If f is defined on a smooth curve C given by (x(t),y(t)), 
, then the line integral of f along C is
If f is continuous, then
and this is independent of the parameterization (provided the curve is traversed just once) as t increases from a to b.
This generalizes easily to curves in space:
line integrals of f along C with respect to x and y:
and
Let 
be a continuous vector field defined on a smooth curve C given
by a vector function 
, . Then the line
integral of F along C is
where 
is the tangent vector (the component of velocity in the
direction tangential to the motion).
If vector field 
, then we can break the
integral into three:
If we change the orientation of the parameterization (t runs from b to a), then
If the curve C is made up of a bunch of smooth curves 
, then the
line integral over C is just the sum of the line integrals over the .
We've seen line integrals before, in the context of the calculation of arc length. The first problem is to parameterize the curve along which you're integrating. Once we have our parameterization, the integral reduces to an ordinary integral of a single variable (usually called t), the variable of the paremeterization. Today with a GPS unit and a watch it's easy to parameterize a curve in space.
Then we simply multiply the ``height'' (f(x(t),y(t))) times the tiny chunk of arc length (ds(t)) to get our integral. We can do this for a curve embedded in the plane, or for a curve in space: the difference is simply one of additional labor.
One of the twists here is that we're going to be computing line integrals with vector fields (e.g. calculating ``work'', in its technical sense of force through distance). It's not big deal: the scalar function we're working with in that case is simply the dot product of the force and displacement vectors.