Answer
$e- \dfrac{1}{e}$
Work Step by Step
$I=\int_{C}e^x dx$
or, $=\int_{-1}^1 [e^{t^3}] \times(3t^2) dt$
or, $=\int_{-1}^1 e^{t^3}(3t^2) dt$
Suppose $p=t^3; dp=3t^2dt$
Thus, $I=\int_{-1}^1 [e^{p} dp]=e-e^{-1}$
Hence, $I=e- \dfrac{1}{e}$
Multivariable Calculus, 7th Edition
by Stewart, James
Published by
Brooks Cole
ISBN 10:
0-53849-787-4
ISBN 13:
978-0-53849-787-9
Chapter 16 - Vector Calculus - 16.2 Exercises - Page 1096: 6
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