Answer
The position vector:
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t} + 1,\frac{1}{3}{t^3} + 1,\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t + 1} \right)$
The velocity vector:
${\bf{v}}\left( t \right) = \left( {{{\rm{e}}^t},{t^2},\frac{1}{2}{t^2} + t + 1} \right)$
Work Step by Step
Find the velocity vector:
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {{{\rm{e}}^t},2t,t + 1} \right){\rm{d}}t = \left( {{{\rm{e}}^t},{t^2},\frac{1}{2}{t^2} + t} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = \left( {1,0,1} \right)$ gives
$\left( {1,0,1} \right) = \left( {1,0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {0,0,1} \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {{{\rm{e}}^t},{t^2},\frac{1}{2}{t^2} + t} \right) + \left( {0,0,1} \right)$
${\bf{v}}\left( t \right) = \left( {{{\rm{e}}^t},{t^2},\frac{1}{2}{t^2} + t + 1} \right)$
Find the position vector:
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {{{\rm{e}}^t},{t^2},\frac{1}{2}{t^2} + t + 1} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},\frac{1}{3}{t^3},\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {2,1,1} \right)$ gives
$\left( {2,1,1} \right) = \left( {1,0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {1,1,1} \right)$
Thus,
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t},\frac{1}{3}{t^3},\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t} \right) + \left( {1,1,1} \right)$
${\bf{r}}\left( t \right) = \left( {{{\rm{e}}^t} + 1,\frac{1}{3}{t^3} + 1,\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t + 1} \right)$
