Let $ [a,b]\in \mathbb{R} $ be a closed interval. Let $ \omega\in \Omega^1([a,b]) $ be a smooth 1-form (by this we mean that $ \omega $ is the restriction to $ [a,b] $ of a smooth 1-form defined on some open interval containing $ [a,b] $.): this can be written as $$ \omega(t)=g(t)d $$ for some smooth function $ g $ on $ [a,b] $. Define $$ \int _{[a,b]} \omega =\int_{a}^{b} g(t)dt $$ Let $ r:[c,d]\to [a,b] $ be a smooth map such that $ r(c)=a $ and $ r(d)=b $. I want to calculate $ r^*\omega $ and show that $$ \int _{[c,d]} r^*\omega = \int _{[a,b]} \omega. $$ The problem is that I don't understand what $ r^* $ is, and how it acts on $ \omega $. I think it's domain is $ \Omega^1([a,b]) $, but I don't know what it does with these 1-forms. Can someone help me out?
EDIT:
I think I got it now. Since $ \omega(t)=g(t)dt $, the pull-back $ r^* $ on $ \omega $ is $$r^*\omega=r^*\omega(t)=r^*g(t)dt=g(r(t))dt$$ Hence, \begin{align*} \int_{[c,d]}r^*\omega&=\int_{c}^{d}g(r(t))dt\\ &=\int_{a}^b g(t)dt\\ &=\int_{[a,b]} \omega \end{align*} where I have used that $r^*\omega=g(r(t))dt$, $ r(c)=a $ and $ r(d)=b $.
Is it correct?