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Stokes agrees with the fundamental theorem of calculus
Statement
Assume . For , orient increasingly. Every smooth on this interval satisfies where the boundary point signs are at and at . This agrees with the Riemann fundamental theorem of calculus.
Facts & Assumptions
The general Stokes theorem: Assume . Let be an oriented smooth -manifold with boundary, , and let . With and the outward-normal-first orientation, An empty boundary contributes zero; in dimension one its integral is a finite signed sum of point values.
Newton–Leibniz needs only continuity on , differentiability on , and a Riemann-integrable extension of the interior derivative: Let . Suppose is continuous on and differentiable on . If is Riemann integrable and then No derivative of at either endpoint is assumed, and the two endpoint values assigned to the integrable extension do not enter the conclusion.
Proof
Given: The objects and hypotheses in the statement above.
The interval is compact, and the outward directions are at and at . Outward-first gives the point signs . Stokes therefore yields the difference , including constant and zero functions.
In its increasing coordinate, , so the left side is the ordinary Riemann integral of . The published FTC applies: is continuous on the closed interval and differentiable inside, and its smooth derivative is Riemann integrable. It gives the same endpoint difference. The condition avoids treating a point as a one-manifold.
Depends on
Used by
Dependency tree · two levels
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Sources
- Lee Example 16.12, p.414 (standard reference, not scraped)