Every integral so far had finite bounds and a finite integrand. This closer breaks both
rules — bounds that run to infinity, functions that blow up — and then hands you the unit's real final
skill: looking at any integral and knowing, in five seconds, which machine to reach for.
An improper integral is an ordinary integral plus a limit — nothing more. Replace the bad endpoint
(∞, or the blow-up point) with a variable, integrate normally, then take the limit. If the limit
exists, the integral converges to that number; if not, it diverges — and "diverges" is a
complete, correct answer.
Infinite region, finite area
The whole subject in one paradox. Two nearly identical curves on [1, ∞):
1/x² and 1/x. Both regions are infinitely long. Push the right edge out and watch the race:
one area saturates — an infinite region holding exactly one unit of paint — while the other grinds
upward past any number you name.
push b →b = 6.3
The race
∫₁ᵇ dx/x² = 1 − 1/b—
its destination1 — converges
∫₁ᵇ dx/x = ln b—
its destination∞ — diverges
left to collect, 1/x² tail—
left to collect, 1/x tail∞ — always
The difference is the tail: 1/x² thins fast enough that what's left beyond b shrinks to
nothing; 1/x doesn't, and the leftovers add up forever. "Converges" means the running total has a
destination.
The mechanics — an integral plus a limit
The move: swap the bad endpoint for a variable, integrate as usual, take the limit. Writing the limit explicitly is a scored step on FRQs — skipping straight to "plugging in ∞" loses the point.
Blow-ups count too: if the integrand has a vertical asymptote at an endpoint, the same move applies — approach the bad point from inside the interval.
Converges = the limit exists (a number). Diverges = it doesn't. Both are final answers.
press step
The work, shown
Pick an integral
The third looks
completely ordinary — finite bounds, innocent integrand. Look closer at x = 0.
The p-test — one dial, two regimes
Every example so far was 1/x to some power. Make the power a dial and the whole
theory appears: ∫₁^∞ dx/xᵖ converges exactly when p > 1, and near zero the rule
flips — ∫₀¹ dx/xᵖ needs p < 1. At infinity you need a thin tail; at zero you need a
mild spike. And p = 1 loses on both ends.
p =2.00
The verdict
regimeat infinity
rule hereconverges iff p > 1
this p—
value (if convergent)—
Hold this dial in your head — in Unit 10 it returns as the p-series test, deciding
Σ1/nᵖ with the exact same knife edge at p = 1.
The trap — the asymptote hiding inside
The most dangerous improper integrals don't look improper. Watch what happens if you
run FTC on ∫₋₁¹ dx/x² without looking at the integrand:
Stop. 1/x² is positive everywhere — its integral cannot be negative. The computation above is fluent, confident, and meaningless.
The diagnosis: there's a vertical asymptote at x = 0, inside the interval. FTC requires a continuous integrand on [a, b] — that hypothesis just failed, silently.
The correct treatment: split at the blow-up and give each half its own limit. Here both halves diverge — the honest answer is that the integral diverges.
The habit this buys you: scan the whole interval for blow-ups before integrating. Five seconds, every time.
a = 0.50
Each half stops a distance a short of the blow-up. Press play and close the gap.
The honest computation
left half: ∫₋₁⁻ᵃ dx/x²—
right half: ∫ₐ¹ dx/x²—
each equals1/a − 1
verdictdiverges
Both halves blow up together — and ONE would have been enough to sink the integral. There's no
cancellation rescue here: these are areas of a positive function, all the same sign, piling up
without bound on both sides of the asymptote.
Which machinery? — the unit, in eight integrals
Topic 6.14 is the real boss fight: not doing the techniques, but choosing
them. Eight integrals, four weapons each. Diagnose before you compute — every wrong answer below is a
technique that looks right.
Integral 1 of 8
Score: 0 / 0
First move?
The tell, and the first move
Pick a weapon to see the diagnosis.
The five-second scan
fingerprint pair?u-sub
unrelated product?parts / tabular
rational? check Q:÷ · ln · PF · arctan
∞ or blow-up?limit first
The closer on one card
Blow-up at an endpoint — approach it from inside
The p-test, both regimes — and p = 1 fails both
Interior asymptote — split first, then limit each piece
The ideas everything else builds on
Integral + limit
That's the entire definition. The limit is written, not implied — it's a
rubric point, and "plugging in ∞" is not a number operation.
Convergence is a destination
Converges = the running total has a limit. Diverges = it doesn't.
Divergence is an answer, not a failure to find one.
The knife edge flips
At ∞ you need p > 1 (thin tail); near 0 you need p
< 1 (mild spike). Same dial, opposite directions — and 1/x loses both.
Scan before you integrate
FTC needs a continuous integrand on the whole interval. Five seconds of
looking for blow-ups beats a fluent, confident, wrong computation.
Your turn
Five problems — limits written out, verdicts justified. Try before revealing.
Quick quiz
Eight fast checks across the whole lesson.
Question 1 of 8
Score: 0 / 0
Common mistakes & exam tips
Plugging ∞ in like a number
Write the limit: lim_{b→∞} [F(x)]₁ᵇ. The exam awards the limit notation specifically — and ∞ − ∞, 1/∞ and friends need limit reasoning, not arithmetic.
Missing the interior blow-up
∫₋₁¹ dx/x² = −2 is the unit's most famous wrong answer. Scan the interval for asymptotes BEFORE touching FTC — continuity on [a, b] is a hypothesis, not a decoration.
p-test direction confusion
p > 1 at infinity, p < 1 near zero. If you remember one anchor, make it 1/x: it diverges in BOTH regimes, so the convergent side is always away from it.
Treating "diverges" as failure
Students who compute lim ln b = ∞ then erase and try another technique. Divergence IS the answer — state it and move on.
Splitting and only checking one half
An interior asymptote splits the integral in two, and BOTH halves must converge. One divergent half sinks the whole thing — no cancelling +∞ against −∞.
Technique tunnel vision
Just learned partial fractions, so everything looks like partial fractions. Run the five-second scan first: fingerprint → u-sub; unrelated product → parts; THEN the rational-function tree.
On the AP exam
Show the limit conversion explicitly: “∫₁^∞ = lim_{b→∞} ∫₁ᵇ” — it's a scored step, every time.
Justify convergence verdicts in words: “converges because lim_{b→∞}(1 − 1/b) = 1” — a bare number doesn't earn the point.
Before integrating ANY rational or root integrand on a closed interval, scan for asymptotes inside it.
Technique selection is tested directly in MC — practice the five-second scan until the diagnosis is faster than the computation.
That's the unit — accumulation built from rectangles, the Fundamental Theorem connecting it to
derivatives, and a complete toolkit for the hunt. Prove it on the
Unit 6 Exam Practice capstone — and when the p-test's knife edge
returns in Unit 10's infinite series, you'll already own it. Back to the
Unit 6 Guide.