## Calculus: Early Transcendentals (2nd Edition)

$$\frac{1}{2}{\bf{i}} + 3{\bf{j}} - 4{\bf{k}}$$
\eqalign{ & \int_0^{\ln 2} {\left( {{e^{ - t}}{\bf{i}} + 2{e^{2t}}{\bf{j}} - 4{e^t}{\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^{\ln 2} {{e^{ - t}}} dt} \right){\bf{i}} + \left( {\int_0^{\ln 2} {2{e^{2t}}} dt} \right){\bf{j}} - \left( {\int_0^{\ln 2} {4{e^t}} dt} \right){\bf{k}} \cr & {\text{evaluate integrals for each component use the basic integration rule}} \cr & \int {{e^{at}}} dt = \frac{{{e^{at}}}}{a} + C \cr & then \cr & = \left( { - {e^{ - t}}} \right)_0^{\ln 2}{\bf{i}} + \left( {{e^{2t}}} \right)_0^{\ln 2}{\bf{j}} - \left( {4{e^t}} \right)_0^{\ln 2}{\bf{k}} \cr & = \left( { - {e^{ - \ln 2}} + {e^0}} \right){\bf{i}} + \left( {{e^{2\left( {\ln 2} \right)}} - {e^0}} \right){\bf{j}} - \left( {4{e^{\ln 2}} - 4{e^0}} \right){\bf{k}} \cr & {\text{simplifying}} \cr & = \left( { - \frac{1}{2} + 1} \right){\bf{i}} + \left( {4 - 1} \right){\bf{j}} - \left( {8 - 4} \right){\bf{k}} \cr & = \frac{1}{2}{\bf{i}} + 3{\bf{j}} - 4{\bf{k}} \cr}