AP Calculus AB and BC

When FTC Part 2 Fails: Integrating Across a Pole

The second part of the Fundamental Theorem of Calculus needs the integrand continuous on the whole closed interval. Apply the formula across a vertical asymptote and it returns an answer that is not merely wrong but impossible: a negative number for an integrand that is positive everywhere.

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. 1

   f is continuous on the whole closed interval [a, b]

   $$f \in C[a, b]$$

   Continuity on the entire interval is what guarantees the integral exists in the first place and that a single antiderivative covers the whole range. The moment the integrand has a pole inside, the interval is really two intervals, and an antiderivative on each side need not join up.

   Drop it and the theorem fails

   $$\int_{-1}^{1} \frac{dx}{x^{2}} \ \ \text{with} \ \ F(x) = -\frac{1}{x}$$

   The integral of 1/x squared from -1 to 1

   Substituting endpoints into $F(x) = -1/x$ gives $F(1) - F(-1) = -1 - 1 = -2$. But the integrand $1/x^2$ is positive at every point where it is defined, so any honest answer must be positive. The formula did not produce an inaccurate number, it produced an impossible one. The integral in fact diverges, because of the pole at $x = 0$ that sits right in the middle of the interval.

2. 2

   F is an antiderivative of f across the whole interval, not just on pieces of it

   $$F'(x) = f(x) \text{ for every } x \in [a, b]$$

   This is the same failure viewed from the other side. $-1/x$ satisfies $F' = 1/x^2$ on $(-\infty, 0)$ and again on $(0, \infty)$, but there is no antiderivative on an interval containing 0, because the function is not even defined there. Writing one symbol for both branches hides the fact that they are two separate functions, and the subtraction $F(b) - F(a)$ silently compares a value from one branch with a value from the other.

   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

1. Let $G(x) = \int_a^x f(t)\, dt$. By Part 1, $G' = f$ on $[a,b]$, using continuity.
2. If $F$ is any other antiderivative then $(F - G)' = 0$ throughout the interval, so by the Mean Value Theorem $F - G$ is constant there.
3. Evaluate that constant at $a$: since $G(a) = 0$, the constant is $F(a)$, so $F(x) = G(x) + F(a)$.
4. Put $x = b$: $F(b) - F(a) = G(b) = \int_a^b f$. The step that fails across a pole is the second one, because the Mean Value Theorem needs a single interval, and a pole cuts the domain in two.

What it does not say