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AP Calculus AB

Board · College Board Units · 1–8 Exam · MCQ + FRQ, scored 1–5 Status · Preview

Overview & units

🗺 AP Calculus AB — College Board Exam Weightings

Unit	Title	Exam %	Key Topics
U1	Limits & Continuity	10–12%	Limit definition, one/two-sided, squeeze theorem, continuity, IVT
U2	Differentiation — Definition & Rules	10–12%	Avg vs. instantaneous rate, limit definition of derivative, power/trig/exp rules
U3	Composite, Implicit & Inverse	9–13%	Chain rule, implicit differentiation, derivatives of inverse trig
U4	Contextual Applications	10–15%	Related rates, straight-line motion, L'Hôpital's rule
U5	Analytical Applications	15–18%	MVT, EVT, increasing/decreasing, concavity, optimisation, 1st & 2nd derivative tests
U6	Integration & Accumulation	17–20%	Riemann sums, FTC Parts 1 & 2, u-substitution, definite integrals
U7	Differential Equations	6–12%	Slope fields, separation of variables, exponential growth/decay
U8	Applications of Integration	10–15%	Area between curves, cross-section & disk/washer volumes, average value

How to use this track

Full per-unit lessons for all eight units are below, followed by exam orientation, worked free-response questions, and a 50-question practice exam. For interactive visual intuition on any concept, the Calculus subject carries matching explorers you can drag and explore.

AP Calculus AB covers single-variable differential and integral calculus across eight units, assessed by a multiple-choice and free-response exam.

Unit 1

Limits & continuity

Unit 2

Differentiation: definition & basic rules

Unit 3

Composite, implicit & inverse

Unit 4

Contextual applications

Unit 5

Analytical applications

Unit 6

Integration & accumulation

Unit 7

Differential equations

Unit 8

Applications of integration

⏱️ AP Exam Format & Scoring Strategy

Section I: Multiple Choice (50% of total score, 45 questions, 105 min):
• Part A (60 min, 30 questions, NO calculator): Average 2 minutes per question. Prioritize standard derivatives, integrals, and graphical interpretations.
• Part B (45 min, 15 questions, Graphing calculator required): Average 3 minutes per question. Use your calculator efficiently for graphing intersections, numerical derivatives (nDeriv), and numerical definite integrals (fnInt).
Section II: Free Response (50% of total score, 6 questions, 90 min):
• Part A (30 min, 2 questions, Graphing calculator required): Typically involves rate in/rate out accumulation or area/volume/motion.
• Part B (60 min, 4 questions, NO calculator): Graph of derivative analysis, separable differential equations, Riemann sum approximations with error, and particle motion. You may continue to write on Part A during this time, but without a calculator.
Composite Score Threshold: A composite score of roughly 68–72% yields a 5, and ~55–60% yields a 4. Maximizing points on setup and theorem statements across every FRQ is the single most effective strategy for reaching a 5.

💡 The 4 Big Ideas of AP Calculus AB

Big Idea 1: Change (CHA): Calculus models how quantities change continuously. Derivatives express instantaneous rates of change, while differential equations relate rates to current states.
Big Idea 2: Limits (LIM): Limits provide the mathematical foundation for both the derivative (limit of difference quotients) and the definite integral (limit of Riemann sums). Continuity and asymptotic behavior depend entirely on limits.
Big Idea 3: Analysis of Functions (FUN): Calculus connects a function $f$ with its first derivative $f'$ (growth, extrema) and second derivative $f''$ (concavity, inflection points) both algebraically and graphically.
Big Idea 4: Integration & Accumulation (INT): The definite integral calculates the accumulated net change of a rate over time. The Fundamental Theorem of Calculus (FTC) unifies differentiation and integration as inverse processes.

⚠️ Authentic AP Calculus AB Exam Traps & Pitfalls

Equating directly to indeterminate forms: Writing $\lim = \frac{0}{0}$ or $\lim = \frac{\infty}{\infty}$ is penalized by AP readers. Always evaluate numerator and denominator limits separately before applying L'Hôpital's Rule.
Omitting hypotheses in theorem justifications: Citing MVT, IVT, or EVT without stating that the function is continuous on $[a, b]$ (and differentiable on $(a, b)$ for MVT) forfeits justification points.
Rounding prematurely: Intermediate rounding in calculator-active problems causes final discrepancies. Always keep full precision in your graphing calculator and round or truncate only the final answer to 3 decimal places.
Confusing absolute extrema with local extrema: The First Derivative Test proves local extrema only. To find absolute extrema on a closed interval, you must use the Candidates Test evaluating all interior critical points and endpoints.

AP Calculus AB Exam Weightings by Unit. Units 5 (Analytical Applications) and 6 (Integration & Accumulation) constitute over one-third of the total exam, followed by Units 4 and 8.

On the AP Calculus AB exam, Section I (Multiple Choice) consists of 45 questions worth 1 point each, and Section II (Free Response) consists of 6 questions worth 9 points each (54 points). Each section accounts for 50% of the composite score. If the composite score is scaled out of 108 total points, what is the weighting multiplier applied to the raw Section I MCQ score and to the raw Section II FRQ score?

Step 1: Determine total points allocated to each section: $\text{Points per section} = \frac{108}{2} = 54\text{ composite points}.$

Step 2: Find the MCQ conversion factor: $\text{MCQ multiplier} = \frac{54\text{ composite points}}{45\text{ raw points}} = 1.20.$

Step 3: Find the FRQ conversion factor: $\text{FRQ multiplier} = \frac{54\text{ composite points}}{54\text{ raw points}} = 1.00.$

Therefore, each multiple-choice question contributes $1.2$ composite points, while each free-response point contributes $1.0$ composite point to the scaled 108-point total.

A student correctly answers 32 out of 45 questions on Section I (MCQ) and earns 34 out of 54 points on Section II (FRQ). Using the standard College Board composite scaling (MCQ raw

\times 1.2

+ FRQ raw

\times 1.0

), calculate the student's composite score and determine if they exceed the typical cutoff of 68 points needed for a top score of 5.

Step 1: Calculate weighted Section I score: $32 \times 1.2 = 38.4\text{ points}.$

Step 2: Calculate weighted Section II score: $34 \times 1.0 = 34.0\text{ points}.$

Step 3: Sum composite points: $\text{Composite score} = 38.4 + 34.0 = 72.4 \approx 72\text{ points}.$

Because $72.4 > 68$ , this composite score comfortably earns an AP score of 5.

📝 Chapter Quiz

Questions Level

Unit 1 · Limits & Continuity

📌 AP Calc AB Unit 1 — Limits & Continuity 10–12% of exam · Limit laws · Squeeze theorem · Continuity & types of discontinuity · IVT

Calculus begins with the limit — the value a function approaches near a point — and with continuity, which guarantees no breaks in a graph.

▲ Reference — Limits & continuity

Limit laws apply term by term. A function is continuous at $a$ when $\lim_{x\to a}f(x)=f(a)$ . Discontinuities are removable, jump, or infinite. The Intermediate Value Theorem: a function continuous on $[a,b]$ attains every value between $f(a)$ and $f(b)$ .

