Answer
$-\dfrac{15}{2}$
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, $\int_C (x-y) dx=(\dfrac{-9}{2})-3=-\dfrac{15}{2}$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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
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