Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 16 - Section 16.2 - Line Integrals - 16.2 Exercise - Page 1142: 31

Answer

$\frac{11}{8}-\frac{1}{e}$

Work Step by Step

We know that: $F(x,y)=(e^{x-1},xy)$ and $ C$ is $r(t)=(t^2,t^3), $$ 0 \leq t \leq1$ $$\int_{c} F.dr= \int_{c} F(r(t)).r'(t) dt $$ $r'(t)=2t i +3t^2 j$ $ \displaystyle\int _{0}^{1} (e^{t^{2}-1}i + t^5 j). (2t i + 3t^2 j)$ $= \displaystyle \int _{0}^{1} (2te^{t^{2}-1} + 3t^7 ) dt$ $= \left(e^{t^{2}-1} + \frac{3}{8}t^8\right) \bigg |_{0}^{1} $ $= \frac{11}{8}-\frac{1}{e}$
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