All 10 Calculus BC units
AP Calculus BC · Unit 6 of 10

Integration & Accumulation of Change

15–20% of the exam7 lessons · 102 min42 terms

What this unit covers

The topics below follow the published Calculus BC course framework for Unit 6. This unit is worth 15–20% of the exam, so budget your time against that rather than against how long the unit takes to teach.

FTCIntegration by partsPartial fractionsImproper integrals

Lessons in this unit

Formulas in Unit 6

Trapezoidal sum, unequal widths
Σ [(f(xᵢ) + f(xᵢ₊₁))/2] · (xᵢ₊₁ − xᵢ) · equal widths: (Δx/2)[f(x₀) + 2f(x₁) + … + 2f(xₙ₋₁) + f(xₙ)]
The h/2 shortcut with doubled interior terms is valid only when every subinterval has the same width.
FTC Part 1 (with the chain rule)
d/dx ∫ₐ^{u(x)} f(t) dt = f(u(x)) · u′(x)
When the upper limit is a function u(x) rather than plain x, multiply by u′(x). This chain-rule twist is heavily tested.
FTC Part 2 (evaluation)
∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f
Any antiderivative works — the constant of integration cancels in the subtraction.
Second Fundamental Theorem, with the chain rule
d/dx ∫ₐˣ f(t) dt = f(x) · d/dx ∫ₐ^(u(x)) f(t) dt = f(u(x)) · u′(x)
g″(x) = f′(x), so the concavity of the accumulation function is the increase or decrease of the integrand.
Definite integral by substitution
∫ₐᵇ f(g(x))·g′(x) dx = ∫_(g(a))^(g(b)) f(u) du
The new limits are the old ones passed through g. Once the limits are converted, never substitute back.
Integration by parts
∫ u dv = u·v − ∫ v du
You pick u (to differentiate) and dv (to integrate); then du = u′ dx and v = ∫dv.
Decomposition over distinct linear factors
P(x) / [(x − a)(x − b)] = A/(x − a) + B/(x − b)
One unknown constant per distinct linear factor. Solve for A and B by clearing denominators and substituting convenient x-values.
Improper integral over an infinite interval
∫ₐ^∞ f(x) dx = lim(b→∞) ∫ₐᵇ f(x) dx
Convert the infinite bound to a finite b, integrate, then let b → ∞. A finite result means convergence.

Every term in Unit 6

All 42 terms we publish for Integration & Accumulation of Change, with definitions. Reading them through is the fastest way to find the ones you cannot define — then drill those in cram mode until you can produce them without the prompt.

