Fundamental Theorem of Calculus/Second Part/Proof 3

Let $\closedint a b$ be the closed (real) interval.

We claim that closed (real) interval is a smooth 1-dimensional oriented manifold.

By Classification of Compact One-Manifolds, every compact connected 1-dimensional manifold is diffeomorphic to either a circle or a closed interval.

Therefore, the closed interval is a 1-dimensional] [[Definition:Topological Manifold|manifold.

By Subset of Real Numbers is Interval iff Connected, since $\closedint a b$ is an interval of $\R$, it follows that $\closedint a b$ is connected.

Let $\F$ be a smooth 0-form with compact support on the $\closedint a b$.

Let the boundary of $\closedint a b$ be:
 * $\partial \closedint a b$

Since the manifold is oriented, and has compact support, the integrals:
 * $\ds \int_{\partial \closedint a b} \F$

and:
 * $\int_{\closedint a b} \rd \F$

are well-defined.

Then, by General Stokes' Theorem:
 * $\ds \int_{\partial \closedint a b} \F = \int_{\closedint a b} \rd \F$

where $\d \F$ is the exterior derivative of 0-form:
 * $\F = \map f x dx$

It follows that:

as required.