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: 6


$e- \dfrac{1}{e}$

Work Step by Step

$I=\int_{C}e^x dx=\int_{-1}^1 e^{t^3}(3t^2) dt$ $=\int_{-1}^1 e^{t^3}(3t^2) dt$ Plug $a=t^3\implies da=3t^2dt$ $=\int_{-1}^1 e^{a} da$ $=e-e^{-1}$ $=e- \dfrac{1}{e}$
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