Parameterize the line segment (call it [tex]C[/tex]) by
[tex]\vec r(t)=(1-t)(4,4,2)+t(5,8,-4)=(4+t,4+4t,2-6t)[/tex]
with [tex]0\le t\le1[/tex]. Then the line integral of [tex]\vec f(x,y)=y\,\vec\imath+x\,\vec\jmath+3\,\vec k[/tex] is
[tex]\displaystyle\int_C\vec f\cdot\mathrm d\vec r=\int_0^1(4+4t,4+t,3)\cdot(1,4,-6)\,\mathrm dt=\int_0^1(2+8t)\,\mathrm dt=6[/tex]
