AP Calculus AB and BC

When FTC Part 1 Fails: The Jump Counterexample

Q: What is the derivative of an integral with a function in the upper limit?

Combine this theorem with the chain rule: \frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x). With both limits varying, differentiate each and subtract.

Q: Is the accumulation function always continuous?

Yes, whenever f is integrable, even if f jumps. Integration smooths: accumulating a jump gives a corner, and accumulating a corner gives a smooth curve. What the continuity hypothesis buys is the stronger claim that the accumulation is differentiable with derivative exactly f.

The first part of the Fundamental Theorem of Calculus needs the integrand to be continuous. Accumulate an integrand that jumps and the accumulation function gets a corner exactly at the jump, so its derivative fails to exist there and cannot equal the integrand.

Fundamental Theorem of Calculus, Part 1

\frac{d}{dx} \int_a^x f(t)\, dt = f(x)

If f is continuous on an interval containing a, then the accumulation function that integrates f from a up to x is differentiable, and its derivative is f itself.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1
The integrand f is continuous on the interval
$f \in C[a, b]$
Continuity is what makes the accumulation smooth. The proof compares $F(x+h) - F(x)$ to $f(x) \cdot h$ , and that comparison is only tight when $f$ does not move much across the small interval. A jump means $f$ moves a fixed amount no matter how small $h$ gets.
Drop it and the theorem fails
$F(x) = \int_{-1}^{x} \operatorname{sgn}(t)\, dt = |x| - 1$
Accumulating the sign function gives F(x) = |x| - 1
The sign function jumps from $-1$ to 1 at the origin. Accumulating it from $-1$ gives $F(x) = |x| - 1$ , which is continuous everywhere, so the integral behaves. But $F$ has a corner at 0, and the check confirms there is no interior point where $F' = 0$ even though $F(-1) = F(1)$ . The derivative simply does not exist at the jump, so it cannot equal $f(0)$ or anything else.
2
The lower limit a is a fixed number, and x is the variable
This is bookkeeping rather than an analytic condition, but reversing it flips the sign, and putting a function in the upper limit brings the chain rule into play: $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\,g'(x)$ . Nothing breaks the theorem here, so no counterexample is claimed; the failure mode is an arithmetic slip rather than a false conclusion.
No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

Write the difference quotient: $\frac{F(x+h) - F(x)}{h} = \frac{1}{h}\int_x^{x+h} f(t)\, dt$ .
The right side is the average value of $f$ on the tiny interval between $x$ and $x + h$ .
Continuity at $x$ means $f$ is nearly constant on that tiny interval, so its average is nearly $f(x)$ , and the approximation improves without limit as $h$ shrinks.
So the difference quotient tends to $f(x)$ , which is exactly $F'(x) = f(x)$ . Every use of the word nearly in step three is continuity doing the work.

What it does not say

Part 1 and Part 2 say the same thing.

Part 1 says differentiating an accumulation returns the integrand. Part 2 says a definite integral can be computed from any antiderivative. They point in opposite directions, and only Part 2 gives you a number.

The accumulation function is only defined when you can find an antiderivative in closed form.

It is defined by the integral itself. $\int_0^x e^{-t^2}\,dt$ has no elementary closed form and is still a perfectly good differentiable function, with derivative $e^{-x^2}$ by this theorem.

A discontinuous integrand cannot be integrated.

It often can. The sign function integrates fine, and its accumulation $|x| - 1$ is continuous. What fails is not the integral but the claim that the accumulation is differentiable with derivative $f$ .

Frequently asked questions

What is the derivative of an integral with a function in the upper limit?

Combine this theorem with the chain rule: $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x))\cdot g'(x)$ . With both limits varying, differentiate each and subtract.

Is the accumulation function always continuous?

Yes, whenever $f$ is integrable, even if $f$ jumps. Integration smooths: accumulating a jump gives a corner, and accumulating a corner gives a smooth curve. What the continuity hypothesis buys is the stronger claim that the accumulation is differentiable with derivative exactly $f$ .