Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 11 - Vectors and Vector-Valued Functions - 11.6 Calculus of Vector-Valued Functions - 11.6 Exercises - Page 815: 59

Answer

$$2{\bf{i}} + 2{\bf{k}}$$

Work Step by Step

$$\eqalign{ & \int_{ - 1}^1 {\left( {{\bf{i}} + t{\bf{j}} + 3{t^2}{\bf{k}}} \right)} dt \cr & {\text{use }}\int_a^b {{\bf{r}}\left( t \right)} dt = \left( {\int_a^b {f\left( t \right)} dt} \right){\bf{i}} + \left( {\int_a^b {g\left( t \right)} dt} \right){\bf{j}} + \left( {\int_a^b {h\left( t \right)} dt} \right){\bf{k}}{\text{ }}\left( {{\text{see page 814}}} \right) \cr & {\text{ then}} \cr & = \left( {\int_{ - 1}^1 {dt} } \right){\bf{i}} + \left( {\int_{ - 1}^1 {tdt} } \right){\bf{j}} + \left( {\int_{ - 1}^1 {3{t^2}dt} } \right){\bf{k}} \cr & {\text{evaluate integrals for each component}} \cr & = \left( t \right)_{ - 1}^1{\bf{i}} + \left( {\frac{{{t^2}}}{2}} \right)_{ - 1}^1{\bf{j}} + \left( {{t^3}} \right)_{ - 1}^1{\bf{k}} \cr & = \left( {1 - \left( { - 1} \right)} \right){\bf{i}} + \left( {\frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{{{\left( { - 1} \right)}^2}}}{2}} \right){\bf{j}} + \left( {{{\left( 1 \right)}^3} - {{\left( { - 1} \right)}^3}} \right){\bf{k}} \cr & {\text{simplifying}} \cr & = \left( 2 \right){\bf{i}} + \left( 0 \right){\bf{j}} + \left( 2 \right){\bf{k}} \cr & = 2{\bf{i}} + 2{\bf{k}} \cr} $$
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