Answer
$e- \dfrac{1}{e}$
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 $a=t^3\implies da=3t^2dt$
$=\int_{-1}^1 e^{a} da$
$=e-e^{-1}$
$=e- \dfrac{1}{e}$
Calculus 8th Edition
by Stewart, James
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
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