Answer
$2$
Work Step by Step
Given: $y=t^2 \implies dy=2t dt$
Now, $\int_C \dfrac{x}{y} dx=\int_1^2 (\dfrac{t}{t^2})(2t) dt$
$\int_1^2 [2] dt=[2t]_1^2$
or, $=4-2$
Thus, $\int_C \dfrac{x}{y} dx=2$
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: 14
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