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: 51

Answer

$\int\textbf{r}(t) dt = \langle \frac{1}{3}e^{3t}, tan^{-1}(t), -\sqrt {2t}\rangle + C$

Work Step by Step

To find the indefinite integral, compute the integral of each component. $\int\textbf{r}(t) dt = \langle \int e^{3t}\ dt,\int \frac{1}{1+t^2}\ dt,\int \frac{-1}{\sqrt {2t}}\ dt\rangle = \langle \frac{1}{3}e^{3t}, tan^{-1}(t), -\sqrt {2t}\rangle + C$
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