FTC with Discontinuous Functions: Continuity vs. Differentiability

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FTC with Discontinuous Functions: Continuity vs. Differentiability

TL;DR

The Fundamental Theorem of Calculus requires the integrand to be continuous on the interval, but differentiability of the antiderivative breaks down at discontinuities. You need to understand which regularity condition (continuity, differentiability, or jump discontinuity) applies so you know whether FTC holds and how to handle integration around problem points.

1. The Mental Model

Here's the key: when a function has a jump or breaks, the FTC can still work if you're smart about what you're integrating. The FTC doesn't fall apart—it just doesn't apply everywhere at once. Think of it like a road with a gap: you can measure distance from point A to the gap, and from the gap to point B separately, but you can't measure the gap itself with a simple ruler. The deeper fact is that continuity guarantees the antiderivative exists and is differentiable, but a discontinuous integrand can still have an antiderivative—it's just not differentiable at the bad point.

2. The Core Material

Understanding the FTC conditions

The classical Fundamental Theorem of Calculus states: if $f$ is continuous on $[a,b]$, then
$$F(x) = \int_a^x f(t) \, dt$$
is differentiable on $[a,b]$ and $F'(x) = f(x)$.

That "continuous" clause matters. It's not a suggestion—it's a requirement for the derivative to exist everywhere. But here's the thing: the theorem is strongest when $f$ is continuous, but weaker versions exist when $f$ is not.

What happens at a jump discontinuity?

Let's say $f(x) = \begin{cases} 1 & x < 0 \\ 3 & x \geq 0 \end{cases}$ on $[-1, 1]$.

The function jumps by 2 at $x=0$. You can still integrate it:
$$F(x) = \int_{-1}^x f(t) \, dt = \begin{cases} x + 1 & -1 \leq x < 0 \\ 2x + 1 & 0 \leq x \leq 1 \end{cases}$$

Notice: $F$ is continuous everywhere, even though $f$ is not. But at $x=0$, the derivative from the left is $\frac{dF}{dx}|_{x=0^-} = 1$ and from the right is $\frac{dF}{dx}|_{x=0^+} = 2$. So $F'(0)$ doesn't exist. The antiderivative gets a "corner" at the discontinuity.

Removable discontinuities are gentler

If $f$ has a removable discontinuity—say, $f(x) = \frac{\sin x}{x}$ for $x \neq 0$ and $f(0) = 1$—then $f$ is discontinuous at only one point, but you can define it to be continuous there. Once you fill in the hole with the right value, FTC applies fully. The antiderivative will be smooth.

Integrability vs. differentiability

Here's the subtle bit: a function can be integrable (you can compute its integral) without being continuous. The Riemann integral still exists for functions with finitely many jump discontinuities. But FTC requires continuity to guarantee that the antiderivative is differentiable. So:

• $f$ continuous on $[a,b]$ ⟹ FTC applies fully, $F$ is differentiable.
• $f$ has jump discontinuities ⟹ $F$ still exists and is continuous, but $F'$ does not exist at the jumps.
• $f$ has a removable discontinuity ⟹ fix it, and FTC applies.

graph TD
    A["Function f on [a,b]"] --> B{Is f continuous everywhere?}
    B -->|Yes| C["FTC applies: F' = f everywhere<br/>F is differentiable"]
    B -->|No| D{Type of discontinuity?}
    D -->|Jump discontinuity| E["F exists & is continuous<br/>F' does NOT exist at jump<br/>FTC fails at that point"]
    D -->|Removable| F["Redefine f at the point<br/>Then f becomes continuous<br/>FTC applies"]
    D -->|Infinite discontinuity| G["Integral may diverge<br/>FTC does not apply<br/>Use improper integrals"]
    style C fill:#90EE90
    style E fill:#FFB6C6
    style F fill:#FFE4B5
    style G fill:#FFB6C6

How to work around discontinuities

When you encounter a discontinuous integrand, break the integral into pieces at the discontinuities and sum them:

$$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx$$

where $c$ is the location of a jump. On each piece $[a, c)$ and $(c, b]$, $f$ is continuous (or you restrict to the continuous parts), so FTC applies locally. The sum gives the total integral, and you can compute $F$ piecewise.

3. Worked Example

Let's integrate $f(x) = \begin{cases} 2x & 0 \leq x < 1 \\ 5 & 1 \leq x \leq 3 \end{cases}$ and find $F(x) = \int_0^x f(t) \, dt$.

For $0 \leq x < 1$:
$$F(x) = \int_0^x 2t \, dt = x^2$$

For $1 \leq x \leq 3$:
$$F(x) = \int_0^1 2t \, dt + \int_1^x 5 \, dt = 1 + 5(x-1) = 5x - 4$$

So $F(x) = \begin{cases} x^2 & 0 \leq x < 1 \\ 5x - 4 & 1 \leq x \leq 3 \end{cases}$.

Check continuity: as $x \to 1^-$, $F(x) = x^2 \to 1$. As $x \to 1^+$, $F(x) = 5x-4 \to 1$. Yes, $F$ is continuous.

Check differentiability: For $x < 1$, $F'(x) = 2x$, so $F'(1^-) = 2$. For $x > 1$, $F'(x) = 5$, so $F'(1^+) = 5$. These don't match, so $F'(1)$ doesn't exist. The graph has a corner.

This is exactly what we expect: $f$ jumps from 2 to 5 at $x=1$, so $F$ is continuous but not differentiable there.

4. Key Takeaways

4.1 Most Important Concepts

1. FTC requires continuity: The classical statement needs $f$ continuous on $[a,b]$ to guarantee $F'(x) = f(x)$ everywhere.

2. The antiderivative always exists for piecewise continuous functions: Even with jump discontinuities, $F(x) = \int_a^x f(t) \, dt$ is well-defined and continuous.

3. Discontinuities create corners in the antiderivative: At a jump, $F$ is continuous but not differentiable—the left and right derivatives differ.

4. Break integrals at discontinuities: Use $\int_a^b = \int_a^c + \int_c^b$ to handle jumps, then apply FTC piecewise.

5. Removable discontinuities are forgivable: Redefine the function at one point to make it continuous, and FTC applies fully.

6. Regularity matters for calculus: Higher regularity (continuity → differentiability → continuity of derivative) lets you use more powerful theorems without restriction.

7. Integration is more forgiving than differentiation: You can integrate discontinuous functions; you can't differentiate them at the bad points.

4.2 Common Misconceptions

4.3 Compare & Contrast

5. Now Try It

Problem: Consider $g(x) = \begin{cases} 3x^2 & 0 \leq x < 2 \\ 12 & 2 \leq x \leq 4 \end{cases}$.

Compute $G(x) = \int_0^x g(t) \, dt$ for all $x \in [0,4]$. Then:

1. Write $G$ as a piecewise function.
2. Check that $G$ is continuous everywhere.
3. Find $G'(x)$ on each piece and explain why $G'(2)$ does not exist.
4. Verify that $\int_0^4 g(t) \, dt = G(4) - G(0)$.

Hint: Compute the two pieces separately at $x=2$, then combine.

For your answer:
- Show all integral calculations with limits.
- State the piecewise formula for $G$.
- Verify left and right derivatives at $x=2$ separately.
- Show the final value $G(4)$.

Success looks like: you have a piecewise-defined antiderivative $G$ with correct formulas on each piece, you've verified it's continuous, and you've shown that the derivative exists and matches $g$ everywhere except at $x=2$, where you explain the corner using left and right derivatives.

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