Unified explanation for parametrization independence Let $\gamma:[a,b]\rightarrow\mathbb{R}^n$ be a $C^1$ curve. We have the following different kinds of line integrals:
(i) $\displaystyle\int_a^bf(\gamma(t))|\gamma'(t)|dt$ where $f:\mathbb{R}^n\rightarrow\mathbb{R}$ is a scalar field;
(ii) $\displaystyle\int_a^bF(\gamma(t))\cdot\gamma'(t)dt$ where $F:\mathbb{R}^n\rightarrow\mathbb{R}^n$ is a vector field;
(iii) $\displaystyle\int_a^bf(\gamma(t))\gamma'(t)dt$ where $n=2$ and $f:\mathbb{C}\rightarrow\mathbb{C}$ is a complex function (identify $\mathbb{C}$ with $\mathbb{R}^2$).
These are all parametrization independent, in the sense that if we replace $\gamma$ by $\eta=\gamma\circ\alpha$, where $\alpha:[c,d]\rightarrow[a,b]$ is a $C^1$ increasing diffeomorphism, then the integral does not change. I wonder if there is some "high-tech" way to explain them all at once.
My attempt: both the dot product $\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}$ and complex multiplication $\mathbb{R}^2\times\mathbb{R}^2\rightarrow\mathbb{R}^2$ are bi-linear. I feel like the following general form is true. If $B:\mathbb{R}^m\times\mathbb{R}^n\rightarrow\mathbb{R}^k$ is a bi-linear map, then $\displaystyle\int_a^bB(f(\gamma(t)),\gamma'(t))dt$ is parametrization independent for $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$; actually if $G:[a,b]\rightarrow\mathcal{L}(\mathbb{R}^n;\mathbb{R}^k)$ is continuous, then $\displaystyle\int_a^bG(t)(\gamma'(t))dt$ is parametrization independent. If this works then it unifies (ii) and (iii), but I don't see how to also cover (i). Could this be explained using differential form and density (I am still trying to understand their definitions)? Are there generalizations to higher dimensions?
 A: Let $\phi : \mathbb R^n \times \mathbb R^n \to \mathbb R^m$ be continuous with the property
$$\phi(x,\lambda y)  = \lambda \phi(x,y) \text{ for all } x, y \in \mathbb R^n \text{ and all } \lambda > 0 \tag{1}.$$
All of your integrals have the form
$$\int_a^b \phi(\gamma(t),\gamma'(t)) dt \in \mathbb R^m \tag{2}. $$
In case (i) we have $\phi(x,y) = f(x)\lvert y \rvert$, in case (ii)  $\phi(x,y) = F(x)\cdot y $ and in case (iii) $\phi(x,y) = f(x)y$ (here we use complex multiplication on $\mathbb R^2 = \mathbb C$).
The integrals $(2)$ are of course defined componentwise, i.e. the $i$-th coordinate of $\int_a^b \phi(\gamma(t),\gamma'(t)) dt$ is given as
$$\int_a^b \phi_i(\gamma(t),\gamma'(t)) dt$$
where $\phi_i : \mathbb R^n \times \mathbb R$ is the $i$-th coordinate function of $\phi$. Therefore the case $m > 1$ is covered by the case $m = 1$.
Recall that for $\mathbb R^2 =  \mathbb C$ the coordinate functions of $\phi$ are denoted as $\Re(\phi)$ and $\Im(\phi)$, thus
$$\int_a^b \phi(\gamma(t),\gamma'(t)) dt = \int_a^b \Re(\phi(\gamma(t),\gamma'(t))) dt + i \int_a^b \Im(\phi(\gamma(t),\gamma'(t))) dt  .$$
So let us consider  $\phi : \mathbb R^n \times \mathbb R^n \to \mathbb R$. If we replace $\gamma$ by $η=γ∘α$, then integration by substitution gives
$$\int_a^b \phi(\gamma(t),\gamma'(t)) dt = \int_c^d \phi(\gamma(\alpha(s)),\gamma'(\alpha(s)))\alpha'(s) ds .$$
But $\alpha'(s) > 0$ because $\alpha$ is an increasing diffeomorphism, thus
$$\int_a^b \phi(\gamma(t),\gamma'(t)) dt = \int_c^d \phi(\gamma(\alpha(s)),\alpha'(s)\gamma'(\alpha(s))) ds = \int_c^d \phi((\gamma \circ \alpha)(s),(\gamma \circ \alpha)'(s)) ds .$$
A: Not an answer, too long for a comment:
I think case $(i)$ cannot be covered by this (nice!) generalization, as it is clearly not linear with respect to $\gamma'$. I also feel like the $B$ construction would be parametrization independent but I do not think that the $G$ construction would be, as it explicitly depends on $t$.
In the lenguage of differential forms $|\gamma'|dt$ corresponds to restricting the ambient measure of $\mathbb{R}^n$ to the curve, under the specific parametrization. And indeed there are generalizations, leading to the theory of integration of manifolds.
Note that $(i)$ does not depend on the orientation of the curve (you might as well use $\gamma(-t)$, while $(ii)$ and $(iii)$ do depend on the orientation!
