I am following a book on Rough Paths (although Rough Paths are not relevant to this question) and would like to integrate the function of a path against that path. I have the following set up:
$X = (X^1,\cdots,X^d): [0,T] \rightarrow \mathbb{R}^d$ is a path and $F:\mathbb{R}^d \rightarrow \mathbb{R}$ is a smooth function. Also $s,t \in [0,T]$ such that $s < t$. We choose an arbitrary $u \in [s,t]$.
I would like to compute the following integral:
$$ \int^t_sF(X_u)dX_u. $$
The book begins by Taylor expanding $F(X_u)$ and then integrating this expansion. This is how its written out in the book:
My issue comes with the first equality. If I do the Taylor expansion, I get
$$ F(X_u) = F(X_s) + DF(X_s)(X_u-X_s)+ \cdots $$
where $DF$ is the gradient of $F$. Then integrating both sides, I get
$$ \int^t_sF(X_u)dX_u = F(X_s)\int^t_sdX_u + DF(X_s)\int^t_s(X_u-X_s)\otimes dX_u+ \cdots $$ $$ \Leftrightarrow \int^t_sF(X_u)dX_u = F(X_s)(X_s-X_t)+ DF(X_s)\int^t_s(X_u-X_s)\otimes dX_u+ \cdots $$
My question is, how am I meant to recover the first equality in the proof in the book?
Many thanks