The function

f(x) = \frac{x^2 - 4}{x - 2} = x + 2

for

x \neq 2

has a removable hole at

(2,4)

. Both one-sided limits equal

4

, so

\lim_{x \to 2} f(x) = 4

despite

f(2)

being undefined.

Limit

\lim_{x\to a}f(x)=L

Squeeze theorem

g\leq f\leq h

and

\lim g=\lim h=L

then

\lim f=L

Continuity

f

continuous at

a

iff

\lim_{x\to a}f(x)=f(a)

Key trig limit

\lim_{x\to0}\dfrac{\sin x}{x}=1

L'Hôpital's rule

\frac{0}{0}

\frac{\pm\infty}{\pm\infty}

\lim\frac{f}{g}=\lim\frac{f'}{g'}

IVT

Continuous

f

[a,b]

attains all intermediate values

Worked example 1.A

Evaluate $\displaystyle\lim_{x\to2}\frac{x^2-4}{x-2}$ .

\lim_{x\to2}\frac{(x-2)(x+2)}{x-2}=\lim_{x\to2}(x+2)=4.

Worked example — ε-δ and IVT

Use the Intermediate Value Theorem to show $f(x)=x^3-x-1$ has a root on $[1,2]$ .

$f(1)=1-1-1=-1<0$ and $f(2)=8-2-1=5>0$ . Since $f$ is continuous on $[1,2]$ and changes sign, by IVT there exists $c\in(1,2)$ with $f(c)=0$ .

⚠ Common traps — Indeterminate forms & theorem hypotheses

L'Hôpital's notation penalty: Never write $= \frac{0}{0}$ or $= \frac{\infty}{\infty}$ on an AP exam! College Board scoring guidelines explicitly penalize equating an expression to an indeterminate form. Always state $\lim_{x\to a} f(x) = 0$ and $\lim_{x\to a} g(x) = 0$ separately before writing L'Hôpital's step.

Always state continuity for IVT: Stating only that $f(a) < k < f(b)$ earns zero points for an IVT justification. You must begin by writing: “Since $f$ is continuous on the closed interval $[a, b]$ ...”

One-sided limits at piece boundaries: For a piecewise function to have a two-sided limit at $x = c$ , the left-hand limit $\lim_{x\to c^-} f(x)$ and right-hand limit $\lim_{x\to c^+} f(x)$ must be finite and equal.

Find

\displaystyle\lim_{x\to0}\dfrac{\sin(3x)}{5x}

Use

\lim_{u\to0}\frac{\sin u}{u}=1

\dfrac{\sin(3x)}{5x}=\dfrac{3}{5}\cdot\dfrac{\sin(3x)}{3x}\to\dfrac{3}{5}

Evaluate

\displaystyle\lim_{x\to 0} \frac{\sqrt{x + 9} - 3}{x}

Multiply numerator and denominator by the conjugate $\sqrt{x + 9} + 3$ :

\lim_{x\to 0} \frac{(\sqrt{x + 9} - 3)(\sqrt{x + 9} + 3)}{x(\sqrt{x + 9} + 3)} = \lim_{x\to 0} \frac{(x + 9) - 9}{x(\sqrt{x + 9} + 3)} = \lim_{x\to 0} \frac{x}{x(\sqrt{x + 9} + 3)} = \lim_{x\to 0} \frac{1}{\sqrt{x + 9} + 3} = \frac{1}{6}.

📝 Chapter Quiz

Questions Level

Unit 2 · Differentiation — Definition & Basic Rules

📌 AP Calc AB Unit 2 — Differentiation 10–12% of exam · Limit definition of derivative · Power/trig/exp/ln rules · Higher-order derivatives

The derivative is the instantaneous rate of change: the slope of the tangent line, defined as a limit of average rates.

Below, $f(x)=\tfrac{x^{3}}{3}-x$ is in indigo and its derivative $f'(x)=x^{2}-1$ in amber. Slide $x$ : the height of the amber curve equals the slope of the indigo one.

⊙ Explorer · A function and its derivativeinteractive

Point x = 1

f(x)—

f '(x) = slope—

Where $f$ is flat ( $f'=0$ ) the amber curve crosses zero — those are the max and min of $f$ .

▲ Reference — The derivative

$f'(x)=\displaystyle\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$ . From it follow the power, constant-multiple, sum, product, and quotient rules.

As the point

Q(x+h, f(x+h))

slides toward

P(x, f(x))

(letting

h \to 0

), the slope of the secant line approaches the slope of the tangent line:

f'(1) = 2

Definition

\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}

Power rule

\dfrac{d}{dx}x^{n}=nx^{n-1}

Product

(uv)'=u'v+uv'

Quotient

\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}

Worked example 2.A

Differentiate $f(x)=3x^{4}-2x$ .

f'(x)=12x^{3}-2.

Worked example — Differentiability from the limit definition

Find $f'(x)$ for $f(x)=x^2-3x$ using the limit definition.

f'(x)=\lim_{h\to0}\frac{(x+h)^2-3(x+h)-(x^2-3x)}{h}=\lim_{h\to0}\frac{2xh+h^2-3h}{h}=\lim_{h\to0}(2x+h-3)=2x-3.

⚠ Common traps — Quotient rule order & non-differentiability

Quotient rule order matters: $\left(\dfrac{u}{v}\right)' = \dfrac{u'v - uv'}{v^2}$ . Reversing the numerator to $uv' - u'v$ introduces a false negative sign across the entire derivative.

Continuity does not imply differentiability: A function can be continuous everywhere but fail to be differentiable at sharp corners (e.g. $f(x) = |x|$ at $x=0$ ), cusps ( $x^{2/3}$ at $x=0$ ), or vertical tangents ( $x^{1/3}$ at $x=0$ ).

Power rule on fractions: When differentiating $f(x) = \dfrac{1}{x^3}$ , write it as $x^{-3}$ first: $f'(x) = -3x^{-4} = -\dfrac{3}{x^4}$ . Do not simply differentiate the denominator in place!

Differentiate

y=x^4\sin x

. (Product rule)

y'=4x^3\sin x+x^4\cos x

Find the derivative of

f(x) = \dfrac{2x - 1}{x^2 + 3}

using the quotient rule.

Apply the quotient rule $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$ :

f'(x) = \frac{2(x^2 + 3) - (2x - 1)(2x)}{(x^2 + 3)^2} = \frac{2x^2 + 6 - (4x^2 - 2x)}{(x^2 + 3)^2} = \frac{-2x^2 + 2x + 6}{(x^2 + 3)^2}.

📝 Chapter Quiz

Questions Level

Unit 3 · Composite, Implicit & Inverse Functions

📌 AP Calc AB Unit 3 — Composite, Implicit & Inverse 9–13% of exam · Chain rule · Implicit differentiation · Inverse trig derivatives

The chain rule differentiates compositions; implicit differentiation handles curves not solved for $y$ ; and inverse-function derivatives complete the toolkit.

