FTC with Discontinuous Functions: Continuity vs. Differentiability

<<>>

FTC with Discontinuous Functions: Continuity vs. Differentiability

TL;DR

The Fundamental Theorem of Calculus requires continuity to link integration and differentiation properly—if a function jumps or breaks, the theorem's guarantees fall apart. Differentiability is even stricter than continuity: a function can be continuous everywhere but fail to have a derivative at certain points (like a sharp corner). Understanding when and why these conditions matter is the key to using calculus correctly.

1. The Mental Model

Here's the thing: the FTC promises that if you differentiate an integral, you get back the original function—but only under certain conditions. Continuity is the minimum requirement; differentiability is stronger. Think of continuity as "no jumps" and differentiability as "no kinks or sharp turns." The FTC works best when your function is smooth enough (continuous, ideally differentiable). When it isn't, the theorem's conclusion may fail, and you need to know why before you trust your answer.

2. The Core Material

2.1 The Fundamental Theorem and Continuity

The Fundamental Theorem of Calculus comes in two parts. Part 1 says: if $f$ is continuous on $[a, b]$, then the function $F(x) = \int_a^x f(t) \, dt$ is differentiable on $(a, b)$, and $F'(x) = f(x)$.

Why does continuity matter? When you compute the derivative of $F$ using the limit definition, you're asking: does $\frac{F(x+h) - F(x)}{h}$ approach a single value as $h \to 0$? This limit exists and equals $f(x)$ precisely because $f$ is continuous at $x$. If $f$ jumps at some point, the numerator doesn't shrink smoothly, and the limit may not exist.

Example: Let $f(x) = \begin{cases} 1 & \text{if } x < 0 \\ 2 & \text{if } x \geq 0 \end{cases}$

Define $F(x) = \int_0^x f(t) \, dt$. For $x > 0$, we have $F(x) = 2x$, so $F'(x) = 2$. But $f(0) = 2$, and as $x \to 0^-$, $F'(x)$ would be $1$ (from the left), not $2$. The discontinuity at $x = 0$ breaks the FTC's conclusion at that point.

2.2 Continuity vs. Differentiability

Continuity means no jumps: $\lim_{x \to a} f(x) = f(a)$. A continuous function is "connected"—you can draw it without lifting your pencil.

Differentiability means the function has a well-defined slope (derivative) everywhere. It's stronger: every differentiable function is continuous, but not every continuous function is differentiable.

Key fact: a function can be continuous at a point but fail to be differentiable there. The classic example is a corner or cusp.

Example: $f(x) = |x|$ at $x = 0$.
- Continuity: $\lim_{x \to 0} |x| = 0 = f(0)$. ✓ It's continuous.
- Differentiability: From the left, the slope is $-1$. From the right, the slope is $+1$. The left and right derivatives don't match, so $f'(0)$ does not exist. ✗ It's not differentiable there.

When you compute $\int_0^a |x| \, dx$ and then differentiate, you get back $|x|$ everywhere except at $x = 0$, where the FTC technically doesn't guarantee $F'(0) = f(0)$ (though in this case it does: $F'(0) = 0$ and $f(0) = 0$). The theorem doesn't break; it just doesn't promise differentiability at non-differentiable points of $f$.

2.3 What Happens When Continuity Fails

If $f$ has a jump discontinuity, the FTC fails. Let's be concrete.

$$f(x) = \begin{cases} 0 & x \in [0, 1) \\ 1 & x \in [1, 2] \end{cases}$$

Define $F(x) = \int_0^x f(t) \, dt$. Then:
- For $0 \le x < 1$: $F(x) = 0$, so $F'(x) = 0$.
- For $1 < x \le 2$: $F(x) = 1 \cdot (x - 1) = x - 1$, so $F'(x) = 1$.
- At $x = 1$: $F$ is continuous (both sides give $F(1) = 0$ from the left, and the right side gives $F(1) = 0$ too). But $F'(1)$ doesn't exist because the left derivative is $0$ and the right derivative is $1$.

