Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 12 - Vector-Valued Functions - Review Exercises - Page 863: 31

Answer

$$2\left( {e - 1} \right){\bf{i}} - 8{\bf{j}} - 2{\bf{k}}$$

Work Step by Step

$$\eqalign{ & \int_0^2 {\left( {{e^{t/2}}{\bf{i}} - 3{t^2}{\bf{j}} - {\bf{k}}} \right)dt} \cr & {\text{Split the integrand }}\left( {{\text{See example 6, page 829}}} \right) \cr & = \left( {\int_0^2 {{e^{t/2}}} dt} \right){\bf{i}} - \left( {\int_0^2 {3{t^2}} dt} \right){\bf{j}} - \left( {\int_0^2 {dt} } \right){\bf{k}} \cr & = \left[ {2{e^{t/2}}} \right]_0^2{\bf{i}} - \left[ {{t^3}} \right]_0^2{\bf{j}} - \left[ t \right]_0^2{\bf{k}} \cr & {\text{Evaluating}} \cr & = 2\left[ {{e^{2/2}} - {e^0}} \right]{\bf{i}} - \left[ {{2^3} - {0^3}} \right]{\bf{j}} - \left[ {2 - 0} \right]{\bf{k}} \cr & = 2\left( {e - 1} \right){\bf{i}} - \left( {8 - 0} \right){\bf{j}} - \left( {2 - 0} \right){\bf{k}} \cr & = 2\left( {e - 1} \right){\bf{i}} - 8{\bf{j}} - 2{\bf{k}} \cr} $$
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