▲ Reference — Chain rule & friends

$\dfrac{d}{dx}f(g(x))=f'(g(x))\,g'(x)$ . Differentiate relations implicitly term by term. For inverses, $(f^{-1})'(b)=\dfrac{1}{f'(a)}$ where $f(a)=b$ .

Implicit differentiation of

x^2 + y^2 = 25

yields

\frac{dy}{dx} = -\frac{x}{y}

. At

P(3,4)

, the tangent slope is

-\frac{3}{4}

, forming tangent line

y - 4 = -\frac{3}{4}(x - 3)

Chain rule

f'(g)\,g'

Exponential

\dfrac{d}{dx}e^{x}=e^{x}

Logarithm

\dfrac{d}{dx}\ln x=\dfrac1x

Sine

\dfrac{d}{dx}\sin x=\cos x

Worked example 3.A

Differentiate $\sin(x^{2})$ .

\frac{d}{dx}\sin(x^{2})=2x\cos(x^{2}).

Worked example — Implicit differentiation

Find $\dfrac{dy}{dx}$ for $x^2+y^2=25$ and the equation of the tangent at $(3,4)$ .

2x+2y\frac{dy}{dx}=0\;\Rightarrow\;\frac{dy}{dx}=-\frac{x}{y}.

At $(3,4)$ : slope $=-\tfrac{3}{4}$ . Tangent: $y-4=-\tfrac{3}{4}(x-3)$ .

⚠ Common traps — Missing chain factor & implicit product rule

Forgetting the inside derivative: The derivative of $f(g(x))$ requires multiplying by $g'(x)$ . A classic error is writing $\frac{d}{dx}[\cos(4x)] = -\sin(4x)$ instead of $-4\sin(4x)$ , or omitting the inner derivative on composite powers like $(3x - 1)^5$ .

Hidden product rule in implicit terms: When differentiating $x^2y$ or $3xy$ with respect to $x$ , apply the product rule: $\frac{d}{dx}(xy) = (1)y + x\frac{dy}{dx}$ . Treating $y$ as a constant is the most common error in implicit differentiation FRQs.

Inverse function derivative argument: $(f^{-1})'(b) = \dfrac{1}{f'(a)}$ where $f(a) = b$ . The derivative $f'$ is evaluated at the input $a$ , not at the output $b$ .

Find

\dfrac{d}{dx}[\arctan(2x)]

Chain rule with

\dfrac{d}{du}[\arctan u]=\dfrac{1}{1+u^2}

\dfrac{2}{1+4x^2}

Given the curve

x^3 + y^3 = 6xy

(the Folium of Descartes), find the slope of the tangent line at the point

(3, 3)

Differentiate implicitly with respect to $x$ :

3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x\frac{dy}{dx}.

Substitute $x = 3$ and $y = 3$ :

3(9) + 3(9)\frac{dy}{dx} = 6(3) + 6(3)\frac{dy}{dx} \implies 27 + 27\frac{dy}{dx} = 18 + 18\frac{dy}{dx}.

9\frac{dy}{dx} = -9 \implies \frac{dy}{dx} = -1.

📝 Chapter Quiz

Questions Level

Unit 4 · Contextual Applications of Differentiation

📌 AP Calc AB Unit 4 — Contextual Applications 10–15% of exam · Related rates · Straight-line motion · L'Hôpital's rule

Derivatives model the real world: velocity and acceleration, related rates, and local linear approximation.

▲ Reference — Rates in context

For motion, $v(t)=s'(t)$ and $a(t)=v'(t)$ . Related rates differentiate a relationship with respect to time. Linearization: $L(x)=f(a)+f'(a)(x-a)$ . L'Hôpital's rule resolves $\tfrac00$ and $\tfrac{\infty}{\infty}$ .

Velocity

v(t) = 3t(t-4)

and acceleration

a(t) = 6(t-2)

. On

(2,4)

, both

v(t) < 0

and

a(t) > 0

(opposite signs), so speed is decreasing; on

(4,\infty)

, both are positive, so speed is increasing.

Motion

v=s',\ a=v'

Related rates

differentiate in

t

Linearization

f(a)+f'(a)(x-a)

L'Hôpital

\lim\dfrac{f}{g}=\lim\dfrac{f'}{g'}

Worked example 4.A

A particle has position $s(t)=t^{3}-6t^{2}$ . When is it at rest?

v(t)=3t^{2}-12t=3t(t-4)=0\ \Rightarrow\ t=0,\ 4.

Worked example — Related rates

A spherical balloon is inflated so its volume increases at 10 cm³/s. How fast is the radius increasing when $r=5$ cm?

V=\tfrac{4}{3}\pi r^3\;\Rightarrow\;\frac{dV}{dt}=4\pi r^2\frac{dr}{dt}.

10=4\pi(25)\frac{dr}{dt}\;\Rightarrow\;\frac{dr}{dt}=\frac{10}{100\pi}=\frac{1}{10\pi}\approx0.032\text{ cm/s}.

⚠ Common traps — Speeding up vs slowing down & early substitution

Speeding up requires matching signs: A particle is speeding up when velocity $v(t)$ and acceleration $a(t)$ have the same sign (both positive or both negative). It is slowing down when they have opposite signs. A negative acceleration does NOT automatically mean slowing down!

Premature variable substitution in related rates: Never plug changing numeric quantities (e.g. $r = 5$ or $h = 4$ ) into your geometric formula before differentiating! Differentiate the equation implicitly with respect to $t$ first, then substitute instantaneous values.

Displacement vs total distance: Displacement over $[a, b]$ is $\int_a^b v(t)\,dt = s(b) - s(a)$ . Total distance travelled requires integrating speed: $\int_a^b |v(t)|\,dt$ .

Use L'Hôpital's rule to evaluate

\displaystyle\lim_{x\to0}\dfrac{e^x-1-x}{x^2}

Form

0/0

. Apply once:

\dfrac{e^x-1}{2x}

. Still

0/0

. Apply again:

\dfrac{e^x}{2}\to\dfrac{1}{2}

A 10-foot ladder rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 2 ft/sec, how fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?

Let $x$ be the distance from the bottom of the ladder to the wall, and $y$ be the height of the top of the ladder on the wall. By the Pythagorean theorem, $x^2 + y^2 = 10^2 = 100$ . When $x = 6$ , $y = \sqrt{100 - 36} = 8$ ft.

Differentiate with respect to time $t$ :

2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \implies x\frac{dx}{dt} + y\frac{dy}{dt} = 0.

Substitute $x = 6$ , $y = 8$ , and $\frac{dx}{dt} = 2$ :

6(2) + 8\frac{dy}{dt} = 0 \implies 12 + 8\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{12}{8} = -1.5\text{ ft/sec}.

The top of the ladder is sliding down at a rate of $1.5\text{ ft/sec}$ .

📝 Chapter Quiz

Questions Level

Unit 5 · Analytical Applications of Differentiation

📌 AP Calc AB Unit 5 — Analytical Applications 15–18% of exam · MVT & EVT · 1st & 2nd derivative tests · Optimisation · Concavity

Derivatives reveal where a function rises, falls and curves, and pin down its maxima and minima.

