What kind of Taylor expansion is this? Let $b \colon  \mathbb{R}^n \to \mathbb{R}^n$ be $\mathcal{C}^{1,1}(\mathbb{R})$, i.e. its derivative $b_x$ is Lipschitz.
Let $0 \leq \lambda \leq 1$ and define $x^\lambda=\lambda x_0+(1-\lambda)x_1$ for $x_0,x_1 \in \mathbb{R}^n$.
In [Yong, Jiongmin, and Xun Yu Zhou. Stochastic controls: Hamiltonian systems and HJB equations] p.188 (4.22) I find the following equality
\begin{aligned}
&\left|\lambda b\left( x_{1}\right)+(1-\lambda) b\left(x_{0}\right)-b\left( x^{\lambda}\right)\right| \\
&=\Big | \lambda \int_{0}^{1} b_{x}\left( x^{\lambda}+\theta(1-\lambda)\left(x_{1}-x_{0}\right)\right) d \theta(1-\lambda)\left(x_{1}-x_{0}\right) \\
&\quad+(1-\lambda) \int_{0}^{1} b_{x}\left( x^{\lambda}+\theta \lambda\left(x_{0}-x_{1}\right)\right) d \theta \lambda\left(x_{0}-x_{1}\right) \Big |
\end{aligned}
Where does it come from?
 A: You have a typo: it should be
$$ x^\lambda := (1-\lambda)x_0+\lambda x_1 = x_0 + \lambda(x_1-x_0) = x_1 + (1-\lambda)(x_0-x_1)$$ so that $x^0=x_0$ and $x^1=x_1$.
Indeed, the book gives my definition and not yours on page 187.
With this fix, it is just the fundamental theorem of calculus + chain rule.
Note that for any constant $C$,
$$b(x+C)-b(x) = \int_0^1 \frac{d}{d\theta} \Big(b(x+C\theta)\Big) d\theta= \int_0^1 b_x(x+C\theta) Cd\theta$$
For the first integral we consider $b(x_1) - b(x^\lambda)$, with $x:=x^\lambda$, $x^1 = x^\lambda+C$ with
$$C=-(1-\lambda)(x_0-x_1)=(1-\lambda)(x_1-x_0).$$
Then the formula immediately gives
\begin{align}
 b(x_1) - b(x^\lambda) 
%&= \int_0^1 b_x(x^\lambda - (1-\lambda)(x_0-x_1)\theta)  d\theta \times -(1-%\lambda)(x_0-x_1)
%\\
&= \int_0^1 b_x(x^\lambda + (1-\lambda)(x_1-x_0)\theta)  d\theta \times (1-\lambda)(x_1-x_0)
\end{align}
Similarly for the second term: set $x:=x^\lambda$, $x_0=x^\lambda+C$ where $C = -\lambda(x_1-x_0)$. This gives
\begin{align}  b(x_0) - b(x^\lambda) 
&=\int_0^1 b_x(x^\lambda + \lambda(x_0-x_1)\theta)d\theta \times  \lambda(x_0-x_1)
\end{align}
Now $\lambda$ times first one plus $(1-\lambda)$ times second gives the identity in question.
