Thomas' Calculus 13th Edition

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: 23



Work Step by Step

Given: $\int_C xy dx +(x+y) dy$ Plug $y^2=x \implies dy =2x dx$ Now, $\int_C xy dx +(x+y) dy=\int_C xy dx + \int_C (x+y) dy$ Thus, $\int_{-1}^2 (2x^2+3x^3) dx=\dfrac{69}{4}$
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