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: 7



Work Step by Step

$I=\int_{C_1}(x+2y) dx+x^2 dy+\int_{C_2}(x+2y) dx+x^2 dy$ $=\int_{0}^1 [2t+2(t)](2 dt)+(2t)^2(1 dt)+\int_{0}^1 [(2+t)+2(1-t)(1 dt) +(2+t)^2(-1 dt)$ $=\int_{0}^1 8t+4t^2 dt+\int_0^1 (4-t) dt-(4+4t+t^2) dt$ $=4t^2+[\dfrac{4}{3}]_0^1-\dfrac{5t^2}{2}-[\dfrac{t^3}{3}]_0^1$ $= \dfrac{16}{3}-\dfrac{17}{6}$ $=\dfrac{5}{2}$
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