# Chapter 16: Integrals and Vector Fields - Section 16.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises 16.2 - Page 955: 24

$1$

#### Work Step by Step

Given: $\int_C (x-y) dx +(x+y) dy$ Now, $\int_C (x-y) dx +(x+y) dy=\int_{0}^1 x dx+\int_{0}^1 (2x-2) dx+\int_{0}^1 y dy$ Thus, $(\dfrac{1}{2})+1-(\dfrac{1}{2})=1$

After 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.