Answer
$$s\left( t \right) = - \frac{1}{2}{e^{ - 2t}} + 4t + \frac{5}{2}$$
Work Step by Step
$$\eqalign{
& v\left( t \right) = {e^{ - 2t}} + 4;{\text{ }}s\left( 0 \right) = 2 \cr
& {\text{Calculating the position function}} \cr
& s\left( t \right) = \int {v\left( t \right)} dt \cr
& s\left( t \right) = \int {\left( {{e^{ - 2t}} + 4} \right)} dt \cr
& s\left( t \right) = - \frac{1}{2}{e^{ - 2t}} + 4t + C \cr
& {\text{Calculating the function for the initial position }}s\left( 0 \right) = 2 \cr
& 2 = - \frac{1}{2}{e^{ - 2\left( 0 \right)}} + 4\left( 0 \right) + C \cr
& 2 = - \frac{1}{2} + C \cr
& C = \frac{5}{2} \cr
& {\text{Thenerefore,}} \cr
& s\left( t \right) = - \frac{1}{2}{e^{ - 2t}} + 4t + \frac{5}{2} \cr
& {\text{Graphing the velocity and position functions}} \cr} $$