Answer
$$\frac{{32}}{3}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 2}^2 {\left( {3t{\bf{i}} + 2{t^2}{\bf{j}} - {t^3}{\bf{k}}} \right)} dt \cr
& {\text{Split the integrand }}\left( {{\text{See example 6, page 829}}} \right) \cr
& = \left( {\int_{ - 2}^2 {3t} dt} \right){\bf{i}} + \left( {\int_{ - 2}^2 {2{t^2}} dt} \right){\bf{j}} - \left( {\int_{ - 2}^2 {{t^3}} dt} \right){\bf{k}} \cr
& = \left[ {\frac{3}{2}{t^2}} \right]_{ - 2}^2{\bf{i}} + \left[ {\frac{2}{3}{t^3}} \right]_{ - 2}^2{\bf{j}} - \left[ {\frac{1}{4}{t^4}} \right]_{ - 2}^2{\bf{k}} \cr
& {\text{Evaluating}} \cr
& = \frac{3}{2}\left[ {{{\left( 2 \right)}^2} - {{\left( { - 2} \right)}^2}} \right]{\bf{i}} + \frac{2}{3}\left[ {{{\left( 2 \right)}^3} - {{\left( { - 2} \right)}^3}} \right]{\bf{j}} - \frac{1}{4}\left[ {{{\left( 2 \right)}^4} - {{\left( { - 2} \right)}^4}} \right]{\bf{k}} \cr
& = \frac{3}{2}\left( 0 \right){\bf{i}} + \frac{2}{3}\left( {16} \right){\bf{j}} - \frac{1}{4}\left( 0 \right){\bf{k}} \cr
& = \frac{{32}}{3}{\bf{j}} \cr} $$