The Mean Value Theorem made visible: the amber secant joins the endpoints, and somewhere between them a tangent (dashed) runs exactly parallel. Slide the right endpoint $b$ .

⊙ Explorer · Mean Value Theoreminteractive

Right endpoint b = 6

secant slope—

tangent at c—

There is a $c$ in $(a,b)$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ — the ringed point.

▲ Reference — Shape of a graph

Mean Value Theorem: some $c$ has $f'(c)=\dfrac{f(b)-f(a)}{b-a}$ . $f$ increases where $f'>0$ and is concave up where $f''>0$ . The first- and second-derivative tests classify critical points; optimization finds extreme values.

f(x) = x^3 - 3x^2

f'(x) = 3x(x-2) = 0

identifies critical points at

x=0

and

x=2

. The second derivative

f''(x) = 6x - 6

changes from negative (concave down) to positive (concave up) at the inflection point

(1,-2)

MVT

f'(c)=\dfrac{f(b)-f(a)}{b-a}

Critical points

f'=0

or undefined

Concavity

sign of

f''

2nd-deriv test

f''>0\Rightarrow

min

Worked example 5.A

Classify the critical points of $f(x)=x^{3}-3x^{2}$ .

$f'(x)=3x^{2}-6x=3x(x-2)$ , so $x=0,2$ . With $f''(x)=6x-6$ : $f''(0)<0$ , $f''(2)>0$ .

\text{Local max at }x=0,\quad \text{local min at }x=2.

Worked example — Optimisation

A farmer has 200 m of fencing to enclose a rectangular field with one side along a river (no fence needed there). What dimensions maximise the area?

Let width $=x$ , length $=y$ . Constraint: $2x+y=200\Rightarrow y=200-2x$ .

A=xy=x(200-2x)=200x-2x^2.

A'=200-4x=0\;\Rightarrow\;x=50\text{ m},\;y=100\text{ m}.\quad A_{\max}=5000\text{ m}^2.

⚠ Common traps — FRQ justification language & inflection points

Strict AP justification phrasing: State “ $f$ has a local minimum at $x = 2$ because $f'(x)$ changes from negative to positive at $x = 2$ ”. Never write “the graph turns around” or “it changes sign” without specifying $f'$ . Referring to an unnamed “it” forfeits justification points.

$f''(c) = 0$ alone is NOT an inflection point: An inflection point requires an actual change in concavity (i.e. $f''(x)$ must change sign across $c$ ). For example, $f(x) = x^4$ has $f''(0) = 0$ , but $f''(x) \ge 0$ everywhere, so there is no inflection point.

Candidates Test for absolute extrema: On a closed interval $[a, b]$ , absolute extrema can occur at critical points OR at endpoints. You must construct a table evaluating $f(x)$ at $x = a$ , $x = b$ , and all interior critical points.

Verify the Mean Value Theorem for

f(x)=x^3

[0,2]

by finding the value

c

f'(x)=3x^2

. MVT:

f'(c)=\dfrac{f(2)-f(0)}{2-0}=\dfrac{8}{2}=4

. So

3c^2=4\Rightarrow c=\dfrac{2}{\sqrt3}=\dfrac{2\sqrt3}{3}\approx1.155\in(0,2)

✓.

Find all local extrema and points of inflection for

f(x) = x^4 - 4x^3

First derivative: $f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$ . Critical points are $x = 0$ and $x = 3$ . Testing signs of $f'$ : for $x < 0$ , $f' < 0$ ; for $0 < x < 3$ , $f' < 0$ ; for $x > 3$ , $f' > 0$ . By the First Derivative Test, $x = 0$ is neither a local max nor min (stationary inflection), and $x = 3$ is a local minimum at $(3, -27)$ .

Second derivative: $f''(x) = 12x^2 - 24x = 12x(x - 2)$ . Setting $f''(x) = 0$ gives $x = 0$ and $x = 2$ . Because $f''$ changes sign at both $x = 0$ and $x = 2$ , the inflection points are $(0, 0)$ and $(2, -16)$ .

📝 Chapter Quiz

Questions Level

Unit 6 · Integration & Accumulation of Change

📌 AP Calc AB Unit 6 — Integration & Accumulation 17–20% of exam · Riemann sums · FTC Part 1 & 2 · u-substitution · Definite integrals

Integration reverses differentiation and accumulates change — the second great engine of calculus.

A definite integral is a limit of Riemann sums. Add rectangles and switch the sample point — the amber estimate marches toward the exact area.

⊙ Explorer · Riemann sumsinteractive

Rectangles n = 6

Sample point

rectangle estimate—

exact integral—

As $n\to\infty$ the estimate converges to $\displaystyle\int_0^4(0.4x^{2}+1)\,dx$ .

▲ Reference — The Fundamental Theorem

Part 1: $\dfrac{d}{dx}\displaystyle\int_a^x f(t)\,dt=f(x)$ . Part 2: $\displaystyle\int_a^b f=F(b)-F(a)$ . Use $u$ -substitution to reverse the chain rule, and Riemann sums to approximate area.

Definite integral

\int_0^3 (2x - x^2)\,dx = -\frac{2}{3}

. The positive area above the

x

-axis on

[0,2]

(

+\frac{4}{3}

) is cancelled by the negative area below the

x

-axis on

[2,3]

(

-2

), giving a negative net signed accumulation.

Power rule

\displaystyle\int x^{n}dx=\dfrac{x^{n+1}}{n+1}+C

FTC

\displaystyle\int_a^b f=F(b)-F(a)

u-substitution

\displaystyle\int f(g)g'\,dx

Accumulation

\displaystyle\int_a^b f

Worked example 6.A

Evaluate $\displaystyle\int_0^2 3x^{2}\,dx$ .

\big[x^{3}\big]_0^2=8.

Worked example — FTC Part 1 & 2

If $g(x)=\int_0^x\sin(t^2)\,dt$ , find $g'(x)$ and evaluate $\int_1^3(2x-x^2)\,dx$ .

FTC Part 1: $g'(x)=\sin(x^2)$ .

\int_1^3(2x-x^2)\,dx=\left[x^2-\frac{x^3}{3}\right]_1^3=\left(9-9\right)-\left(1-\frac{1}{3}\right)=0-\frac{2}{3}=-\frac{2}{3}.

⚠ Common traps — FTC Part 1 chain rule & limits of u-substitution

FTC Part 1 with a function in the upper limit: If $g(x) = \int_a^{u(x)} f(t)\,dt$ , then by the chain rule: $g'(x) = f(u(x)) \cdot u'(x)$ . A frequent error is omitting the factor $u'(x)$ ; for example, $\frac{d}{dx}\int_1^{x^3} \cos(t)\,dt = \cos(x^3) \cdot 3x^2$ , NOT $\cos(x^3)$ .

Changing limits during definite $u$ -substitution: Always convert the limits of integration from $x$ to $u$ when changing variables: $\int_{x=a}^{x=b} f(g(x))g'(x)\,dx = \int_{u=g(a)}^{u=g(b)} f(u)\,du$ . Writing $x$ -limits on a $u$ -integral is mathematically incorrect and loses notation points.

