Home ›
AP Calculus AB/BC ›
Unit 6 ›
Visual Review
AP Calculus AB/BC Unit 6 Visual ReviewA topic-by-topic visual walkthrough of Unit 6: Integration and Accumulation of Change — Riemann sums, the Fundamental Theorem of Calculus, and antidifferentiation techniques.
← Back to Unit 6 hub
TOPIC 6.1
Accumulations of Change
Add up a rate to get a total
If you know a RATE of change, the total
accumulated change is the AREA under
the rate curve.
Speed over time → distance traveled.
Integration reverses differentiation.
Area under a rate = accumulated change
∫ₐᵇ (rate) dt = total change on [a,b]
A car at 60 mph for 2 h travels 120 mi —
the area of a 60×2 rectangle.
Area BELOW the axis counts as
NEGATIVE change.
The definite integral is a net total
The signed area under f gives the NET accumulated change — positive above, negative below the x-axis.
The area under a rate curve = accumulated change (integration undoes rates).
The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.2
Approximating Areas: Riemann Sums
rectangles approximate the area
Sum rectangle areas
Σ f(xᵢ)·Δx, Δx = (b−a)/n
LEFT, RIGHT, or MIDPOINT endpoints set
each rectangle's height.
Over- or under-estimate
For an INCREASING f: LEFT sum under-,
RIGHT sum over-estimates.
More rectangles (larger n) → better estimate.
▭ Estimate area with Σ f(xᵢ)Δx using left, right, or midpoint rectangles.
The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.3
The Definite Integral as a Limit
∫ₐᵇ f(x) dx = lim(n→∞) Σ f(xᵢ)·Δx
Exact area = limit of Riemann sums
As the number of rectangles → ∞ (width
→ 0), the sum becomes the exact area.
The ∫ sign is an elongated "S" for SUM.
a, b are the limits of integration;
dx marks the variable.
Signed area
Area ABOVE the x-axis is positive;
area BELOW is negative.
The definite integral gives the NET
signed area.
Geometry (triangles/rectangles) can evaluate simple ones.
∫ The definite integral is the limit of Riemann sums — exact net signed area.
The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.4
FTC & Accumulation Functions
g(x) = ∫ₐˣ f(t) dt ⟹ g′(x) = f(x)
the derivative of an accumulation function is the integrand
FTC Part 1 (derivative form)
Differentiation and integration are
INVERSE operations.
With a variable upper limit u(x), chain:
d/dx ∫ₐ^u(x) f dt = f(u)·u′
e.g. d/dx ∫₀^(x²) sin t dt = sin(x²)·2x
Reading g(x)
g(x) accumulates the area under f from
a up to x.
g increasing where f > 0; g has a max
where f crosses + → −.
g concave up where f is increasing.
d/dx ∫ₐˣ f(t)dt = f(x) — differentiation undoes integration (chain the upper limit).The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.5
Behavior of Accumulation Functions
g(x) = ∫ₐˣ f(t) dt when f (the integrand) is…
g increasing f > 0 (above the axis)
g decreasing f < 0 (below the axis)
g local max / min f crosses zero (+→− / −→+)
g concave up f increasing (f′ > 0)
Compute values as areas
g(x) equals the NET area of f from a to x. Add area above the axis, subtract area below. g(a) = 0.
g is increasing where f>0 ; g has an extremum where f changes sign.The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.6
Properties of Definite Integrals
∫ₐᵃ f dx = 0 ∫ₐᵇ f dx = −∫ᵇₐ f dx
∫ₐᵇ f dx + ∫ᵇᶜ f dx = ∫ₐᶜ f dx (split at b)
∫ₐᵇ [f ± g] dx = ∫f ± ∫g ∫ₐᵇ k·f dx = k∫f
// swapping the limits flips the sign
Use them to combine or split integrals
Given ∫₀³f=10 and ∫₀⁵f=4, then ∫₃⁵f = 4 − 10 = −6. These properties are constant FRQ tools.
Integrals are linear and additive over intervals ; swapping limits flips the sign.
The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.7
FTC & Evaluating Definite Integrals
∫ₐᵇ f(x) dx = F(b) − F(a), where F′ = f
FTC Part 2 (evaluation form)
Find an ANTIDERIVATIVE F, then subtract
its values at the limits.
∫₁³ 2x dx = [x²]₁³
= 9 − 1 = 8
No +C needed — it cancels in the subtraction.
Net change theorem
∫ₐᵇ F′(x) dx = F(b) − F(a)
The integral of a rate gives the NET
CHANGE of the quantity.
final = initial + ∫(rate).
Core of accumulation FRQs.
