Answer
$1$
Work Step by Step
Since, we have $\int_C (x-y) dx +(x+y) dy$
Now,
$\int_C (x-y) dx +(x+y) dy=\int_{0}^1 x dx+\int_{0}^1 (2x-2) dx+\int_{0}^1 y dy$
$=\dfrac{1}{2}+1-\dfrac{1}{2}$
or, $\int_C (x-y) dx +(x+y) dy=1$
University Calculus: Early Transcendentals (3rd Edition)
by Hass, Joel R.; Weir, Maurice D.; Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0321999584
ISBN 13:
978-0-32199-958-0
Chapter 15 - Section 15.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises - Page 838: 24
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