Area vs value of an integral: A definite integral evaluates net signed area (regions below the $x$ -axis count negatively). If a question asks for total geometric area, you must integrate the absolute value $\int_a^b |f(x)|\,dx$ .

Evaluate

\displaystyle\int\frac{3x^2}{x^3+1}\,dx

Let

u=x^3+1

du=3x^2\,dx

. Integral

=\int\dfrac{du}{u}=\ln|u|+c=\ln|x^3+1|+c

Let

F(x) = \displaystyle\int_2^{x^2} \sqrt{1 + t^3}\,dt

. Find

F'(x)

using the Fundamental Theorem of Calculus and the Chain Rule.

By the Fundamental Theorem of Calculus Part 1 combined with the Chain Rule, if $F(x) = \int_a^{u(x)} f(t)\,dt$ , then $F'(x) = f(u(x)) \cdot u'(x)$ :

F'(x) = \sqrt{1 + (x^2)^3} \cdot \frac{d}{dx}[x^2] = \sqrt{1 + x^6} \cdot 2x = 2x\sqrt{1 + x^6}.

📝 Chapter Quiz

Questions Level

Unit 7 · Differential Equations

📌 AP Calc AB Unit 7 — Differential Equations 6–12% of exam · Slope fields · Separation of variables · Exponential growth & decay

A differential equation links a function to its derivatives; solving one means recovering the family of curves that fit.

▲ Reference — Separable equations

Slope fields sketch solution curves. A separable equation $\dfrac{dy}{dx}=g(x)h(y)$ solves as $\displaystyle\int\frac{dy}{h(y)}=\int g(x)\,dx$ . Exponential change obeys $y=y_0e^{kt}$ .

Slope field for

\frac{dy}{dx} = x - y

with particular solution curve through

(0,1)

. Each segment depicts the local derivative direction

\frac{dy}{dx}

at that grid point.

Separable

\displaystyle\int\dfrac{dy}{h(y)}=\int g\,dx

Exponential

y=y_0e^{kt}

Slope field

plot

y'

directions

Particular

fix

C

from a point

Worked example 7.A

Solve $\dfrac{dy}{dx}=2y$ with $y(0)=3$ .

y=3e^{2x}.

Worked example — Slope field & separation

Solve $\dfrac{dy}{dx}=2xy$ with $y(0)=3$ .

\frac{dy}{y}=2x\,dx\;\Rightarrow\;\ln|y|=x^2+C\;\Rightarrow\;y=Ae^{x^2}.

Initial condition: $3=Ae^0=A$ . Solution: $y=3e^{x^2}$ .

⚠ Common traps — Separating variables & the initial constant +C

Separation must be multiplicative: An equation is separable only if it can be written as $\frac{dy}{dx} = g(x)h(y)$ . An expression like $\frac{dy}{dx} = x - y$ cannot be separated by adding or subtracting terms across sides.

Always write $+C$ on the integration step: Writing $\ln|y| = x^2 + C$ at the exact moment of anti-differentiation is mandatory. On AP FRQ scoring guidelines, omitting $+C$ here caps the question at at most 1 point out of 5 or 6!

Exponents and the constant: $\ln|y| = x^2 + C \implies |y| = e^{x^2 + C} = e^C e^{x^2} = A e^{x^2}$ . Writing $y = e^{x^2} + C$ is a fatal algebraic error.

A population grows at a rate proportional to its size. If

P(0)=500

and

P(2)=800

, find

P(t)

and determine when

P=2000

P=500e^{kt}

800=500e^{2k}\Rightarrow k=\frac{\ln(8/5)}{2}\approx0.2350

2000=500e^{kt}\Rightarrow t=\frac{\ln4}{k}\approx5.90

years.

Find the particular solution to the differential equation

\dfrac{dy}{dx} = 2x(y - 1)

satisfying the initial condition

y(0) = 4

Separate variables:

\frac{1}{y - 1}\,dy = 2x\,dx.

Integrate both sides:

\ln|y - 1| = x^2 + C \implies |y - 1| = e^{x^2 + C} = A e^{x^2}.

Substitute $x = 0, y = 4$ :

|4 - 1| = 3 = A e^0 \implies A = 3.

Since $y(0) = 4 > 1$ , $y - 1 = 3e^{x^2}$ , so the particular solution is:

y(x) = 3e^{x^2} + 1.

📝 Chapter Quiz

Questions Level

Unit 8 · Applications of Integration

📌 AP Calc AB Unit 8 — Applications of Integration 10–15% of exam · Area between curves · Disk/washer volumes · Average value of a function

Definite integrals measure areas between curves, volumes of solids, and average values.

▲ Reference — Area, volume, average

Area between curves: $\displaystyle\int(\text{top}-\text{bottom})\,dx$ . Volumes by disks/washers about an axis. Average value: $\dfrac{1}{b-a}\displaystyle\int_a^b f$ . Solids with known cross-sections integrate the section area.

The region bounded between

y = x

and

y = x^2

[0,1]

has area

\int_0^1 (x - x^2)\,dx = \frac{1}{6}

. The top curve is

y = x

and the bottom curve is

y = x^2

Area

\displaystyle\int(\text{top}-\text{bot})

Disk

\pi\displaystyle\int R^{2}\,dx

Washer

\pi\displaystyle\int (R^{2}-r^{2})

Average value

\dfrac{1}{b-a}\displaystyle\int_a^b f

Worked example 8.A

Find the area between $y=x$ and $y=x^{2}$ on $[0,1]$ .

\int_0^1\big(x-x^{2}\big)\,dx=\frac12-\frac13=\frac16.

Worked example — Disk/washer volume

Find the volume of the solid formed by revolving $y=\sqrt{x}$ about the $x$ -axis on $[0,4]$ .

V=\pi\int_0^4(\sqrt{x})^2\,dx=\pi\int_0^4 x\,dx=\pi\left[\frac{x^2}{2}\right]_0^4=\pi\cdot8=8\pi.

⚠ Common traps — Washer radius squaring & average value denominator

Square each radius individually: In the washer method, volume is $\pi \int (R^2 - r^2)\,dx$ , NOT $\pi \int (R - r)^2\,dx$ . Subtracting radii before squaring calculates the volume of an entirely different, incorrect solid.

Revolving around lines other than the axes: When revolving around $y = k$ , the radius is $\text{distance} = |y_{\text{curve}} - k|$ . Never assume the radius is simply $f(x)$ when the axis of rotation has shifted!

Average value vs average rate of change: Average value of $f$ on $[a, b]$ is $\dfrac{1}{b - a}\int_a^b f(x)\,dx$ . Average rate of change of $f$ on $[a, b]$ is $\dfrac{f(b) - f(a)}{b - a}$ . Mixing up these two definitions is a frequent loss of points on both MC and FRQ sections.

Find the area between

y=x^2

and

y=2x

Intersect:

x^2=2x\Rightarrow x=0,2

. Area

=\int_0^2(2x-x^2)\,dx=\left[x^2-\frac{x^3}{3}\right]_0^2=4-\frac{8}{3}=\frac{4}{3}

The base of a solid is the region in the

xy

-plane enclosed by the circle

x^2 + y^2 = 4

. Cross sections perpendicular to the

x

-axis are squares. Find the volume of the solid.

