High School Question about the fundamental theorem of calculus

[Chenkel]

Hello everyone, I've been brushing up on some calculus and had some new questions come to mind.

I notice that most proofs of the fundamental theorem of calculus (the one stating the derivative of the accumulation function of f is equal to f itself) only use a limit where the derivative is approached from the right for calculating the derivative of the accumulation function.

Generally speaking I've been mostly satisfied with the proofs but I'm wondering about a derivation of this derivative when approaching x from the left. The derivative of the accumulation function regardless of if the derivative is approached from the left or from the right should be equal, right?

When can we trust a derivative approached from the right is the same as the derivative when the derivative is approached from the left?

If anyone can elucidate me on this matter I will appreciate it.

Thank you and let me know what you think!

 

[andrewkirk]

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.

 

[Chenkel]

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.

Using what you wrote I'm able to see the how the left approach derivative of the accumulation function with the mean value theorem is equal to the function in the integrand.

Thank you for the insight!