Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.2 - Integrals of Vector Functions; Projectile Motion - Exercises 13.2 - Page 753: 5

Answer

$$(\ln 4) \mathbf{i}+(\ln 4) \mathbf{j}+(\ln 2) \mathbf{k}$$

Work Step by Step

Given $$ \int_{1}^{4}\left(\frac{1}{t} \mathbf{i}+\frac{1}{5-t} \mathbf{j}+\frac{1}{2 t} \mathbf{k}\right) d t $$ Then \begin{align*} \int_{1}^{4}\left(\frac{1}{t} \mathbf{i}+\frac{1}{5-t} \mathbf{j}+\frac{1}{2 t} \mathbf{k}\right) d t&=[\ln t]\bigg|_{1}^{4} \mathbf{i}+[-\ln (5-t)]\bigg|_{1}^{4} \mathbf{j}+\left[\frac{1}{2} \ln t\right]\bigg|_{1}^{4} \mathbf{k}\\ &=(\ln 4) \mathbf{i}+(\ln 4) \mathbf{j}+(\ln 2) \mathbf{k} \end{align*}
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