Let $v:=u+x\sigma_x+y\sigma_y+z\sigma_z$ be a multi-vector.
How does one take the derivate with respect to x?
$$ \frac{dv}{dx} = \sigma_x $$
- correct?
How does one take the total derivative?
$$ dv=du+\sigma_xdx+\sigma_ydy + \sigma_zdz $$
- correct?
How does one take the integral with respect to dx? $$ \int v dx = \int udx+ \int x\sigma_xdx+\int y\sigma_ydx+ \int z\sigma_zdx\\ =ux+\frac{1}{2}x^2\sigma_x+yx\sigma_y+zx\sigma_z+c $$
- correct?
How does one take the integral over the volume of the multi-vector? $$ V=\int (u+x\sigma_x+y\sigma_y+z\sigma_z)(\sigma_x \sigma_y \sigma_z)dxdydz\\ =\int (u\sigma_x \sigma_y \sigma_z+x\sigma_x\sigma_x \sigma_y \sigma_z+y\sigma_y\sigma_x \sigma_y \sigma_z+z\sigma_z\sigma_x \sigma_y \sigma_z)dxdydz\\ =\int (u\sigma_x \sigma_y \sigma_z+x \sigma_y \sigma_z-y\sigma_x \sigma_z+z\sigma_x \sigma_y)dxdydz\\ =uxyz\sigma_x \sigma_y \sigma_z+x \sigma_y \sigma_z-y\sigma_x \sigma_z+z\sigma_x \sigma_y+c\\ =uxyz\sigma_x \sigma_y \sigma_z+x \sigma_y \sigma_z+y\sigma_z \sigma_x+z\sigma_x \sigma_y+c $$
- correct?
The last one has me confused. I do not understand why the volume depends on the area terms $x \sigma_y \sigma_z+y\sigma_z \sigma_x+z\sigma_x \sigma_y$ and on a scalar $c$ and pseudoscalar $uxyz\sigma_x \sigma_y \sigma_z$. Why is the volume not a scalar?
Those are all correct. For the last, you are free to use a scalar volume element, should you choose.
There are however, some circumstances where you'd want a trivector volume element as you have done. In particular, the expression of the fundamental theorem of calculus for volume integrals can be naturally expressed using a trivector volume element.
That theorem, for $\mathbb{R}^{3}$, stated roughly, is:
Given mulitvector functions $ F, G$, a trivector valued volume element $ d^3 \mathbf{x} = d\mathbf{x}_1 \wedge d\mathbf{x}_2 \wedge d\mathbf{x}_3 = du dv dw\, \mathbf{x}_u \wedge \mathbf{x}_v \wedge \mathbf{x}_w $, where $\mathbf{x}_a = \partial \mathbf{x}/\partial a$, one has:
$$\int_V F d^3\mathbf{x} \stackrel{ \leftrightarrow }{\boldsymbol{\nabla}} G= \int_{\partial V} F d^2 \mathbf{x} G,$$ where $ \stackrel{ \leftrightarrow }{\boldsymbol{\nabla}}$ acts bidirectionally on $F,G$, and $ \partial V $ is the boundary of the volume $ V $, and the surface integral over the $ d^2 \mathbf{x} $ bivector area element has a counterclockwise orientation.
Spelled out more explicitly, that surface integral has the form
$$\begin{aligned}\int_{\partial V} F d^2 \mathbf{x} G&=\int {\left.{{\left( {F d\mathbf{x}_1 \wedge d\mathbf{x}_2 G} \right)}}\right\vert}_{{\Delta w}} \\ &\qquad +\int {\left.{{\left( {F d\mathbf{x}_2 \wedge d\mathbf{x}_3 G} \right)}}\right\vert}_{{\Delta u}} \\ &\qquad +\int {\left.{{\left( {F d\mathbf{x}_3 \wedge d\mathbf{x}_1 G} \right)}}\right\vert}_{{\Delta v}}.\end{aligned}$$
The statics solution of Maxwell's equation is a nice physical example of an an oriented multivector integral. Given a multivector electromagnitic field $$ F = \mathbf{E} + I c \mathbf{B},$$ and a multivector current $$J = \frac{1}{{\epsilon}} \left( { \rho - \frac{1}{{c}} \mathbf{J} } \right),$$ the enclosed multivector current is related to the field by $$\int_{\partial V} d^2\, \mathbf{x} F = \int_V d^3 \mathbf{x} \, J.$$
Of course, you can express this using scalar integrals too $$\int_{\partial V} dA\, \hat{\mathbf{n}} F = \int_V dV\, J,$$ where the normal is related to the surface area element by the unit pseudoscalar using $d^2\, \mathbf{x} = dA I \hat{\mathbf{n}} $, and where the volume element satisfies $d^3 \, \mathbf{x} = I dV$.
Both of these multivector integral variations pack a lot of information, as they can be decomposed into scalar, vector, bivector and trivector grades, and yield the conventional "enclosed current relations":
$$\begin{aligned}\int_{\partial V} dA\, \hat{\mathbf{n}} \cdot \mathbf{D} &= \int_V dV\, \rho \\ \int_{\partial V} dA\, \hat{\mathbf{n}} \times \mathbf{H} &= \int_V dV\, \mathbf{J} \\ \int_{\partial V} dA\, \hat{\mathbf{n}} \times \mathbf{E} &= 0 \\ \int_{\partial V} dA\, \hat{\mathbf{n}} \cdot \mathbf{B} &= 0.\end{aligned}$$