AP Calculus AB

How to Study for AP Calculus AB: Unit-by-Unit Plan

Study by exam weight, not chapter order. Units 5 and 6 are the heaviest, each 15-20% of the multiple-choice, so start there, then Units 1, 2, 4, and 8. Before computing anything, name the method a problem needs; that recognition step is what the exam rewards and what solvers cannot do for you.

Read the exam before you study for it

AP Calculus AB is passable with steady work. In May 2025, 64.2% of the 286,722 students who took it scored a 3 or higher, and 20.3% earned a 5, with a mean score of 3.21. Those odds tilt in your favor when your studying matches how the exam is built rather than how a textbook is ordered.

The exam runs 3 hours 15 minutes and splits into two sections across four parts. Section I is 45 multiple-choice questions; Section II is 6 free-response questions. Each section has a no-calculator part and a calculator part, and that split should shape how you practice.

Section / part	Questions	Calculator	Time	Weight
I, Part A (MC)	30	Not permitted	60 min	33.3%
I, Part B (MC)	15	Required	45 min	16.7%
II, Part A (FRQ)	2	Required	30 min	16.7%
II, Part B (FRQ)	4	Not permitted	60 min	33.3%

Practice the way the exam is weighted

The two no-calculator parts (Section I Part A and Section II Part B) are 66.7% of your score. Most of your practice should be pencil and paper, so the machine never becomes a crutch you cannot use on exam day.

Weight your time by what the exam weights

The College Board publishes a weighting band for each unit's share of the multiple-choice section. Two units, 5 and 6, sit at the top at 15-20% each. Spend the most time there, then on the 10-15% units, and give lighter targeted review to the 5-10% units.

Unit	Topic	AB weight	Study emphasis
1	Limits and Continuity	10-15%	High
2	Differentiation: Definition and Fundamental Properties	10-15%	High
3	Differentiation: Composite, Implicit, and Inverse Functions	5-10%	Medium
4	Contextual Applications of Differentiation	10-15%	High
5	Analytical Applications of Differentiation	15-20%	Highest
6	Integration and Accumulation of Change	15-20%	Highest
7	Differential Equations	5-10%	Medium
8	Applications of Integration	10-15%	High

Read the arithmetic. Units 5 and 6 together can be as much as 40% of the multiple-choice weight. Add Units 1, 2, 4, and 8 and you have the large majority of the exam. Units 3 and 7 are lighter, but they are cheap points because their methods are narrow: master the handful of moves and you rarely miss them.

Study each unit for its recognition trigger

The goal for each unit is not just to execute a method but to recognize when a problem calls for it. Here is the trigger to master in each unit, with the CED topic numbers so you can drill the exact skill.

Unit 1 (10-15%): Given a limit, try direct substitution first. A number is the answer. A $\frac{0}{0}$ result is a signal that tells you the tool: factor and cancel for polynomials, multiply by the conjugate for radicals, combine fractions for compound quotients. Topic 1.7 (Selecting Procedures for Determining Limits) tests exactly this choice.
Unit 2 (10-15%): Know the derivative rules cold, including the derivatives of $\sin x$ , $\cos x$ , $e^x$ , and $\ln x$ (Topic 2.7). The trigger is the shape of the function: a product needs the product rule, a quotient needs the quotient rule.
Unit 3 (5-10%): The chain rule is your response to a composition, a function nested inside another (Topic 3.1). Implicit differentiation is your response when $y$ is tangled with $x$ and cannot be isolated (Topic 3.2). Topic 3.5 asks you to select among the derivative procedures.
Unit 4 (10-15%): This unit is word problems. Two changing quantities linked by an equation, asked for a rate, is a related rate (Topics 4.4-4.5). A $\frac{0}{0}$ or $\frac{\infty}{\infty}$ limit that resists algebra is L'Hospital's Rule (Topic 4.7). Position, velocity, and acceleration language is straight-line motion (Topic 4.2).
Unit 5 (15-20%): The heaviest analytical unit. Increasing and decreasing and local extrema point to the first derivative; concavity and inflection points point to the second. Absolute max or min on a closed interval $[a, b]$ is always the Candidates Test (Topic 5.5). Maximize or minimize a real quantity is optimization (Topics 5.10-5.11).
Unit 6 (15-20%): The other heaviest unit, and the trigger is the integrand. A basic power or known form integrates directly; an integrand holding a function together with a constant multiple of its own derivative is u-substitution (Topic 6.9). Topic 6.14 is the selection skill itself. Note that integration by parts and partial fractions (Topics 6.11-6.12) are BC only, so AB antidifferentiation is substitution or algebraic rearrangement.
Unit 7 (5-10%): A differential equation $\frac{dy}{dx}$ that separates onto opposite sides is solved by separation of variables (Topics 7.6-7.7). A described growth or decay situation is an exponential model (Topic 7.8). Slope fields (Topics 7.3-7.4) are read, not solved. Euler's method and logistic models are BC only.
Unit 8 (10-15%): Integration applied. Area between curves integrates top minus bottom (Topics 8.4-8.6). Volume of a solid of revolution is the disk or washer method (Topics 8.9-8.12); a solid with known cross sections integrates the cross-sectional area (Topics 8.7-8.8). Average value of a function has its own formula (Topic 8.1).

