Flux from cube: Relation between the vector field to that of the opposite side

This is an exerpt from Feynman's lecture:

. . .We wish to find the flux of a vector field $C$ through the surface of the cube. . . First consider the face having edges $\Delta y \quad \& \quad \Delta z$. The flux outward on this face is the negative of the x-component of $C$ ,integrated over the surface. This flux is $-\int C_x dy dz$ . Since, we are considering a small cube, we can approximate this integral by the value of $C_x$ at the center of the face- which we call point (1) - multiplied by the area of the face. Therefore flux out of the face $$= - C_x(1) \Delta y \Delta z$$. Similarly , flux from the opposite face $$= C_x(2) \Delta y \Delta z$$. Now, $C_x(1) \quad \& \quad C_x(2)$ are, in general, slightly different. If $\Delta x$ is small enough we can write $$C_x(2) = C_x(1) + \dfrac{\partial C_x}{\partial x} .\Delta x$$. There are, of course , more terms but they will involve ${\left(\Delta x \right)}^2$ and higher power & so will be negligible. . .

My questions are :

1) Why are $C_x(1)$ & $C_x(2)$ slightly different? What did Feynman want to mean?

2) Both the values can be related by Fundamental theorem of Calulus ie. their difference is equal to the rate of change of $C$ multiplied by $\Delta x$. From where does any higher power of $\Delta x$ ie. ${\left(\Delta x \right)}^2$ ,as told by Feynman, emanate?

1) They are slightly different in the sense that the function $C(x)$ (or $\vec C(\vec r)$) has different values at different places in space, and so even though the two opposite faces of the cube are very close the value of the function $C$ at the two faces will be slightly different.
2) Only if the two points are infinitesimally close or the function $C$ is linear can you say exactly that the difference is equal to the rate of change of $C$ times $\Delta x$, but if neither of these is the case we have to allow for it by, for example as Feynman has done, using a Taylor series expansion of the unknown $C$ about a certain point, here the point labelled (1). This will give higher orders in $\Delta x$ as he has stated.
• The fundamental theorem of calculus is not really relevant here, and you can only say that the change in $C$ is exactly the same as the gradient times the change in $x$ in the limit when the change in $x$ goes to 0 (i.e. infinitesimal), OR if you happen to know that the function is linear, in which case it doesn't matter about the size of $\Delta x$. We allow for the difference to be finite first and use a Taylor expansion (example later) to capture (in principle) all the dependence on $\Delta x$, and then take the limit, which results in a linear form. – danimal Mar 31 '15 at 12:14
• The Taylor expansion of a function $f(x)$ about a point $x=a$ is given by the (infinite) series: $$f(x) = f(a)+f'(a)(x-a) + {f''(a)\over 2!}(x-a)^2+{f'''(a)\over 3!}(x-a)^3+...$$ For this problem we use $f=C$ and $x-a=\Delta x$. – danimal Mar 31 '15 at 12:19