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I start with 1-form $\omega=f\,dx$ on $\left[0,1\right]$ where $f\left(0\right)=f\left(1\right)$ and a $g:\left[0,1\right]\to R$ with $g\left(0\right)=g\left(1\right)$ and I want to integrate $\omega-\lambda \, dx=dg$ on $\left[0,1\right]$. So I write

$\int\limits_0^1 \omega-\lambda \,dx = \int\limits_0^1 \left(f-\lambda\right)\,dx =\int\limits_0^1 f\,dx-\lambda$ and $\int\limits_0^1 \omega-\lambda \, dx =\int\limits_0^1 dg = g\left(1\right)-g\left(0\right)=0$ and find out that $$\lambda=\int\limits_0^1 f\,dx.$$

Is this correct? What would happen if I parametrized the path from $0$ to $1$ another way? Is $\int\limits_0^1 dg=g\left(1\right)-g\left(0\right)$ legal - I'm using dg as a differential form and maybe I'm supposed to check some precondition?

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up vote 1 down vote accepted

The theorem in vector calculus said: For an exact 1-form $\omega$, $C$ and $C'$ are two parametrized curves with the same starting point and ending point, then $$ \int_C \omega = \int_{C'} \omega $$ And in one dimension case, the boundary between the calculus of differential forms and the regular calculus we learned in college is kinda blurry, the exactness and closedness of a differential form start to play a vital role in 2-dimensional spaces, where Green's theorem, Stokes theorem, etc kick in.

Now in your calculation you made the assumption that $\lambda$ is a real number, also $\displaystyle \int_0^1 dg = g\Big\vert^1_0$ holds when $g$ has well-defined point values and a well-defined differentiation, which means $g\in C^1([0,1])$ would suffice, and you also use the fact that $fdx$ is an exact 1-form, which implies the antiderivative of $f$ exists.

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