Parameterize the line segments by
[tex]\vec r(t)=(1-t)\langle1,1,1\rangle+t\langle2,2,2\rangle=\langle1+t,1+t,1+t\rangle[/tex]
and
[tex]\vec s(t)=(1-t)\langle2,2,2\rangle+t\langle3,6,8\rangle=\langle2+t,2+4t,2+6t\rangle[/tex]
both with [tex]0\le t\le1[/tex]. Then
[tex]\vec r'(t)=\langle1,1,1\rangle[/tex]
[tex]\vec s'(t)=\langle1,4,6\rangle[/tex]
so that the work done by [tex]\vec F(x,y,z)=\langle\sin(x+y),xy,x^2z\rangle[/tex] over the respective line segments is given by
[tex]\displaystyle\int_{C_1}\vec F\cdot\mathrm d\vec r=\int_0^1\langle\sin(2+2t),(1+t)^2,(1+t)^3\rangle\cdot\langle1,1,1\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1\sin(2+2t)+(1+t)^2+(1+t)^3\,\mathrm dt=\boxed{\frac{73+6\cos2-6\cos4}{12}}[/tex]
and
[tex]\displaystyle\int_{C_2}\vec F\cdot\mathrm d\vec s=\int_0^1\langle\sin(4+5t),(2+t)(2+4t),(2+t)^2(2+6t)\rangle\cdot\langle1,4,6\rangle\,\mathrm dt[/tex]
[tex]=\displaystyle\int_0^1\sin(4+5t)+4(2+t)(2+4t)+6(2+t)^2(2+6t)\,\mathrm dt=\boxed{\frac{3695+3\cos4-3\cos9}{15}}[/tex]
(both measured in Newton-meters)
