Answer
$$\frac{{{e^2} - 1}}{2}{\bf{i}} - \left( {\frac{1}{e} - 1} \right){\bf{j}} + \frac{1}{2}{\bf{k}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {\left( {{e^{2t}}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}}} \right)} dt \cr
& {\text{Integrating the components yields}} \cr
& \int_0^1 {\left( {{e^{2t}}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}}} \right)} dt = \left[ {\frac{{{e^{2t}}}}{2}} \right]_0^1{\bf{i}} + \left[ { - {e^{ - t}}} \right]_0^1{\bf{j}} + \left[ {\frac{{{t^2}}}{2}} \right]_0^1{\bf{k}} \cr
& {\text{Use the vector form of the Fundamental Theorem of Calculus}} \cr
& \int_a^b {{\bf{r}}\left( t \right)} dt = \left[ {{\bf{R}}\left( t \right)} \right]_a^b = {\bf{R}}\left( b \right) - {\bf{R}}\left( a \right) \cr
& {\text{then}} \cr
& \int_0^1 {\left( {{e^{2t}}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}}} \right)} dt = \left[ {\frac{{{e^{2\left( 1 \right)}}}}{2} - \frac{{{e^{2\left( 0 \right)}}}}{2}} \right]{\bf{i}} - \left[ {{e^{ - 1}} - {e^0}} \right]_0^1{\bf{j}} + \left[ {\frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^2}}}{2}} \right]{\bf{k}} \cr
& {\text{simplify}} \cr
& \int_0^1 {\left( {{e^{2t}}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}}} \right)} dt = \left[ {\frac{{{e^2}}}{2} - \frac{1}{2}} \right]{\bf{i}} - \left[ {{e^{ - 1}} - 1} \right]_0^1{\bf{j}} + \left[ {\frac{1}{2} - 0} \right]{\bf{k}} \cr
& \int_0^1 {\left( {{e^{2t}}{\bf{i}} + {e^{ - t}}{\bf{j}} + t{\bf{k}}} \right)} dt = \frac{{{e^2} - 1}}{2}{\bf{i}} - \left( {\frac{1}{e} - 1} \right){\bf{j}} + \frac{1}{2}{\bf{k}} \cr} $$
