If the Laplace transform of a signal y(t) is $$Y\left( s \right) = {1 \over {s\left( {s - 1} \right)}},$$ then its final value is

The Laplace transform of i(t) tends to
$$I\left( s \right)\,\, = \,{2 \over {s\left( {1 + s} \right)}}$$

As $$t \to \infty $$ , the value of i(t) tends to

Given that $$L\left[ {f\left( t \right)} \right]\, = \,$$ $${{s + 2} \over {{s^2} + 1}},$$ $$$L\left[ {g\left( t \right)} \right] = {{{s^2} + 1} \over {\left( {s + 3} \right)\left( {s + 2} \right)}},$$$ $$$h\left( t \right) = \int\limits_0^t {f\left( \tau \right)\,g\left( {t - \tau } \right)\,d\tau ,} $$$ $$L\left[ {h\left( t \right)} \right]$$ is

$$If\,\,L\left[ {f\left( t \right)} \right]\, = \,F\left( s \right),$$ then $$L\left[ {f\left( {t - T} \right)} \right]$$ is equal to