\right)}{s(s+4)} \]
Step 4: Cross-check with the frequency-domain shortcut.
This matches $X(s)Y(s)$ computed directly from $X(s)=\dfrac{e^{-2s}-e^{-3s}}{s}$ and $Y(s)=\dfrac{1}{s+4}$, confirming the same result by two independent routes.
\[ \boxed{Z(s)=\dfrac{e^{-3s}\left(e^{s}-1\right)}{s(s+4)}} \]
