Question about the fundamental theorem of calculus

andrewkirk said:

A derivative calculating by approaching from the right (positive perturbations) is called a right derivative.
A derivative calculating by approaching from the left (negative perturbations) is called a left derivative.
A function is differentiable at a point only if both of those exist and equal each other. If they are not equal, we call it semi-differentiable. An example of semi-differentiability is the absolute value function zt ##x=0##.

A semi-differentiable function may only be differentiable in one of the two directions, eg the Heaviside step function is right- but not left-differentiable.

You are correct that the FTOC proof needs to cover both left- and right-differentiability, and some proof presentations only cover the latter.

If you work through the wikipedia proof, you'll see that the only place where it effectively is restricted to right-differentiability is where it uses the mean value theorem for integration. There it assumes ##[x_1,x_1+\Delta x]## is a non-empty interval, which requires that ##\Delta x > 0##.

So another paragraph is needed that proves the same intermediate result

$$\int_{x_1}^{x_1 + \Delta x} f(t) \,dt = f(c)\cdot \Delta x$$

is also true when ##\Delta x <0## (it is trivially true when ##\Delta x = 0##).

By convention of the notation for integrals, this is the same as proving that:

$$\int^{x_1}_{x_1 + \Delta x} f(t) \,dt = f(c)\cdot (-\Delta x)$$

so that is easily proven just by using the mean value theorem for integration on the interval ##[x_1+\Delta x, x_1]##.

The rest of the proof then follows without adjustment, simply allowing that now ##\Delta x## may be positive, negative or zero.