Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 14 - Calculus of Vector-Valued Functions - 14.2 Calculus of Vector-Valued Functions - Exercises - Page 720: 43

Answer

$\left\langle1,2,-\frac{1}{3 } \sin 3\right\rangle.$

Work Step by Step

We have \begin{align} \int_{0}^{1}\left\langle 2t, 4t, -\cos 3t\right\rangle dt &=\left\langle t^2, 2t^2, -\frac{1}{3} \sin 3t \right\rangle|_{0}^{1}\\ &=\left\langle1,2,-\frac{1}{3 } \sin 3\right\rangle-\left\langle0,0,0\right\rangle\\ &=\left\langle1,2,-\frac{1}{3 } \sin 3\right\rangle. \end{align}
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