Consider the vector field:
F(x,y)=2xyi+x^(2)j
Integrate F over a path starting at (0,0) and ending at (2,2).
The path is
x(t) = t
y(t) = t
r(t) = ti + tj
r' = i+j
F(r(t)) = 2(t)(t)i + (t^2)j = 2t^2 i + t^2 j
So we have the integral over t in [0,2] of
∫F•dr = ∫[0,2] 3t^2 dt
= t^3 [0,2]
= 8