So $F'(x) = \begin{cases} 0 & x < 1 \\ \text{undefined} & x = 1 \\ 1 & x > 1 \end{cases}$

This is not equal to $f(x)$. The FTC promised: if $f$ is continuous, then $F' = f$. Here, $f$ is discontinuous at $x = 1$, and indeed $F' \ne f$ at that point.

graph TD
    A["Is f continuous on [a,b]?"]
    A -->|YES| B["FTC Part 1 applies:<br/>F'(x) = f(x) for x in (a,b)"]
    A -->|NO| C["FTC does NOT guarantee<br/>F'(x) = f(x) at discontinuities"]
    B --> D["F is differentiable"]
    C --> E["F may not be differentiable<br/>at jump points"]
    F["Is f differentiable<br/>at x = c?"]
    D --> F
    F -->|YES| G["Smooth, no corners"]
    F -->|NO| H["Possible corner/cusp<br/>F still continuous,<br/>but F' may not exist at c"]

3. Worked Example

Let's apply the FTC and check continuity/differentiability carefully.

Problem: Let $f(x) = \begin{cases} x & 0 \le x < 2 \\ 5 & 2 \le x \le 4 \end{cases}$

Define $F(x) = \int_0^x f(t) \, dt$. Find $F(x)$ explicitly, identify where $F$ is differentiable, and verify the FTC.

Solution:

For $0 \le x < 2$:
$$F(x) = \int_0^x t \, dt = \frac{x^2}{2}$$

For $2 \le x \le 4$:
$$F(x) = \int_0^2 t \, dt + \int_2^x 5 \, dt = 2 + 5(x - 2) = 5x - 8$$

So $F(x) = \begin{cases} \frac{x^2}{2} & 0 \le x < 2 \\ 5x - 8 & 2 \le x \le 4 \end{cases}$

Check continuity at $x = 2$:
- Left: $\lim_{x \to 2^-} \frac{x^2}{2} = 2$
- Right: $\lim_{x \to 2^+} (5x - 8) = 2$
- Value: $F(2) = 2$

$F$ is continuous. ✓

Check differentiability at $x = 2$:
- Left derivative: $\frac{d}{dx}\left[\frac{x^2}{2}\right]\big|_{x=2^-} = 2$
- Right derivative: $\frac{d}{dx}[5x - 8]\big|_{x=2^+} = 5$

Left and right derivatives don't match, so $F'(2)$ does not exist. $F$ is not differentiable at $x = 2$. ✗

Why? Because $f$ is discontinuous at $x = 2$ (it jumps from $2$ to $5$). The FTC says: if $f$ is continuous, then $F' = f$. Since $f$ is discontinuous at $x = 2$, the theorem doesn't guarantee that $F'(2) = f(2)$. And indeed it doesn't!

For $x \ne 2$: $F'(x) = f(x)$. ✓

4. Key Takeaways

4.1 Most Important Concepts

1. Continuity is the FTC's requirement: If $f$ is continuous on $[a, b]$, then $F(x) = \int_a^x f(t) \, dt$ is differentiable and $F'(x) = f(x)$.

2. Discontinuity breaks the FTC: If $f$ has a jump discontinuity at some point $c$, then $F'(c)$ may not equal $f(c)$, and $F$ may not be differentiable at $c$.

3. Differentiability is stronger than continuity: Every differentiable function is continuous, but a continuous function can have corners or cusps where it's not differentiable.

4. Continuity survives integration: If $f$ is discontinuous at $c$, then $F(x) = \int_a^x f(t) \, dt$ is still continuous at $c$, but may not be differentiable there.

5. Corners don't break the FTC requirement: A function like $|x|$ is continuous everywhere but not differentiable at $x = 0$. If $f(x) = |x|$ and $F(x) = \int_0^x |t| \, dt$, then $F' = f$ everywhere except possibly at $x = 0$.

6. The FTC works "almost everywhere": Even if $f$ is discontinuous at finitely many points, the FTC holds at all continuous points of $f$.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Your task (20 minutes):

Consider $f(x) = \begin{cases} -1 & -1 \le x < 0 \\ 0 & x = 0 \\ 1 & 0 < x \le 1 \end{cases}$

1. Write down $F(x) = \int_{-1}^x f(t) \, dt$ explicitly for each region.
2. Check continuity of $F$ at $x = 0$. Is $F$ continuous?
3. Compute the left and right derivatives of $F$ at $x = 0$. Does $F'(0)$ exist?
4. Verify the FTC: For what values of $x$ does $F'(x) = f(x)$ hold?
5. Explain why the FTC doesn't guarantee $F'(0) = f(0)$.

Success looks like: You've computed $F$ piecewise, shown that $F$ is continuous at $x = 0$, calculated that the left and right derivatives at $0$ are $-1$ and $1$ respectively (so $F'(0)$ doesn't exist), and written a sentence explaining that $f$ is discontinuous at $x = 0$, so the FTC's hypothesis isn't met there.

<<>>