Convolution in Continious time
The $u(t-2)$ term needs to be included because the upper limit of the definite integral is $t-2$, not $\infty$:

\begin{align*}
h(t) &amp;= e^{-(t-2)}u(t-2)\\
y(t) &amp;= x(t) * h(t)= \int_{-\infty}^{\infty} x(t')h(t-t') dt'\\
 &amp;= \int_{-\infty}^{\infty} x(t')e^{-(t-t'-2)}u(t-t'-2) dt'\\
u(t-t'-2)&amp;= \begin{cases} 1 &amp; t'&lt;t-2 \\ 0 &amp; \text{otherwise} \end{cases}\\
y(t) &amp;= \int_{-\infty}^{t-2} x(t')e^{-(t-t'-2)}(1) dt' + \int_{t-2}^{\infty} x(t')e^{-(t-t'-2)} (0) dt'\\
&amp;= \int_{-\infty}^{t-2} x(t')e^{-(t-t'-2)} dt'.
\end{align*}