Chapter 15 - Section 15.1 - Line Integrals - Exercises - Page 826: 9

$\sqrt 2$

Since, we have $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ or, $ds=\sqrt{1^2+(-1)^2+0^2} dt \implies ds= \sqrt 2 dt$ Now, the line integral is: $\int_C (x+y) ds=\int_0^1 (t+1-t) \sqrt 2 dt$ or, $=\sqrt 2\int_0^1 dt$ or, $=\sqrt 2$