Chapter 15 - Section 15.2 - Vector Fields and Line Integrals: Work, Circulation, and Flux - Exercises - Page 839: 27

$\dfrac{25}{6}$

Since, we have $\int_C F \cdot T ds-\int_C F. dr=\int_C xy dx +(y-x) dy$ Now, $\int_C xy dx +(y-x) dy=\int_C x(2x-1) dx +(2x-1-x) (2 dx) $ $=\int_{1}^{2} 2x^2 +x-2 dx$ $=[\dfrac{2x^3}{3}+\dfrac{x^2}{2}-2x]_1^2$ or, $=\dfrac{16}{3}+2-4-\dfrac{2}{3} -\dfrac{1}{2} -2$ or, $=\dfrac{25}{6}$