Answer
FALSE.
Work Step by Step
Since,$$ds = \sqrt{{(dx)}^2+{(dy)}^2}=\sqrt{2}dt$$ $$ \int_{C} xyds = \sqrt{2}\int_{0}^{1}{t^2}dt$$.
Thus the statement is $ \fbox{false}$.
Calculus 10th Edition
by Larson, Ron; Edwards, Bruce H.
Published by
Brooks Cole
ISBN 10:
1-28505-709-0
ISBN 13:
978-1-28505-709-5
Chapter 15 - Vector Analysis - 15.2 Exercises - Page 1064: 83
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