For any $x \in [-2, 2]$ , the top boundary of the circle is $y = \sqrt{4 - x^2}$ and the bottom is $y = -\sqrt{4 - x^2}$ . The side length of the square cross section is:

s(x) = 2\sqrt{4 - x^2}.

The area of each square cross section is:

A(x) = [s(x)]^2 = 4(4 - x^2) = 16 - 4x^2.

Integrating along the $x$ -axis from $-2$ to $2$ :

V = \int_{-2}^2 (16 - 4x^2)\,dx = 2 \int_0^2 (16 - 4x^2)\,dx = 2\left[16x - \frac{4}{3}x^3\right]_0^2 = 2\left(32 - \frac{32}{3}\right) = 2\left(\frac{64}{3}\right) = \frac{128}{3}.

📝 Chapter Quiz

Questions Level

Exam format

● Section I — Multiple choice · 50%

45 questions, 1 hr 45 min total. Part A: 30 questions, 60 min, no calculator. Part B: 15 questions, 45 min, graphing calculator required.

● Section II — Free response · 50%

6 questions, 1 hr 30 min total. Part A: 2 questions, 30 min, calculator. Part B: 4 questions, 60 min, no calculator.

The composite score is reported on the 1–5 scale. Free-response questions reward clear setup and justification, not just the final number — show the integral or derivative you evaluate.

Section I-A

30 MCQ · No calculator · 60 min

Section I-B

15 MCQ · Calculator · 45 min

Section II-A

2 FRQ · Calculator · 30 min

Section II-B

4 FRQ · No calculator · 60 min

● Scoring

MCQ: 45 points raw (1 pt each, no penalty). FRQ: 54 points (6 problems × 9 pts). Total 99 pts scaled to 1–5. Score of 3+ is generally considered passing.

⚠ Critical College Board exam rules & timing traps

The 3-decimal rounding rule: Any numerical answer derived on calculator-active sections (Section I Part B and Section II Part A) must be reported to at least 3 places after the decimal point (rounded or truncated). Never round intermediate values during your calculations!

Setup is required on FRQs: Writing down only a numerical answer obtained from your calculator with no calculus setup earns zero credit. Always write the definite integral, derivative equation, or limit before reporting the calculator's result.

Strict calculator containment: In Section II Part B (the final 60 minutes), you must put your graphing calculator away. You may continue working on or checking Questions 1 and 2, but without the calculator.

Exam strategy — Pacing & maximizing partial credit

Pacing target: Budget approximately 2 minutes per multiple-choice question in Part A (60 min / 30 Q) and 3 minutes per question in Part B (45 min / 15 Q). On Section II, spend approximately 15 minutes per Free Response Question.

Harvesting independent points: If part (b) of an FRQ asks you to use the answer from part (a), but you were unable to solve (a), clearly state an assumed value or symbolic representation for (a) and carry out the correct calculus method in (b). AP readers award full follow-through points for correct logic!

AP Calculus AB Exam Structure and Time Allocation (minutes). Section I comprises 105 minutes across two multiple-choice parts; Section II comprises 90 minutes across two free-response parts.

In Section I Part A of the AP Calculus AB exam, students must complete 30 non-calculator multiple-choice questions in 60 minutes. If a student spends 90 seconds on average on each of the first 20 questions, how many minutes remain on average for each of the final 10 questions?

Step 1: Calculate total time available in seconds: $60\text{ minutes} \times 60\text{ seconds/minute} = 3600\text{ seconds}.$

Step 2: Calculate time spent on the first 20 questions: $20 \times 90\text{ seconds} = 1800\text{ seconds (30 minutes)}.$

Step 3: Calculate remaining time: $3600 - 1800 = 1800\text{ seconds (30 minutes)}.$

Step 4: Calculate average time per remaining question: $\frac{1800\text{ seconds}}{10\text{ questions}} = 180\text{ seconds} = 3.0\text{ minutes per question}.$

Section II Part A consists of 2 free-response questions in 30 minutes where a graphing calculator is required. A student uses numerical integration (fnInt) to evaluate

\int_1^4 \sqrt{x^3 + 2}\,dx

. The calculator displays

12.784619

. According to College Board AP scoring conventions, what acceptable rounded or truncated value should the student report, and what setup must be written in the exam booklet?

Step 1: State the required mathematical setup: Writing only the numeric result earns 0 points. The student must write the definite integral: $\int_1^4 \sqrt{x^3 + 2}\,dx.$

Step 2: Apply the 3-decimal rule: AP guidelines accept answers rounded or truncated to 3 decimal places:

Truncated: $12.784$
Rounded: $12.785$

Both $12.784$ and $12.785$ receive full credit.

📝 Chapter Quiz

Questions Level

✶

Worked FRQs

Show all work

Partial credit awarded for correct method even with arithmetic errors

State theorems

Write MVT, IVT, FTC explicitly when applying them

Units

Always include correct units in applied problems

Rounding

Round to 3 decimal places only at the final step

Velocity

v(t) = (t-1)(t-3)

for particle motion. The particle is moving right when

v(t) > 0

[0,1) \cup (3,\infty)

, and moving left when

v(t) < 0

(1,3)

. Total distance requires integrating the speed

|v(t)|

FRQ 1 · Particle motion

A particle moves along a line with velocity $v(t)=t^2-4t+3$ for $t\geq 0$ .

(a) When is the particle at rest? $v(t)=(t-1)(t-3)=0 \Rightarrow t=1,\ 3$ .

(b) On $[0,3]$ , when is it moving left? $v<0$ on $(1,3)$ , so it moves left there (and right on $[0,1)$ ).

(c) Total distance travelled on $[0,3]$ :

\int_0^3 |v(t)|\,dt=\Big|\!\int_0^1 v\,\Big|+\Big|\!\int_1^3 v\,\Big|=\tfrac43+\tfrac43=\tfrac83.

FRQ 2 · Area & volume

Let $R$ be the region bounded by $y=\sqrt{x}$ and $y=x$ .

(a) Area of $R$ (here $\sqrt{x}\geq x$ on $[0,1]$ ):

\int_0^1\big(\sqrt{x}-x\big)\,dx=\Big[\tfrac{2}{3}x^{3/2}-\tfrac{x^2}{2}\Big]_0^1=\tfrac23-\tfrac12=\tfrac16.

(b) Volume when $R$ is revolved about the $x$ -axis (washers):

\pi\int_0^1\big[(\sqrt{x})^2-x^2\big]\,dx=\pi\int_0^1\big(x-x^2\big)\,dx=\frac{\pi}{6}.

FRQ 3 · Rate In / Rate Out Accumulation Model

Water enters a municipal reservoir at a rate modelled by $R(t) = 200 + 40\sin\left(\frac{\pi t}{12}\right)$ gallons/hour and is pumped out at a constant rate of $L(t) = 180$ gallons/hour for $0 \leq t \leq 24$ hours. At $t = 0$ , the reservoir contains $1200$ gallons.

