Answer
$\dfrac{-15}{2}$
Work Step by Step
As we are given that $x=t$
thus, $dx=dt$
Now, $\int_C (x-y) dx=\int_0^3 [t-(2t+1) ] dt=\int_0^3 [t-2t-1) ] dt$
or, $=[\dfrac{-t^2}{2}-t]_0^3$
or, $=\dfrac{-9}{2}-3$
or, $=\dfrac{-15}{2}$
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: 13
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