University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 12 - Section 12.2 - Integrals of Vector Functions; Projectile Motion - Exercises - Page 654: 2



Work Step by Step

We need to integrate the given integral as follows: $[\int_1^2 (6-6t)dt]i+[\int_1^2 3 \sqrt t dt] j+[\int_0^2 (4/t^2)dt]= (6t-3t^2)_1^2i+(2t^{3/2})_1^2j+(\dfrac{-4}{t})_1^2k$ or, $(6t-3t^2)_1^2i+(2t^{3/2})_1^2j+(\dfrac{-4}{t})_1^2k=(0-3)i+(2^{5/2}-2)j+[-2-(-4)] k$ Thus, $(0-3)i+(2^{5/2}-2)j+[-2-(-4)] k=-3i+(2^{5/2}-2)j+2k$
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