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Suppose $M$ is a smooth manifold and $f$ is a real valued smooth function on $M$. Set $N:=f^{-1}([0,1])$ and suppose $N$ is a compact submanifold of $M$. Let $\mu$ be a volume form on $M$ and $v$ a volume preserving vector field on $M$, so that $L_v\mu=0$, where $L$ denotes the Lie derivative. Suppose in addition that $i_v\mu = d\alpha$ is exact.


$$\int_N df(v) \mu = \int_{\partial N}f i_v\mu$$

by Stokes' theorem. As $f$ is constant on each connected component on $\partial N$, the integral on the right-hand side is zero.

Now I thought Stokes' theorem was meant to generalize the fundamental theorem of calculus. Suppose I take $M=\mathbb{R}$ and $\mu = dt$ and $v = \partial_t$ then $L_{\partial_t}dt=0$ and $i_{\partial_t}dt = 1$ (which is exact!). Then if $f^{-1}([0,1])=[a,b]$ (say), with $f(a)=0$ and $f(b)=1$ with $0<f(t)<1$ for all $t\in (a,b)$, then one has

$$\int_a^b f'(t)dt = f(b) - f(a) = 1-0\ne0.$$

What gives?

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If $f$ is constant on $\partial [0,1]$, then $f(0) = f(1)$. – Raeder Jun 7 '11 at 14:53
closely related:… – t.b. Jun 7 '11 at 14:59
Since $f$ is constant on each connected component you need to evaluate the rhs on the boundary of each connected component individually - not just the rightmost boundary of the rightmost interval and the leftmost boundary of the leftmost interval, as you appear to be doing here. – Chris Taylor Jun 7 '11 at 16:07
Why is the integral of a constant function zero? – Tim van Beek Jun 8 '11 at 9:46
I refer you to Jesse Madnick's exposition on this subject: " By the way, the Fundamental Theorem of Calculus ∫baf′(x)=f(b)−f(a) is also a special case of the Generalized Stokes Theorem, and relates the integral over an interval [a,b] to the values of the function on the boundary -- the boundary being the two points x=a and x=b." – Jesse Madnick 11 hours ago – Tom Au Jun 28 '11 at 14:57

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