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: 2

Answer

$$-3 \mathbf{i}+(4 \sqrt{2}-2) \mathbf{j}+2 \mathbf{k}$$

Work Step by Step

Given $$ \int_{1}^{2}\left[(6-6 t) \mathbf{i}+3 \sqrt{t} \mathbf{j}+\left(\frac{4}{t^{2}}\right) \mathbf{k}\right] d t $$ Then \begin{align*} \int_{1}^{2}\left[(6-6 t) \mathbf{i}+3 \sqrt{t} \mathbf{j}+\left(\frac{4}{t^{2}}\right) \mathbf{k}\right] d t&=\left(6 t-3 t^{2}\right)\bigg|_{1}^{2} \mathbf{i}+ 2 t^{3 / 2} \bigg|_{1}^{2} \mathbf{j}- 4 t^{-1} \bigg|_{1}^{2} \mathbf{k}\\ &=-3 \mathbf{i}+(4 \sqrt{2}-2) \mathbf{j}+2 \mathbf{k} \end{align*}
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