Level set representation of 1-forms in univariate functions This question is in connection to an excellent youtube playlist by @eigenchris on Tensor Calculus.

The idea of a differential $\mathrm d$ operator as turning a scalar field $f$ ($0$-form) into a covector field $\mathrm df$ ($1$-form) is nicely explained in this presentation, based on contour maps of a function $f(x,y).$ The scalar field is the value of $f$ for every pair in $(x,y),$ and the covector field is represented by the level set curves. Given a vector at any point, the covector field would produce the directional derivative, $\mathrm df(\vec v).$
Similarly the actual $x$ and $y$ independent variables are scalar fields with each point attached to their own values:

The explanation is clear within the confines of a multivariable function, although clearly the level sets representing $\mathrm dx$ and $\mathrm dy$ on the right column of the plot above are completely random vertical and horizontal lines.
In Wikipedia a scalar field is defined as 

a scalar field on a region U is a real or complex-valued function or distribution on U.

and a region as

a subset of $ \mathbb {R} ^{n}$ or $\mathbb {C} ^{n}$ that is open (in the standard Euclidean topology), connected and non-empty. 


Presumably, then, the construct applied to a unvariable function $f(x)$ would be as follows: The scalar field $x$ is simply the $x$ axis with $\mathrm dx$ best represented by a homogeneously changing color gradient on a line from blue to red:

understanding the real line (or the domain of the function on the real line) as the region $U$ (?). Here $y$ would not be a smooth color gradient as in the graph above, because it will be dependent on $x$ through the action of $f.$
The question is:

How to picture (draw or interpret) the level sets of the covector field $\mathrm df$ (in this univariate setting, I presume it is the same as $\mathrm dy$) in this case, where we can't resort to contour plots?

Would it make sense to picture it like this...

in the case of

for example?

And what would be the equivalent to the vector fed to $\mathrm df$ to obtain the directional derivative in the univariate setting? The "velocity" vector, $\vec v,$ would necessarily have a single dimension in $x,$ but it can't be $\mathrm dx$ (a covector). If it is $\frac{\partial}{\partial x},$ then $\mathrm df\left(\frac{\partial}{\partial x}\right)=\mathrm dy\left(\frac{\partial}{\partial x}\right)=\frac{\partial y}{\partial x}=\frac{\mathrm dy}{\mathrm dx}$ - the derivative of $y$ wrt $x.$ 


For example, the function $f(x) = x^3,$ which in the interval $[1,10],$ has as many level sets as real numbers in the interval $[1,10].$ Therefore the idea of "counting the level sets pierced by a vector" when integrating from $1$ to $10,$ as suggested on this slide on the same series of presentations:

would presumably translate into $\color{red}{f(x)dx=x^3 dx},$ with the number of counted level sets pierced by the arrow being infinite, as we can clearly increase the number of level sets represented in the diagram below (from $5$ to $10$ to $20$ from left to right):

The actual values at each level set would eventually reproduce the entire $y=x^3$ function in that segment, and multiplying each value, $\color{red}{x^3},$ times a small interval of $x,$ i.e $\color{red}{dx},$ would indeed bring us back to the Riemann integral. But that clearly defeats the idea of just counting the level sets pierced by the arrow.

So what would be a good way to reconcile these two ideas?

The $\int_1^{10}x^3 dx \approx 2499,$ while adding the level set lines on the plot to the left results in crossing $6$ level sets, each separate by $166.5,$ which would result in $6\times 166.5=999.$

Wrap up after EDIT of @eigenchris answer:
x = [1:.1:10.1];               % From 1 to 10.1 at intervals of 0.1
y = [1:.1:10.1];               % From 1 to 10.1 at intervals of 0.1
[xx, yy] = meshgrid (x, y);    % Cartesian product of x and y
z = 1/4 * xx.^4;               % F(x) = x^4
max = 1/4 * (11)^4;            % Making the last level set coincide with F(x = 10).
v = [100:100:max];             % Level sets at intervals of 100.
[C,h] = contour(xx,yy,z,v,'ShowText','on'); % Contour plot
set(gca,'ytick',[]);           % Avoids labels on the y axis 
colormap hsv;                  % Color scheme.
xlabel ("x");                  % x label
ylabel ("");                   % y label blank (univariate function).
title ("F = 1/4 x^4");         % Plot title


How do you go from here to $2499.75$ piercing the level sets? In the plot the level sets are separated by $100$ and there are $25$ level sets, resulting in $100 \times 25 = 2500.$ Since we started at $F(1)=\frac 1 4 (1)^4 =\frac 1 4,$ we end up with $2500 - \frac 1 4 = 2499.75.$
 A: I agree with everything in the first half of your question (before the line break). The scalar fields you have drawn look correct.
I disagree with the second half. First $\int_1^{10} x^3 dx = [\frac{1}{4}x^4]^{10}_1 = \frac{1}{4} [10000-1] = 2499.75$ 
.

EDIT:
I made a mistake in my previous answer (I apologize... it's been a while since I've thought about this).
So the integrand you have defined is $f(x)dx = x^3dx$.
Now, according to the differential expansion law, $dF = \frac{dF}{dx}dx$. So in your case: $$dF = \frac{dF}{dx}dx = f(x)dx = x^3dx$$
This implies that :
$$F(x) = \frac{1}{4}x^4 + c$$
(We can throw away the $+c$ .)
Since we are integrating $dF$, NOT $df$, we want to plot the level sets of $F(x) = \frac{1}{4}x^4$, NOT the level sets of $f(x) = x^3$.
I believe this should lead to the correct answer of $2499.75$.
A: The covector stacks (in green below) correspond to stack representation of $x^3 dx,$ as a vector field below, $\begin{bmatrix}x^3\\0 \end{bmatrix},$ using this applet. Please note that the limits of integration could not be made more precise even slowing down the mouse, nor could the window be extended to integrate from $1$ to $10$ as in the OP, so I integrated from $\sim 1$ to $\sim 6$ exactly as the the circulation along a straight line, which parallels exactly the line integrals through a vector field:

Notice how the same plot can be displayed as a traditional vector (arrow) field:

The idea is exactly like in the fundamental theorem of line integrals expression
$$\int_a^b \nabla f(\vec r(t))\cdot \vec r'(t) dt$$
with $\vec r'(t) = \frac{d}{dx}x=1$ and $\nabla f(\vec r(t)) \equiv \frac{ d}{ dx}(\frac 1 4 x^4)=x^3.$
If we consider $f(x,y)=xy,$ the gradient is a vector field $\begin{bmatrix} y\\x \end{bmatrix},$ the integral along a curved segment between two points can also bee seen as a covector field:

The reason why it worked in the last edit of the OP with the level sets of $\frac 1 4 x^4$ as suggested in the @eigenchris answer it because it simply a way of summing the results of $\int_{a_1}^{a_n} x^3 dx$ in the partitions of the interval from $a_1=1$ to $a_n=10$ in the plot; however, it doesn't provide the needed insight.
Another way to think about it is with $f=\frac 1 4 x^4$ being a $0$-form, and $df=x^3 dx$ being its $1$-form (represented as green stacks): the level sets of the original function $f= \frac 1 4 x^4$ are $df=\frac {df}{dx}  dx= x^3 dx,$ which is concordant with the last edit in the OP reflecting the accepted answer: the level sets are of $\frac 1 4 x^4,$ while the stacks (covector field) vary at each point according with the derivative $x^3.$
