(i) Let $\omega: \mathbb R^n \to \Lambda_1(\mathbb R^n)$ be a continuous differential 1-form. Suppose that for every smooth embedding $\Phi: \mathbb R \to \mathbb R^n$,
$\begin{aligned} \int_{\Phi|_{[0,1]}} \end{aligned} \omega=0. $
Prove that $\omega = 0$.
(ii) Let $f: \mathbb R^n \to \mathbb R$ be a continuously differentiable function. Define $\omega:= df: \mathbb R^n \to \Lambda_1(\mathbb R^n).$ Prove that for every smooth embedding $\Phi: \mathbb R \to \mathbb R^n$,
$\begin{aligned} \int_{\Phi|_{[a,b]}} \end{aligned}\omega=f(\Phi(b))-f(\Phi(a)). $
(iii) Let $f: \mathbb R^n \to \mathbb R$ be a continuously differentiable function and let $\omega: \mathbb R^n \to \Lambda_1(\mathbb R^s)$ be a continuous differential 1-form. Suppose that for every smooth embedding $\Phi:\mathbb R \to \mathbb R^n$,
$\begin{aligned} \int_{\Phi|_{[0,1]}} \end{aligned} \omega=f(\Phi(1))-f(\Phi(0)). $
Prove that $\omega = df$.
These are probably very simple I'm sure I'm overthinking each of these, but I'm quite stuck. I'm guessing I would use Stokes Theorem, I considered using the fact I can prove
For any continuous differential 1-form $\omega$ on $\Phi(U)$ and any smooth embedding $\Psi:\mathbb R \to U$, $\begin{aligned} \int_{\Phi\omicron\Psi|_{[0,1]}} \omega= \int_{\Psi|_{[0,1]}}\Phi^*(\omega)\end{aligned}$.
But this may be completely wrong. Any help/ideas appreciated!
(i) Let us write $$ \omega = \omega_1(x^1, \dots , x^n) dx^1 + \dots + \omega_n (x^1, \dots , x^n) dx^n,$$ where $\omega_1, \dots , \omega_n $ are smooth functions $\mathbb R^n \to \mathbb R$.
To show, for instance, that $\omega_1$ vanishes at the point $(3, \dots , 0)$, you could use the fundamental theorem of calculus: $$ \omega_1(3, \dots, 0) = \frac d {dx^1} \left(\int_0^{x^1} \omega_1(u, x^2, \dots, x^n) du \right)|_{x^1 = 3} = \frac d {dx^1} \left( \int_{[(0, \dots, 0),(x^1, \dots , 0)]} \omega \right)|_{x^1 = 3} = 0. $$ Here, the notation $\int_{[(0, \dots, 0),(x^1, \dots , 0)]} $ means that we are integrating over the straight line segment from $(0, \dots, 0)$ to $(x^1, \dots , 0)$. This kind of integral also vanishes because it says so in the question!
(ii) This is the special case of Stoke's theorem that applies to line integrals. You may also have heard people call it the "gradient theorem" of vector calculus.
(iii) Apply (i) to the form $\omega - df$, using (ii) to evaluate integrals for $df$.