(a) Amount of water entering: The total water added on $[0, 8]$ is $\int_0^8 R(t)\,dt \approx 1845.672 \text{ gallons}$ .

(b) Rate of change of volume: At $t = 6$ , the net rate of change is $V'(6) = R(6) - L(6) = \left[200 + 40\sin\left(\frac{\pi}{2}\right)\right] - 180 = 240 - 180 = 60 \text{ gal/hr}$ . Because $V'(6) > 0$ , the amount of water in the reservoir is increasing at $t = 6$ .

(c) Absolute Extrema (Candidates Test): The total volume is $V(t) = 1200 + \int_0^t [R(u) - L(u)]\,du$ . Critical points occur where $V'(t) = R(t) - L(t) = 0 \Rightarrow 200 + 40\sin\left(\frac{\pi t}{12}\right) = 180 \Rightarrow \sin\left(\frac{\pi t}{12}\right) = -0.5 \Rightarrow \frac{\pi t}{12} = \frac{7\pi}{6} \Rightarrow t = 14$ . Evaluate candidates: $V(0) = 1200$ , $V(14) \approx 1895.348$ , and $V(24) \approx 1680$ . By the Candidates Test, the maximum volume is approximately $1895.348 \text{ gallons}$ at $t = 14 \text{ hours}$ .

⚠️ AP Reader Scoring Traps & FRQ Writing Conventions

The Pronoun 'It' Penalty: Never write 'it is increasing because it is positive'. AP readers do not assume what 'it' refers to. Always explicitly name the function and derivative: 'The function $f$ is increasing on $(a, b)$ because $f'(x) > 0$ .'
Candidates Test Format: To justify an absolute extremum on a closed interval $[a, b]$ , you must explicitly evaluate and compare $f(x)$ at all interior critical numbers AND both boundary endpoints. Showing that $f'$ changes sign is only valid for local extrema, not absolute extrema on an interval with endpoints!
Separate Units Point: AP free-response parts frequently include an independent point just for correct units (e.g. $\text{ft/sec}^2$ vs $\text{ft/sec}$ ). Always check the problem stem for 'using correct units, explain the meaning of...' and provide units even if you had difficulty with earlier arithmetic.
Never Round Intermediate Calculations: Store full decimal values in your calculator. Only round or truncate to 3 decimal places at your final stated answer.

A particle moves along a horizontal line so that its position is given by

s(t) = t^3 - 6t^2 + 9t + 2

for

t \ge 0

. On what open time interval(s) is the particle moving to the left, and what is its acceleration when

t = 2

Step 1: Find velocity $v(t)$ : $v(t) = s'(t) = 3t^2 - 12t + 9 = 3(t^2 - 4t + 3) = 3(t - 1)(t - 3).$

Step 2: Determine where $v(t) < 0$ : The roots of $v(t)$ are $t = 1$ and $t = 3$ . Testing intervals: for $t \in (1, 3)$ , $v(t) < 0$ . Thus the particle is moving to the left on $(1, 3)$ .

Step 3: Find acceleration $a(t)$ and evaluate at $t = 2$ : $a(t) = v'(t) = 6t - 12.$ $a(2) = 6(2) - 12 = 0.$

Let

R

be the region in the first quadrant bounded by the graph of

y = 4 - x^2

and the coordinate axes. Find the volume of the solid generated when

R

is revolved about the horizontal line

y = -1

Step 1: Identify boundaries and radii of revolution: The region $R$ extends from $x = 0$ to $x = 2$ where $4 - x^2 = 0$ . Revolved about $y = -1$ , each cross-section perpendicular to the $x$ -axis is a washer with:

Outer radius: $R(x) = (4 - x^2) - (-1) = 5 - x^2$
Inner radius: $r(x) = 0 - (-1) = 1$

Step 2: Set up the volume integral (washer method): $V = \pi \int_0^2 \left([R(x)]^2 - [r(x)]^2\right)\,dx = \pi \int_0^2 \left((5 - x^2)^2 - 1^2\right)\,dx.$

Step 3: Expand the integrand: $(25 - 10x^2 + x^4) - 1 = 24 - 10x^2 + x^4.$

Step 4: Integrate and evaluate: $V = \pi \left[24x - \frac{10}{3}x^3 + \frac{1}{5}x^5\right]_0^2 = \pi \left(48 - \frac{80}{3} + \frac{32}{5}\right) = \pi \left(\frac{720 - 400 + 96}{15}\right) = \frac{416\pi}{15}.$

📝 Chapter Quiz

Questions Level

⊘

Exam practice

Differentiate $f(x)=x^2\sin x$ .

MCQ

Product rule:

f'(x)=2x\sin x+x^2\cos x.

Evaluate $\displaystyle\int_1^{2} 3x^2\,dx$ .

MCQ

\big[x^3\big]_1^{2}=8-1=7.

At which $x$ does $f(x)=x^3-3x^2+2$ have a local maximum?

FRQ

$f'(x)=3x^2-6x=3x(x-2) \Rightarrow x=0,2$ . With $f''(x)=6x-6$ : $f''(0)=-6<0$ .

\text{Local maximum at } x=0.

Evaluate $\displaystyle\lim_{x\to 3}\frac{x^2-9}{x-3}$ .

MCQ

Factor and cancel: $\tfrac{(x-3)(x+3)}{x-3}$ .

\lim_{x\to 3}(x+3)=6.

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sin 5x}{x}$ .

MCQ

5\lim_{x\to 0}\frac{\sin 5x}{5x}=5.

Evaluate $\displaystyle\lim_{x\to\infty}\frac{3x^2+2}{x^2-1}$ .

MCQ

Ratio of leading coefficients:

3.

Evaluate $\displaystyle\lim_{x\to 0}\frac{\sqrt{x+9}-3}{x}$ .

FRQ

Conjugate $\sqrt{x+9}+3$ :

\lim_{x\to 0}\frac{1}{\sqrt{x+9}+3}=\frac16.

Evaluate $\displaystyle\lim_{x\to 0}\frac{1-\cos x}{x}$ .

MCQ

Form $\tfrac00$ ; L'Hôpital gives $\tfrac{\sin x}{1}\to 0$ :

0.

Differentiate $f(x)=4x^5-2x^2$ .

MCQ

f'(x)=20x^4-4x.

Differentiate $y=\sin 2x$ .

MCQ

y'=2\cos 2x.

Differentiate $y=e^{3x}$ .

MCQ

y'=3e^{3x}.

Differentiate $y=\ln\!\big(x^2+1\big)$ .

MCQ

y'=\frac{2x}{x^2+1}.

Differentiate $y=\big(x^2+1\big)^4$ .

MCQ

Chain rule:

y'=8x\big(x^2+1\big)^3.

Differentiate $y=x\sin x$ .

MCQ

y'=\sin x+x\cos x.

Differentiate $y=\dfrac{x-1}{x+1}$ .

MCQ

Quotient rule:

y'=\frac{2}{(x+1)^2}.

