Answer
$2$
Work Step by Step
As we are given that $y=t^2$
thus, $dy=2t dt$
Now, $\int_C \dfrac{x}{y} dx=\int_1^2 [\dfrac{t}{t^2}(2t dt)=\int_1^2 [2] dt$
or, $=[2t]_1^2$
or, $=4-2$
or, $=2$
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: 14
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
