Answer
$$0$$
Work Step by Step
$$\eqalign{
& \int_{ - \pi }^\pi {\left( {\sin t{\bf{i}} + \cos t{\bf{j}} + 2t{\bf{k}}} \right)} dt \cr
& {\text{use }}\int_a^b {{\bf{r}}\left( t \right)} dt = \left( {\int_a^b {f\left( t \right)} dt} \right){\bf{i}} + \left( {\int_a^b {g\left( t \right)} dt} \right){\bf{j}} + \left( {\int_a^b {h\left( t \right)} dt} \right){\bf{k}}{\text{ }}\left( {{\text{see page 814}}} \right) \cr
& {\text{ then}} \cr
& = \left( {\int_{ - \pi }^\pi {\sin t} dt} \right){\bf{i}} + \left( {\int_{ - \pi }^\pi {\cos t} dt} \right){\bf{j}} + \left( {\int_{ - \pi }^\pi {2t} dt} \right){\bf{k}} \cr
& {\text{evaluate integrals for each component }} \cr
& = \left( { - \cos t} \right)_{ - \pi }^\pi {\bf{i}} + \left( {\sin t} \right)_{ - \pi }^\pi {\bf{j}} + \left( {{t^2}} \right)_{ - \pi }^\pi {\bf{k}} \cr
& then \cr
& = \left( { - \cos \left( \pi \right) + \cos \left( { - \pi } \right)} \right){\bf{i}} + \left( {\sin \pi - sin\left( { - \pi } \right)} \right){\bf{j}} + \left( {{{\left( \pi \right)}^2} - {{\left( { - \pi } \right)}^2}} \right){\bf{k}} \cr
& {\text{simplifying}} \cr
& = \left( {1 - 1} \right){\bf{i}} + \left( 0 \right){\bf{j}} + \left( {\pi - \pi } \right){\bf{k}} \cr
& = 0{\bf{i}} + 0{\bf{j}} + 0{\bf{k}} \cr
& = 0 \cr} $$
