Let $$x[n] = 2^{-n}u[n + 1]$$ $$h[n] = 3^{n}u[-n+2]$$If we perform convolution, we get $$y[m]=x[n]*h[n] = \sum_{l=-\infty}^{+\infty}x[l]h[m-l] = \cases{\frac{27\times2^{3 - m}}{5} \ m \ge 2 \\ \frac{4\times3^{2+ m}}{5} \ m\le 1}$$And $y[3] = \frac{27}{5} = 5.4$. Similarly, let $$x(t) = 2^{-t}u(t + 1)$$ $$h(t) = 3^{t}u(-t+2)$$We have $$y(z)=x(t)*h(t) = \int_{-\infty}^{+\infty}x(\tau)h(z-\tau) = \frac{3\times2^{1 - z} (6 + (-6 + 6^z)u(1 - z))}{\ln(6)}$$And $y(3) = \frac{9}{\ln(36)} \approx 2.511$ which is obviously less than $5.4$. Why this can happen? Integration should yield greater result since all of the values are positive.
Edit: Note that by $u[n]$ and $u(t)$, I mean $$u(t) = \cases{1 \ \ \ \ t\ge 0 \\ 0 \ \ \ \ t\lt 0}$$ $$u[n] = \cases{1 \ \ \ \ n\ge 0 \\ 0 \ \ \ \ n\lt 0}$$