The exercise $14.13.21$ gives us 2 equivalent functions, where $Y(t) = X[u(t)]$, for $c \le t \le d$. The function $u$ has a continuous derivative in $[c, d]$, and we need to prove that: $\tag{1}\int_{u(c)}^{u(d)}{\| X'(u) \|du} = \int_{c}^{d}\|Y'(t)\|dt$
I started as follows. $Y'(t) = X'[u(t)]u'(t)$. Therefore:
$$\int_{c}^{d}\|Y'(t)\|dt = \int_{c}^{d}\|X'[u(t)]u'(t)\|dt = \int_{c}^{d}\sqrt{[X'[u(t)]u'(t)]^2}dt = \int_{c}^{d}\sqrt{[X'[u(t)]^2} |u'(t)|dt = \int_{c}^{d}\|X'[u(t)]\||u'(t)|dt$$
If we take that $u' \ge 0$, we have equality $1$, but if it's negative, we get: $\tag{2}\int_{u(c)}^{u(d)}{\| X'(u) \|du} = -\int_{c}^{d}\|Y'(t)\|dt$
Is there any reason for $u$ to be non-negative? Did I make a mistake somewhere?
Thanks!