∫ₐᵇ f dx = F(b) − F(a) — evaluate an antiderivative at the limits and subtract.The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.8
Basic Antiderivatives
// reverse each derivative rule; add + C
∫ xⁿ dx = xⁿ⁺¹/(n+1) + C (n ≠ −1)
∫ 1/x dx = ln|x| + C ∫ eˣ dx = eˣ + C
∫ cos x dx = sin x + C ∫ sin x dx = −cos x + C
∫ sec²x dx = tan x + C
Always add + C for indefinite integrals
An indefinite integral is a FAMILY of antiderivatives differing by a constant. Solve for C using an initial condition.
Antiderivatives reverse derivatives: ∫xⁿ dx = xⁿ⁺¹/(n+1) + C .
The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.9
Integrating Using Substitution
let u = inner function, du = u′ dx — the reverse chain rule
Worked example
∫ 2x·(x²+1)³ dx
u = x²+1, du = 2x dx
= ∫ u³ du = u⁴/4 + C
= (x²+1)⁴ / 4 + C
Substitute back to x at the end.
Definite integrals
CHANGE the limits to u-values, then you
don't have to convert back.
x: a→b becomes u: u(a)→u(b)
Look for a function and (a multiple of) its
derivative both present.
u-substitution reverses the chain rule: set u = inner, du = u′dx.The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.10
Long Division & Completing the Square
Long division first
When deg(top) ≥ deg(bottom), DIVIDE
before integrating.
∫ x²/(x+1) dx
= ∫ (x − 1 + 1/(x+1)) dx
= x²/2 − x + ln|x+1| + C
Completing the square
Turns a quadratic denominator into a
form matching an arctan/arcsin integral.
x² + 4x + 5 = (x+2)² + 1
∫ dx/((x+2)²+1) = arctan(x+2)+C
Match ∫dx/(u²+a²) = (1/a)arctan(u/a).
Rewrite into a form you know
Both techniques transform a hard integrand into standard antiderivative forms before integrating.
Divide when top-heavy; complete the square for arctan/arcsin forms.The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.11
BC ONLY
Integration by Parts
∫ u dv = u·v − ∫ v du
Worked example
∫ x·eˣ dx
u = x (du=dx), dv = eˣ dx (v=eˣ)
= x·eˣ − ∫ eˣ dx
= x·eˣ − eˣ + C
Reverses the product rule.
Choosing u: LIATE
Pick u by priority order:
L-og, I-nv trig, A-lg, T-rig, E-xp
The earlier type becomes u (so it
simplifies when differentiated).
Sometimes apply it twice, or solve for the integral.
∫u dv = uv − ∫v du (BC) — choose u by LIATE.The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.12
BC ONLY
Linear Partial Fractions
1/((x−a)(x−b)) = A/(x−a) + B/(x−b)
Split, then integrate each piece as a log
∫ 1/(x²−1) dx = ∫ [ ½/(x−1) − ½/(x+1) ] dx
= ½ ln|x−1| − ½ ln|x+1| + C
Factor the denominator, write one fraction per DISTINCT LINEAR factor, and solve for the
constants A, B (cover-up method or matching coefficients).
Each 1/(x−a) integrates to ln|x−a|. Central to the LOGISTIC model (7.9).
Partial fractions (BC): split into A/(x−a) + B/(x−b), each integrates to a log.The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.13
BC ONLY
Improper Integrals
∫ₐ^∞ f dx = lim(b→∞) ∫ₐᵇ f dx
Rewrite as a limit
Improper = infinite limit of integration OR
an infinite discontinuity in the interval.
∫₁^∞ 1/x² dx = lim [−1/x]₁ᵇ
= lim (−1/b + 1) = 1
Converges to a finite value.
Converge or diverge
If the limit is FINITE → CONVERGES to that
value.
If the limit is ±∞ or DNE → DIVERGES.
∫₁^∞ 1/x dx diverges (→ ∞)
Compare with ∫1/xᵖ: converges iff p > 1.
Improper integrals (BC) = limits; a finite limit converges, otherwise diverges.The Review Hub · AP Calculus AB/BC Unit 6
TOPIC 6.14
Selecting Antidifferentiation Techniques
Recognize the form → use this technique
matches a basic rule antiderivative directly (6.8)
inner function & its derivative u-substitution (6.9)
product of unlike types (x·eˣ) integration by parts — BC (6.11)
rational, factorable denom. partial fractions — BC (6.12)
Simplify first, then match
Try algebra (expand, divide, rewrite) before a heavier technique — the simplest approach that works is best.
Match the integrand's form to the technique: basic, u-sub, parts, or partial fractions.The Review Hub · AP Calculus AB/BC Unit 6
▤ Show all slides
How to use the visual review
Spend 30 seconds per slide before clicking next. Look at the diagram, then ask yourself: "Could I set up this sum or evaluate this integral from memory?"
Use the fullscreen button () on desktop for the best experience. Use arrow keys to navigate. Tap "Show all slides" to jump around.
This is great for review the night before the exam — fast, visual, and covers every idea you need to recognize in Unit 6.