To convolve two functions we have the following:
$ y(t) = h(t) * x(t) = \int_{-\infty}^\infty x(t)h(t-\tau)d\tau $
Plugging in functions for $ x(t)= e^{-t}u(t) $ and $ h(t)=u(t-1) $ we get:
$ = \int_{-\infty}^\infty e^{-\tau}u(\tau)u(t-1-\tau)d\tau $
We now flip and shift x(t) and for t>1 we find
$ = \int_{1}^t e^{-(t-\tau)}d\tau $
Evaluating this for t>1
$ \begin{align} &= \int_{1}^{t} e^{-\tau}d\tau \\ &= -e^{-t} - (-e^{0}) \\ &= 1 - e^{-(t-1)} \end{align} $