Differentiate $y=\sqrt{3x+1}$ .

MCQ

y'=\frac{3}{2\sqrt{3x+1}}.

Differentiate $y=\tan x$ .

MCQ

y'=\sec^2 x.

Find $\dfrac{dy}{dx}$ if $x^2+xy=4$ .

FRQ

Implicit: $2x+\big(y+x\,y'\big)=0$ .

\frac{dy}{dx}=-\frac{2x+y}{x}.

Find $f''(x)$ for $f(x)=x^4$ .

MCQ

$f'=4x^3$ , $f''=12x^2$ .

12x^2.

Differentiate $y=e^{x}\cos x$ .

MCQ

y'=e^{x}\big(\cos x-\sin x\big).

Find the tangent line to $y=x^3$ at $x=1$ .

FRQ

Point $(1,1)$ ; slope $y'=3x^2=3$ :

y=3x-2.

Find the critical points of $f(x)=x^3-6x^2+9x$ .

FRQ

$f'(x)=3x^2-12x+9=3(x-1)(x-3)$ :

x=1,\ 3.

At what $x$ does $f(x)=x^2-4x+1$ attain its minimum?

MCQ

$f'(x)=2x-4=0$ :

x=2 \ (\text{min value } -3).

On what interval is $f(x)=x^2-6x$ increasing?

MCQ

$f'(x)=2x-6>0$ :

x>3.

A rectangle has perimeter 40. What is the maximum possible area?

FRQ

$l+w=20$ , $A=l(20-l)$ , maximised at $l=10$ :

10\times 10=100.

A circle's radius grows at 3 cm/s. How fast is its area changing when $r=4$ cm?

FRQ

$\dfrac{dA}{dt}=2\pi r\dfrac{dr}{dt}=2\pi(4)(3)$ :

24\pi\ \text{cm}^2/\text{s}.

Evaluate $\displaystyle\lim_{x\to 0}\frac{e^{2x}-1}{x}$ .

MCQ

L'Hôpital: $\tfrac{2e^{2x}}{1}\to 2$ :

2.

For what $x$ is $f(x)=x^3$ concave up?

MCQ

$f''(x)=6x>0$ :

x>0.

Find the inflection point of $f(x)=x^3-3x$ .

FRQ

$f''(x)=6x=0\Rightarrow x=0$ , and $f(0)=0$ :

(0,0).

A cube's edge grows at 2 cm/s. How fast is the volume changing when the edge is 5 cm?

FRQ

$V=s^3$ , $\dfrac{dV}{dt}=3s^2\dfrac{ds}{dt}=3(25)(2)$ :

150\ \text{cm}^3/\text{s}.

Evaluate $\displaystyle\int\big(8x^3-3x^2\big)\,dx$ .

MCQ

2x^4-x^3+C.

Evaluate $\displaystyle\int_0^{2} x^3\,dx$ .

MCQ

\Big[\tfrac{x^4}{4}\Big]_0^{2}=4.

Evaluate $\displaystyle\int_1^{3} 2x\,dx$ .

MCQ

\big[x^2\big]_1^{3}=9-1=8.

Evaluate $\displaystyle\int_0^{\pi}\sin x\,dx$ .

MCQ

\big[-\cos x\big]_0^{\pi}=1+1=2.

Evaluate $\displaystyle\int e^{2x}\,dx$ .

MCQ

\tfrac12 e^{2x}+C.

Evaluate $\displaystyle\int_1^{e^2}\frac{1}{x}\,dx$ .

MCQ

\big[\ln x\big]_1^{e^2}=2.

Evaluate $\displaystyle\int_0^{\pi/4}\sec^2 x\,dx$ .

MCQ

\big[\tan x\big]_0^{\pi/4}=1.

Find $\dfrac{d}{dx}\displaystyle\int_0^{x}\sin\!\big(t^2\big)\,dt$ .

FRQ

By the Fundamental Theorem (Part I):

\sin\!\big(x^2\big).

Evaluate $\displaystyle\int_0^{2}\big(3x^2+1\big)\,dx$ .

MCQ

\big[x^3+x\big]_0^{2}=8+2=10.

Find the average value of $f(x)=2x$ on $[0,4]$ .

MCQ

\frac14\int_0^{4}2x\,dx=\frac14(16)=4.

Evaluate $\displaystyle\int 3x^2\big(x^3+1\big)^5\,dx$ .

MCQ

Let $u=x^3+1$ , $du=3x^2\,dx$ :

\frac{\big(x^3+1\big)^6}{6}+C.

Evaluate $\displaystyle\int x\,e^{x}\,dx$ .

FRQ

By parts:

e^{x}(x-1)+C.

Evaluate $\displaystyle\int 2x\cos\!\big(x^2\big)\,dx$ .

MCQ

Let $u=x^2$ :

\sin\!\big(x^2\big)+C.

Evaluate $\displaystyle\int x\ln x\,dx$ .

FRQ

By parts, $u=\ln x,\ dv=x\,dx$ :

\frac{x^2}{2}\ln x-\frac{x^2}{4}+C.

Find the area between $y=4x$ and $y=x^2$ on $[0,4]$ .

FRQ

On $[0,4]$ , $4x\ge x^2$ :

\int_0^{4}\big(4x-x^2\big)\,dx=\Big[2x^2-\tfrac{x^3}{3}\Big]_0^{4}=\frac{32}{3}.

Find the area under $y=3x^2$ from $0$ to $2$ .

MCQ

\int_0^{2}3x^2\,dx=\big[x^3\big]_0^{2}=8.

The region under $y=\sqrt{x}$ on $[0,9]$ is revolved about the $x$ -axis. Find the volume.

FRQ

Disks with $R^2=x$ :

\pi\int_0^{9}x\,dx=\pi\Big[\tfrac{x^2}{2}\Big]_0^{9}=\frac{81\pi}{2}.

Find the average value of $f(x)=x^3$ on $[0,2]$ .

MCQ

\frac12\int_0^{2}x^3\,dx=\frac12(4)=2.

Solve $\dfrac{dy}{dx}=4y$ with $y(0)=2$ .

FRQ

Exponential model with $k=4$ , $y_0=2$ :

y=2e^{4x}.

Find the general solution of $\dfrac{dy}{dx}=x^2$ .

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y=\frac{x^3}{3}+C.

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Mirrors the College Board AP Calculus AB exam from May 2027: 42 MC (Part A 29 no-calc / Part B 13 calc) + 6 FRQs. 4 answer choices. Scored 1–5.

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AP Calculus AB — Past Exam Questions Free-response questions & scoring guidelines (College Board) AP Calculus AB Sample Questions PDF Official sample MCQ & FRQ packet

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AP Calculus AB

Overview & units

Unit 1 · Limits & Continuity

Unit 2 · Differentiation — Definition & Basic Rules

Unit 3 · Composite, Implicit & Inverse Functions

Unit 4 · Contextual Applications of Differentiation

Unit 5 · Analytical Applications of Differentiation

Unit 6 · Integration & Accumulation of Change

Unit 7 · Differential Equations

Unit 8 · Applications of Integration

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