Answer
$$\left( {{e^2} + 1} \right){\bf{i}} + 2\left( {{e^2} + 1} \right){\bf{j}} - \left( {{e^2} + 1} \right){\bf{k}}$$
Work Step by Step
$$\eqalign{
& \int_0^2 {t{e^t}\left( {{\bf{i}} + 2{\bf{j}} - {\bf{k}}} \right)} dt \cr
& {\text{use }}\int_a^b {{\bf{r}}\left( t \right)} dt = \left( {\int_a^b {f\left( t \right)} dt} \right){\bf{i}} + \left( {\int_a^b {g\left( t \right)} dt} \right){\bf{j}} + \left( {\int_a^b {h\left( t \right)} dt} \right){\bf{k}}{\text{ }}\left( {{\text{see page 814}}} \right) \cr
& {\text{ then}} \cr
& = \left( {\int_0^2 {t{e^t}} dt} \right){\bf{i}} + \left( {\int_0^2 {2t{e^t}} dt} \right){\bf{j}} - \left( {\int_0^2 {t{e^t}} dt} \right){\bf{k}} \cr
& = \left( {\int_0^2 {t{e^t}} dt} \right){\bf{i}} + 2\left( {\int_0^2 {t{e^t}} dt} \right){\bf{j}} - \left( {\int_0^2 {t{e^t}} dt} \right){\bf{k}} \cr
& {\text{factoring}} \cr
& = \left[ {{\bf{i}} + 2{\bf{j}} - {\bf{k}}} \right]\left( {\int_0^2 {t{e^t}} dt} \right) \cr
& {\text{Solving }}\int {t{e^t}} dt{\text{ using integration by parts}} \cr
& \,\,\,\,\,\,\,u = t,\,\,\,\,\,du = dt \cr
& \,\,\,\,\,dv = {e^t}dt\,\,\,\,\,\,v = {e^t} \cr
& \,\,\,\,\,\int {t{e^t}dt} = t{e^t} - \int {{e^t}dt} \cr
& \,\,\,\,\,\int {t{e^t}dt} = t{e^t} - {e^t} + C \cr
& {\text{then}} \cr
& = \left[ {{\bf{i}} + 2{\bf{j}} - {\bf{k}}} \right]\left( {\int_0^2 {t{e^t}} dt} \right) \cr
& = \left( {{\bf{i}} + 2{\bf{j}} - {\bf{k}}} \right)\left( {t{e^t} - {e^t}} \right)_0^2 \cr
& {\text{evaluate }} \cr
& = \left( {{\bf{i}} + 2{\bf{j}} - {\bf{k}}} \right)\left[ {\left( {2{e^2} - {e^2}} \right) - \left( {2{e^0} - {e^0}} \right)} \right] \cr
& = \left( {{\bf{i}} + 2{\bf{j}} - {\bf{k}}} \right)\left[ {\left( {2{e^2} - {e^2}} \right) - \left( {2{e^0} - {e^0}} \right)} \right] \cr
& {\text{simplifying}} \cr
& = \left( {{\bf{i}} + 2{\bf{j}} - {\bf{k}}} \right)\left( {{e^2} + 1} \right) \cr
& = \left( {{e^2} + 1} \right){\bf{i}} + 2\left( {{e^2} + 1} \right){\bf{j}} - \left( {{e^2} + 1} \right){\bf{k}} \cr} $$
