and |

Let be a function continuous at every point of .

Then and . we have:

First let be the same path as in previous example. If , then . and we have:

Now we compute the integral of along the segment from to on the axis; a parametrization of the segment is given by , for . Thus we have:

Note that the two paths have the same origin and the same endpoint, but the integrals are different. We will understand this phenomenon later (v.i. Corollary 3.4).

- The integral is independent of the choice of the parameterization.
- .
- , when and are two paths such that the endpoint of and the origin of are identical.
- If and are opposite paths, .

Noah Dana-Picard 2007-12-24