Using differential forms to define line integrals

Question

I saw some similar questions and answers but they often included some information or mathematics I haven't learned/read so I'm hoping to get a somewhat simpler answer.

Let $\beta:[a,b] \to \mathbb{R}^3$ be a smooth curve. So the three component functions of $\beta(t)$ are all smooth. Let $p = \beta(a)$ and $q=\beta(b)$. Show that for some function $f:\mathbb{R}^3 \to \mathbb{R}$,

$$\int_{a}^{b} \beta^*df=f(q)-f(p)$$

$\beta^*df$ is the pullback though I'm not so sure what sort of a differential form $df$ is. Now, $\beta^*df=d(\beta^*f)=d(f \circ \beta)$ but I'm not sure where to go from here. I suspect there is a chain rule though.

Thank you in advance.

Answer 1

Chain rule is not needed. Using

$$\beta^* df = d(f\circ \beta) = (f\circ \beta)'(t)dt,$$

we have

$$\int_a^b \beta ^* df = \int_a^b (f\circ \beta)'(t) dt = (f\circ \beta) (b) - (f\circ \beta)(a) = f(q) - f(p).$$

Answer 2

From the first fundamental theorem of calculus: $$\int_{x_1}^{x_2} dF(x) = F(x_2)-F(x_1)$$ Your integral becomes: $$\int_{\beta(a)}^{\beta(b)} df(\beta)= f(\beta(b))-f(\beta(a))= f(q)-f(p)$$