Fundamental Integral theorem for functionals Given a functional $L: X\to \mathbb{R}$.  It is possible for two $x,x_h \in X$, to write
$$L(x)-L(x_h)=\int_0^1 L'(x_h+s(x-x_h))(x-x_h) ds$$
where $L'(\cdot)(v)$ is the directional in $v$. Why does this have to be tested with $x-x_h$ ?
Greetings.

Let
$\gamma(s) = x_h + (x - x_h)s, \; s \in [0, 1]; \tag 1$
we note that
$\gamma(0) = x_h, \; \gamma(1) = x_h + (x - x_h) = x; \tag 2$
then
$L(x) - L(x_h) = L(\gamma(1)) - L(\gamma(0)) = \displaystyle \int_0^1 \dfrac{dL(\gamma(s))}{ds} \; ds; \tag 3$
now by the chain rule,
$\dfrac{dL(\gamma(s))}{ds} = L'(\gamma(s)) \dfrac{d\gamma(s)}{ds}, \tag 4$
and 
$\dfrac{d\gamma(s)}{ds} = \dfrac{x_h + s(x - x_h)}{ds} = x - x_h; \tag 5$
therefore
$\displaystyle \int_0^1 \dfrac{dL(\gamma(s))}{ds} \; ds = \int_0^1 L'(\gamma(s)) \dfrac{d\gamma(s)}{ds} \; ds = \int_0^1 L'(\gamma(s)) (x - x_h) \; ds; \tag 6$
thus (3) becomes
$L(x) - L(x_h) = \displaystyle \int_0^1  L'(\gamma(s)) (x - x_h) \; ds = \int_0^1  L'(x_h + s(x - x_h)) (x - x_h) \; ds. \tag 7$
This derivation of the formula given in the text of the question shows that the directional derivative of $L(\cdot)$ in the direction of $\gamma'(s) = d\gamma(s) / ds = x - x_h$, i.e. $L'(\gamma(s)) (x - x_h)$, naturally introduces the "test" against $x - x_h$, that is the factor of $(x - x_h)$ in the integrand, by the use of the chain rule where $\gamma'(s) = x - x_h$ for all $s$; $x - x_h$ arises from the factor $\gamma'(s)$.
