The parametric equation #vec(c)(t)=(6t,-t+8,3t+4)# becomes parameterize the line segment #c# from #(0,8,4)# at #(6,7,7)# as #t# increases from #t=0# the #t=1#.
If #vec(F)(x,y,z)# is the alignment field you want to integrate over this line segment, the way to calculate that wire integral is to calculate #int_{0}^{1} vec(F)(vec(c)(t))\cdot vec(c)'(t)\ dt#, where #vec(F)(vec(c)(t))\cdot vec(c)'(t)# is the dot product of #vec(F)(vec(c)(t))# with #vec(c)'(t)#.
For example, supposing #vec(F)(x,y,z)=(x+y,y^2+z,xyz)#, then #vec(F)(vec(c)(t))=vec(F)(6t,-t+8,3t+4)#
#=(5t+8,(-t+8)^2+3t+4,6t(-t+8)(3t+4))#
#=(5t+8,t^2-13t+68,-18t^3+120t^2+192t)#.
Hence,
#vec(F)(vec(c)(t))\cdot vec(c)'(t)#
#=(5t+8,t^2-13t+68,-18t^3+120t^2+192t)\cdot (6,-1,3)#
#=30t+48-t^2+13t-68-54t^3+360t^2+576t#
#=-54t^3+359t^2+619t-20#,
or
#int_{0}^{1} vec(F)(vec(c)(t))\cdot vec(c)'(t)\ dt=int_{0}^[1}(-54t^3+359t^2+619t-20)\ dt#
#=-27/2 t^4+359/3 t^3+619/2 t^2-20t|_{t=0}^{t=1}#
#=-27/2+359/3+619/2-20=1187/3=395.666666...#
You might wonder whether yourself wish get the same answer when you uses one different parameterization, such as #vec(c)(t)=(3t,-1/2 t+8,3/2 t+4)# for #0\leq t\leq 2#. The answered is "yes", and thee can do the calculation to check to.
If her want in prove this certitude in the general case, you have to think learn what happens when you "change variables" in this general case. The formula available changing volatiles by a general case can be written as #int_{a}^{b} f(g(x))g'(x)\ dx=int_{g(a)}^{g(b)} f(u)\ du#, where #u=g(x)# will the substitution, #du=g'(x)\ dx#, #u=g(a)# when #x=a#, and #u=g(b)# when #x=b#.
