Question about the Definition of Differential Forms Question: Why does the definition of a differential form guarantee that when we do integration using differential forms, it is the same as the usual Riemann integral (before we introduce the concept of differential form)?
Example: If I want to prove Stoke’s Theorem, and I define the “differential form”, then the “differential form” may have certain rules of calculation. Then I use the “differential form” to prove the Stoke’s Theorem. 
But why is Stoke’s Theorem in the form of “differential forms” is equivalent to the usual Stoke’s Theorem (which we learn in multivariable calculus) ? 
 A: " But why Stoke’s Theorem in the form of “differential form” is equivalent to the usual Stoke’s Theorem (we learn in multivariable calculus) ?"
I can answer this question.
Take the generalized stokes theorem:
$$\int\limits_{\partial M} \omega   = \int\limits_M {d\omega } $$
The multivariable calc version of stokes theorem relates the integral over a 2-manifold to an integral over its boundary so we will want to choose omega to be a 1-form.
We can make it: $$\omega  = {F_x}dx + {F_y}dy + {F_z}dz$$
We then take the exterior derivative:
$$d\omega  = d{F_x} \wedge dx + d{F_y} \wedge dy + d{F_z} \wedge dz$$
Recalling that $$d{F_x} = {{\partial {F_x}} \over {\partial x}}dx + {{\partial {F_x}} \over {\partial y}}dy + {{\partial {F_x}} \over {\partial z}}dz$$
and using the symmetry property of differential forms
$$d{x^i} \wedge d{x^j} =  - d{x^j} \wedge d{x^i}$$
We get
$$d\omega  = \left( {{{\partial {F_z}} \over {\partial y}} - {{\partial {F_y}} \over {\partial z}}} \right)dy \wedge dz + \left( {{{\partial {F_x}} \over {\partial z}} - {{\partial {F_z}} \over {\partial x}}} \right)dz \wedge dx + \left( {{{\partial {F_y}} \over {\partial x}} - {{\partial {F_x}} \over {\partial y}}} \right)dx \wedge dy$$
Now if you look at these results you should notice $$\omega  = \vec F \bullet d\vec r$$ and $$d\omega  = \nabla  \times \vec F \bullet d\vec A$$
Which gives us the multivariable calculus version of Stokes Theorem:
$$\int\limits_{\partial M} {\vec F \bullet d\vec r}  = \int\limits_M {\nabla  \times \vec F \bullet d\vec A} $$
A: I think is by definition.
Suppose that $\omega=f(x) dx$ then
$$\int_M w \overset{\mathrm{def}}{=} \int_M f(x) dx.$$
So, you have to demostrate that the other tools of differential form, i.e. $d$, wedge product, hodge operator etc match well with this definition of integral. An example of this proof is the above answer
