Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 16 - Vector Calculus - 16.2 Line Integrals - 16.2 Exercises - Page 1124: 11


$\dfrac{\sqrt {14}}{12}(e^6-1)$

Work Step by Step

$I=\int_{0}^{1} (t) e^{(2t) \cdot (3t)} \sqrt{{{(\dfrac{dx}{dt})^{2}}+{(\dfrac{dy}{dt})^{2}}}}dt$ $=\int_{0}^{1} (t) e^{(2t) \cdot (3t)} \sqrt {14} dt$ $=\sqrt {14} \int_{0}^{1} (t) e^{6t^2} dt$ Plug $6t^2=a ; da=12t dt$ $=\sqrt {14} \int_{0}^{6} \dfrac{e^{a}}{12} da$ $=\dfrac{\sqrt {14}}{12}[e^a] _{0}^{6}$ $=\dfrac{\sqrt {14}}{12}(e^6-1)$
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