Integration by parts
∫u dv = uv − ∫v du. Choose u by LIATE — logarithmic, inverse trig, algebraic, trigonometric, exponential — taking the earliest type as u.
Partial fraction decomposition
For proper rational functions with factorable denominators: split into A/(x − a) + B/(x − b), solve for constants, integrate to logarithms.
Repeated integration by parts
Needed when one application still leaves a product. Tabular integration organizes the work when u differentiates to zero.
Integration by parts returning the original
For ∫eˣ sin x dx the original integral reappears; solve algebraically for it rather than continuing.
When partial fractions applies
The numerator degree must be lower than the denominator degree. Otherwise divide first, then decompose the remainder.
Improper integral with an infinite limit
Replace the infinity with b, integrate, then take the limit as b→∞. Converges if that limit is finite.
Improper integral with a discontinuity
Split at the discontinuity and take one-sided limits. Integrating straight through an infinite discontinuity gives a wrong answer that often looks reasonable.
Trigonometric substitution qualitatively
For √(a² − x²) use x = a sin θ; for √(a² + x²) use x = a tan θ. Beyond the AB toolkit, and it converts a radical into a trig identity.
Fundamental Theorem of Calculus part 1
d/dx ∫ₐˣ f(t)dt = f(x); with a variable upper limit g(x), multiply by g′(x).
Fundamental Theorem of Calculus part 2
∫ₐᵇ f = F(b) − F(a) for any antiderivative F.
u-substitution with definite limits
Either convert the limits to u values or convert back to x before evaluating. Mixing the two is a common error.
Riemann sum approximations
Left sums underestimate an increasing function and right sums overestimate it; the trapezoid rule overestimates a concave-up one.
Net change theorem
∫ₐᵇ f′(x)dx = f(b) − f(a): integrating a rate gives the total change.
Interpreting an integral in context
State what accumulated, over what interval, with units — "8 liters of water entered the tank between t = 0 and t = 4 minutes".
Technique decision order
Substitution first, then parts, then partial fractions. Trying them out of order wastes the most time on parts, which takes longest to abandon.
LIATE
Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential — choose u as the earliest type present.
Why LIATE works
Logs and inverse trig simplify when differentiated and are painful to integrate, so they belong in u; exponentials integrate as easily as they differentiate.
Check substitution before committing to parts
∫x·e^(x²)dx looks like parts and is a one-line substitution, because 2x is present up to a constant.
Partial fractions preconditions
Denominator factors into distinct linear factors AND numerator degree is lower. Otherwise divide first.
Solving for A and B fastest
Clear denominators, then substitute the roots one at a time — each substitution kills one unknown.
Partial fractions produce logarithms
∫A/(x − a)dx = A·ln|x − a| + C. The absolute values matter where the argument is negative.
Outside the AP scope
Repeated factors and irreducible quadratics are not tested. Encountering one means the intended method was something else.
Improper: infinite interval
∫₁^∞ f means lim(b→∞) ∫₁^b f. Write the limit — the rubric awards it separately.
Improper: infinite integrand
A discontinuity inside the interval also makes an integral improper; take the limit approaching the bad point.
Never substitute ∞
Writing "−1/(2·∞²)" loses the setup point even when the final number is right. ∞ is not a value.
p-integral at infinity
∫₁^∞ dx/xᵖ converges exactly when p > 1, with value 1/(p − 1).
p-integral near zero
∫₀¹ dx/xᵖ converges exactly when p < 1 — the condition reverses, because the trouble is at the origin.
At infinity shrink fast, near zero blow up slowly
The mnemonic that keeps the two p-integral conditions straight.
What the integral test does not give
It settles convergence only. When both converge the sum and the integral are NOT equal.
FTC Part 2 with a chain rule
d/dx ∫ₐ^(u(x)) f = f(u(x))·u′(x). Substitute the upper limit, then multiply by its derivative.
Variable in the lower limit
The sign flips: d/dx ∫ₓ^b f = −f(x), because swapping limits negates an integral.
Do not integrate when asked to differentiate
Part 2 exists so you need not. The exam includes integrands with no elementary antiderivative to enforce this.
g′ = f and g″ = f′
For g(x) = ∫ₐˣ f, every Unit 5 question about f becomes a question about g shifted one derivative.
g(a) = 0 always
The integral from a to a is zero, and this is often the only value of g supplied.
Left sum with an increasing function
Underestimates. Monotonicity — not concavity — decides left and right sums.
Trapezoid and concavity
Concave up puts each chord above the curve, so the trapezoidal sum overestimates. Concave down reverses it.
Midpoint runs opposite the trapezoid
On a concave-up function the midpoint rule underestimates while the trapezoid overestimates.
Unequal widths are normal
Table data rarely has equal spacing, and the (h/2) shortcut is valid only when it does.
Reversal property
∫ᵇᵃ f = −∫ₐᵇ f. The property most often forgotten, and the one recombination questions are built on.
Integral of a constant
∫ₐᵇ c dx = c(b − a). The most-dropped term when recombining given integral values.
No product rule for integrals
∫fg is not ∫f·∫g. If a question appears to need that, the intended method is substitution.
u-substitution changes the limits
Convert the limits to u values, or convert back to x before evaluating. Mixing them is a silent error.

What examiners penalize here

Practice Calculus BC

Our practice bank is drawn from across the whole course rather than filtered to one unit, which is closer to how the exam asks anyway — it will not tell you which unit a question is testing.

Questions about this unit

How much of the AP Calculus BC exam is Unit 6?

Unit 6, Integration & Accumulation of Change, is worth 15–20% of the Calculus BC multiple-choice section according to the published course framework. Across all 10 units that makes it one of the heaviest units on the exam, and worth front-loading.

What topics are covered in Calculus BC Unit 6?

Integration & Accumulation of Change covers FTC, Integration by parts, Partial fractions and Improper integrals. We publish 42 terms with definitions for this unit, all of them on this page.

How should I study Calculus BC Unit 6?

Read the 7 lessons below first — about 100 minutes — then drill the 42 terms in cram mode until you can produce each definition from memory rather than just recognize it. Recognition is what makes a unit feel finished when it is not. Finish with practice questions and read the explanation for every one you get right by elimination as well as the ones you miss.

All 10 units of AP Calculus BC

  1. Unit 1 · Limits & Continuity
  2. Unit 2 · Differentiation: Definition & Fundamental Properties
  3. Unit 3 · Differentiation: Composite, Implicit & Inverse Functions
  4. Unit 4 · Contextual Applications of Differentiation
  5. Unit 5 · Analytical Applications of Differentiation
  6. Unit 6 · Integration & Accumulation of Change
  7. Unit 7 · Differential Equations
  8. Unit 8 · Applications of Integration
  9. Unit 9 · Parametric, Polar & Vector-Valued Functions
  10. Unit 10 · Infinite Sequences & Series

Unit names, topics and exam weights follow the published College Board course framework for AP Calculus BC. AP® is a trademark registered by the College Board, which does not endorse this site.