Compute y(t) = x(t) * h(t) given x(t) = u(t−3) − u(t−5) and h(t) = e⁻³tu(t).

The convolution $y(t)$ is given by:

$$y(t) = \int_{-\infty}^{\infty} x(\tau) h(t-\tau) d\tau$$

Given signals are $x(t) = u(t-3) - u(t-5)$ and $h(t) = e^{-3t}u(t)$.
This means $x(\tau) = 1$ for $3 \le \tau < 5$ and 0 otherwise.
Also, $h(t-\tau) = e^{-3(t-\tau)}u(t-\tau)$, which is $e^{-3(t-\tau)}$ for $t-\tau \ge 0$ (i.e., $\tau \le t$) and 0 otherwise.