Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 16: Integrals and Vector Fields - Section 16.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises 16.2 - Page 955: 13



Work Step by Step

Given: $x=t \implies dx=dt$ Now, $\int_C (x-y) dx=\int_0^3 [t-(2t+1) ] dt$ or, $\int_0^3 [t-2t-1) ] dt=[\dfrac{-t^2}{2}-t]_0^3$ Thus, $\int_C (x-y) dx=(\dfrac{-9}{2})-3=-\dfrac{15}{2}$
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