Fundamental Theorem of Calculus/Second Part/Proof 3

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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-[[Definition:Dimension (Topology)|dimensional] 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.

This needs considerable tedious hard slog to complete it. In particular: Prove that the closed (real) interval is a smooth and oriented manifold To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Finish}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

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.

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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:

\(\ds \ds \int_{\closedint a b} \map f x dx\)

\(=\)

\(\ds \int_{\closedint a b} \rd F\)

\(\ds \)

\(=\)

\(\ds \ds \int_{\partial \closedint a b } F\)

by General Stokes' Theorem

\(\ds \)

\(=\)

\(\ds \int_{\set{a}^- \cup \set{b}^+} F\)

Definition of Boundary of Manifold

\(\ds \)

\(=\)

\(\ds \map F b - \map F a\)

as required.

$\blacksquare$