Chapter 15 - Section 15.3 - Path Independence, Conservative Fields, and Potential Functions - Exercises - Page 850: 14

$-2$

We know that the line equation is defined as: $r(t)=r_0+kt$ Thus, $r_0+kt=(1,1,1)+t \lt 2, 4, -2 \gt=\lt 1+2t, 1+4t, 2-2t \gt$ So, $x=1+2t \implies dx= 2 dt$; $y=1+4t \implies dy=4 dt$ and $z=2-2t \implies dz=-2dt$ We need to substitute all the values in the given integral. $\int_{0}^1(1+2t)(2 -2t)2 dt+(1+2t)(2-2t)4d+(1+2t)(1+4t)-2dt=-2$