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: 24



Work Step by Step

Given: $\int_C (x-y) dx +(x+y) dy$ Now, $\int_C (x-y) dx +(x+y) dy=\int_{0}^1 x dx+\int_{0}^1 (2x-2) dx+\int_{0}^1 y dy$ Thus, $(\dfrac{1}{2})+1-(\dfrac{1}{2})=1$
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