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: 30

Answer

$$\frac{1}{2}{\bf{i}} + \frac{2}{3}{\bf{j}} + 2{\bf{k}}$$

Work Step by Step

$$\eqalign{ & \int_0^1 {\left( {t{\bf{i}} + \sqrt t {\bf{j}} + 4t{\bf{k}}} \right)} dt \cr & {\text{Split the integrand }}\left( {{\text{See example 6, page 829}}} \right) \cr & = \left( {\int_0^1 t dt} \right){\bf{i}} + \left( {\int_0^1 {\sqrt t } dt} \right){\bf{j}} + \left( {\int_0^1 {4t} dt} \right){\bf{k}} \cr & = \left[ {\frac{1}{2}{t^2}} \right]_0^1{\bf{i}} + \left[ {\frac{2}{3}{t^{3/2}}} \right]_0^1{\bf{j}} + \left[ {2{t^2}} \right]_0^1{\bf{k}} \cr & {\text{Evaluating}} \cr & = \frac{1}{2}\left[ {{{\left( 1 \right)}^2} - {{\left( 0 \right)}^2}} \right]{\bf{i}} + \frac{2}{3}\left[ {{{\left( 1 \right)}^{3/2}} - {{\left( 0 \right)}^{3/2}}} \right]{\bf{j}} + 2\left[ {{{\left( 1 \right)}^2} - {{\left( 0 \right)}^2}} \right]{\bf{k}} \cr & = \frac{1}{2}\left( 1 \right){\bf{i}} + \frac{2}{3}\left( 1 \right){\bf{j}} + 2\left( 1 \right){\bf{k}} \cr & = \frac{1}{2}{\bf{i}} + \frac{2}{3}{\bf{j}} + 2{\bf{k}} \cr} $$
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