Old Peano theorem (demonstration is missing details) Let $A, B :[a,b]\subset\mathbb{R}\to\mathbb{R}^2$ be two functions of class $C^{1}([a,b])$ such that two segments (or intervals) $[A(t_1),B(t_1)]$ and $[A(t_2), B(t_2)]$ never intersect for $t_1, t_2\in [a,b], \ t_1\neq t_2$. Prove that the area of the region described by the segment $[AB]$, $R=\{[A(t),B(t)] \ |\ t\in[a,b]\}$, can be computed by the formula:
$$\text{Area}(R)=\int_{a}^b \displaystyle\Big |\det\begin{pmatrix}x(t) & y(t) \\ x'(t) & y'(t) \end{pmatrix}\Big |\ \text{d}t$$
where $B(t)-A(t)=(x(t),y(t)), \ t\in[a,b]$.
P.S. For $U,V\in\mathbb{R}^2$, by the segment $[U,V]$ I mean the set $\{\lambda U+(1-\lambda)V,\ \lambda\in [0,1]\}$.
Question: Can there be found a similar formula in the space? (when $A, B :[a,b]\subset\mathbb{R}\to\mathbb{R}^3)$
I found this propetry in a book of Giuseppe Peano, Applicazioni geometriche del calcolo infinitesimale, 1887, page 238, but I see that his proof is not complete, and I found some really hard problems in trying to fullfil it. You can find the book here: https://archive.org/details/applicazionigeo01peangoog
Note that this property is a nice generalization of Mamikon Theorem (see the book New Horizons in Geometry,
by Tom M. Apostol or simply search on the internet), found long time ago by Peano. This is why I'm asking for a generalization in $\mathbb{R}^3$, because Mamikon theorem works for space curves too. For details, see that article, pages 27-28: http://dolecki.perso.math.cnrs.fr/Peano=vulgarization_1307.pdf
 A: $\newcommand{\bR}{\mathbb{R}}$
I think that there is something wrong with the formula you wrote.  Suppose that
$$ A(t)=(t,0),\;\;B(t)=(t,1),\;\;\;t\in [0,1]. $$
In this case the segment $[A(t), B(t)]$ is the vertical segment from $(t,0)$ to $(t,1)$. The region $R$ is thus the unit square $[0,1]\times [0,1] \subset\bR^2$. We have $x=0$, $y=1$ so the integrand you wrote is $0$.
Here is one way to compute the area  of the region $R$ in the general case.  We introduce  new coordinates $(t,s)$ on $R$ as follows. Let
$$A(t)= ( p(t), q(t) )\;\;\; t\in [a,b]. $$
Every point $(u,v)$ in $R$ can be uniquely described by the parameters $t,s$
$$ u= p(t)+s x(t),\;\;v=q(t)+s y(t),   t\in [a,b], s\in [0,1]. $$
Geometrically  to locate a point in $R$ we first need to understand on which of the segments $[A(t), B(t)]$ it lies. This is the role of the parameter $t$. The parameter $s$    fixes the position of the point  on this segment.
We have
$$ du\wedge dv= \bigl(\dot p dt+s \dot{x} dt+xds\bigr)\wedge \bigl(\dot{q}dt+s\dot{y} dt +y ds\bigr) $$
$$=\bigl(\; (\dot{p}+s\dot{x})y-(\dot{q}+s\dot{y})x\;\bigr) dt\wedge ds $$
$$ =\underbrace{\bigl(\dot{p}y-\dot{q}x +s(\dot{x}y-\dot{y}x)\;\bigr)}_{\rho(t,s)} \;dt\wedge ds. $$
We deduce that 
$$ {\rm Area}\;(R) =\pm\int_{[a,b]\times [0,1]}\rho(t,s) dt dt= \pm\int_a^b\bigl(\;(\dot{p}y-\dot{q}x ) +\frac{1}{2}(\dot{x}y-\dot{y}x)\;\bigr). $$
Note that in the special case described earlier we have $p=t$, $t=0$, $x=0$, $y=1$ $a=0$, $b=1$ and the above formula yields the correct answer.
Perhaps the generalization  you are looking for is the coarea formula. It states that for a  Riemann manifold $(M, g)$ and  $f: M\to\bR$ we have
$$ {\rm vol}\;(M,g) =\int_{\bR} \left(\int_{f^{-1}(t)} \frac{1}{|\nabla^g f(p)|} dV_{f^{-1}(t)}\right) dt. $$
Above $f^{-1}(t)$ is a smooth  hypersurface for  almost all $t$ and $dV_{f^{-1}(t)}$ is the volume element on this hypersurface  defined by the  induced metric.  In the case at  hand, use the function $f:R\to\bR$ described in the coordinates  by $f(t,s)=t$.
Edit 1. In $n$ dimensions $A(t)=(a^1,\dotsc, a^n(t))$, $X(t)=B(t)-A(t)=(x^1(t),\dotsc, x^n(t))$. Parametrize  the surface $R$ as above
$$ u^k=a^k(t)+sx^k(t),\;\;k=1,\dotsc, n,\;\;t\in [a,b],\;\; s\in [0,1]. $$
We have
$$du^k =(\dot{a}^k+s\dot{x}^k) dt +x^kds,\;\; k=1,\dotsc, n, $$
$$ (du^k)^2=  (\dot{a}^k+s\dot{x}^k)^2 dt^2 +2(\dot{a}^k+s\dot{x}^k)x^k dtds+(x^k)^2 ds^2,\;\;k=1,\dotsc, n. $$
The induced metric on $R$  is then
$$\sum_k (du^k)^2= \underbrace{\left(\sum_k(\dot{a}^k+s\dot{x}^k)^2 \right)}_{E}\,dt^2+ 2\;\underbrace{\left(\sum_k (\dot{a}^k+s\dot{x}^k)x^k\right)}_{F}\;dtds+\underbrace{\left(\sum_k (x^k)^2\right)}_{G}\; ds^2. $$
The area element on $R$ is then
$$dS=\sqrt{EG-F^2} dt ds. $$
