Chapter 16: Integrals and Vector Fields - Section 16.1 - Line Integrals - Exercises 16.1 - Page 943: 9

$\sqrt 2$

As we know that $ds=\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2+(\dfrac{dz}{dt})^2} dt$ Here, $ds=\sqrt{1^2+(-1)^2+0^2} dt \implies ds= \sqrt 2 dt$ Line integral: $\int_C (x+y) ds=\int_0^1 (t+1-t) \sqrt 2 dt=\sqrt 2\int_0^1 dt$ Thus, $\int_C (x+y) ds=\sqrt 2$