Stokes' theorem generalized the FTC part 2. Is there a known generalization for part 1? Stokes' theorem generalizes the fundamental theorem of calculus (part 2) using differential forms. Is there a known generalization of part 1?
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In case anyone is unaware, The fundamental theorem of calculus part 1 states that the derivative of the map $t \mapsto \int_{a}^{t} f(s) ds$ is equal to 
$f(t)$. From this, it easily follows that if $F' = f$, then $\int_{a}^{b} f(x) dx = F(b) - F(a)$ (part 2).
Stokes' theorem ($\int_{\Sigma} d \omega = \oint_{\partial \Sigma} \omega$) generalizes part 2 is analogous to part 2 in that in both cases one does a
calculation on the boundary.
But is there an analogous version of part 1? This question comes from my previous question in which I did such a calculation.
In the GTC part 1, we consider a perametrized set of intervals $[0,t]_{t \in R}$. So the generalization ought to consist of a set of (hyper-)surfaces $\{\Sigma_{t}\}_{t \in R}$ in $R^N$. And thus, we wish to calculate the derivative of the mapping $t \mapsto \int_{\Sigma_t} \omega$.
Suppose there exists a smooth $\phi (r,s): U\times R \to R^N$ ($U$ open subset of $R^{N-1}$)such that the restriction of $\phi$ on $[0,1]\times[0,t]$ parametrizes $\Sigma_{t}$. Then for fixed $s$, the map $r \mapsto \phi(r,s)$ perametrizes a subsurface $\sigma_{s}$ of $\Sigma_{s}$ whose dimension is one less ($\sigma_{s} \subset \partial \Sigma_{s}$). I believe that the derivative of the map $t \mapsto \int_{\Sigma_{t}} \omega$ is equal to $\oint_{\sigma_t} \omega_{t}$, where $\omega_{t}$ is some differential form that represents $\omega$ "evaluated at" $\sigma_{t}$.
Am I in the right direction?
 A: Sorry to have just seen this question 7 years late. An answer is in the Poincar'e Lemma. See Spivak's Calculus on Manifolds for example. I wrote an exposition for the vector field version in 3 dimensions that is on my web page bterrell.net.
A: Consider the following equivalent formulation of the first fundamental theorem:  given a function $f$ defined on an interval $I = [a,b]$.  Let $G$ be the 1d Green's function for differentiation:  $G(s) = \text{sgn } s/2$.  Then an antiderivative $F$ for the function $f$ can be found by, for $x \in I$,
$$F(x) G(x-t) |_{t=a}^b = \int_a^b f(t) G(x-t) \, dt + \int_a^b F(t) \delta(x-t) \, dt = -F(x) + \int_a^b f(t) G(x-t) \, dt$$
This is just integration by parts, right?  Using the expression of the Green's function given above, we get
$$-\frac{1}{2} [F(b) + F(a)] = -F(x) + \frac{1}{2} \left( \int_a^x f(t) \, dt + \int_x^b f(t) \, dt \right)$$
Suppose we already take the second fundamental theorem to be true--this corresponds, in the general multivariate case, to already having Stokes' theorem proved.  Then we can use the second theorem to evaluate those integrals on the right in terms of some antiderivative $H$:
$$F(x) = \frac{1}{2} [F(b) + F(a)] + \frac{1}{2} [H(x) - H(a) + H(x) - H(b)] = \frac{1}{2} [F(b) + F(a) - H(b) - H(a)] + H(x)$$
$F$ and $H$ differ by a constant, and so $F$ is also an antiderivative, just with a different constant of integration depending on the boundary values of $H$ (and of course, if we take $F=H$, we get a truism).
It is not difficult to prove, from here, that $I(x) = \int_a^x f(t) \, dt$ is also an antiderivative, as it differs from $F$ by another constant.
Of course, you could prove that more or less directly with the second fundamental theorem, so why did I go through this cumbersome construction using a Green's function?
Because it generalizes easily to higher dimensions.  For instance, consider a vector field $E$.  If the field is curlless, we can construct its gradient antiderivative $\phi$ using a similar procedure:
$$\oint_{\partial V'} \phi(x') G(x-x') \cdot dS' = -\phi(x) + \int_{V'} E(x') \cdot G(x-x') \, dV'$$
where now we use the vector Green's function $G(s) = s/4\pi |s|^3$.  The surface integral term, like in the 1d case, merely distinguishes different antiderivatives.
I've tried to answer this question in the sense of whether there's an explicit construction for an antiderivative in higher dimensional spaces. It does not, however, have the same form as that typically used in the first fundamental theorem.