Method selection is the skill the exam actually tests

The CED builds method selection into the course on purpose. Topics 1.7, 3.5, and 6.14 are each named Selecting Procedures or Selecting Techniques, and they introduce no new formulas. They test one thing: can you look at a problem and name the right tool before you compute? A solver will crank out any integral you hand it. The exam rewards the student who knows which integral this is and why.

Turn that into a habit. For every practice problem, write one sentence before any algebra: this is a $\frac{0}{0}$ limit with a radical, so I use the conjugate, or this asks for absolute extrema on a closed interval, so I use the Candidates Test. Students who skip this step burn time on the no-calculator parts, where 66.7% of the score lives and there is no machine to fall back on.

Where AB students lose points

Most lost points on this exam are not hard concepts. They are habits. Drill these out before May.

Dropping the $+ C$ on an indefinite integral, or forgetting to change the limits of integration after a u-substitution in a definite integral (Topic 6.9).
Leaving out the differential, writing an integral with no $dx$ . Free-response readers score notation, and Communication and Notation is assessed only on the free-response section.
Not justifying. Writing that $f$ has a local maximum at $x = 2$ earns nothing without the reason: because $f'$ changes from positive to negative there. The free-response section weights justification heavily.
Skipping units in applied answers. A rate problem answered as 12 instead of 12 liters per minute loses the point.
Reaching for the calculator by reflex. Practice Section I Part A and Section II Part B entirely by hand so the exam feels the same as your prep.

Worked examples

Worked example

Choosing a Limit Method by Its Indeterminate Form

Evaluate $\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$ .

Start with the default move: substitute $x = 4$ . The numerator is $\sqrt{4} - 2 = 0$ and the denominator is $4 - 4 = 0$ , so the limit has the indeterminate form $\frac{0}{0}$ . A number would have been the answer; $\frac{0}{0}$ is a signal, not a result.
Read the form to pick the tool. The obstacle is a radical in the numerator, so multiply the numerator and denominator by the conjugate $\sqrt{x} + 2$ .
Multiply: $\frac{\sqrt{x} - 2}{x - 4} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{(\sqrt{x})^2 - 2^2}{(x - 4)(\sqrt{x} + 2)} = \frac{x - 4}{(x - 4)(\sqrt{x} + 2)}$ .
Cancel the common factor $x - 4$ , which is valid because $x \to 4$ means $x \neq 4$ : the expression becomes $\frac{1}{\sqrt{x} + 2}$ .
Now direct substitution works: $\frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$ .

The limit equals 1/4. The recognition step (0/0 with a radical means use the conjugate) is what the exam tests; the algebra is routine once you have chosen the method.

Worked example

Recognizing When an Integral Calls for u-Substitution

Find $\int 2x\sqrt{x^2 + 1}\,dx$ .

Scan the integrand for a function nested inside another. Here $x^2 + 1$ sits inside the square root. Check whether its derivative appears as a factor: $\frac{d}{dx}(x^2 + 1) = 2x$ , and $2x$ is indeed multiplying the root. An inside function together with its derivative is the trigger for u-substitution (Topic 6.9).
Let $u = x^2 + 1$ . Then $\frac{du}{dx} = 2x$ , so $du = 2x\,dx$ .
Rewrite the whole integral in terms of $u$ . The $2x\,dx$ becomes $du$ and $\sqrt{x^2 + 1}$ becomes $\sqrt{u}$ : $\int \sqrt{u}\,du = \int u^{1/2}\,du$ .
Integrate with the power rule for antiderivatives, raising the exponent by one and dividing: $\int u^{1/2}\,du = \frac{u^{3/2}}{3/2} + C = \frac{2}{3}u^{3/2} + C$ .
Substitute back $u = x^2 + 1$ : $\frac{2}{3}(x^2 + 1)^{3/2} + C$ .

The antiderivative is (2/3)(x^2+1)^(3/2) + C. On AB, an integral like this is substitution or nothing: integration by parts and partial fractions are BC only, so if basic rules and a u-sub do not crack an AB integral, recheck your algebra.

Worked example

Using the Candidates Test for Absolute Extrema

Find the absolute maximum and minimum values of $f(x) = x^3 - 3x^2 + 1$ on the closed interval $[-1, 3]$ .

Recognize the phrasing. Absolute max and min on a closed interval names one method: the Candidates Test (Topic 5.5). By the Extreme Value Theorem, because $f$ is continuous on $[-1, 3]$ the extrema exist, and they can occur only at critical points or endpoints.
Find the critical points. Differentiate: $f'(x) = 3x^2 - 6x = 3x(x - 2)$ . Set $f'(x) = 0$ : $3x(x - 2) = 0$ gives $x = 0$ and $x = 2$ . Both lie in $[-1, 3]$ , and $f'$ exists everywhere, so these are the only critical points.
List the candidates: the two critical points $x = 0$ and $x = 2$ , plus the two endpoints $x = -1$ and $x = 3$ .
Evaluate $f$ at each. $f(-1) = (-1)^3 - 3(-1)^2 + 1 = -1 - 3 + 1 = -3$ . $f(0) = 0 - 0 + 1 = 1$ . $f(2) = 2^3 - 3(2)^2 + 1 = 8 - 12 + 1 = -3$ . $f(3) = 3^3 - 3(3)^2 + 1 = 27 - 27 + 1 = 1$ .
Compare the four outputs. The largest is $1$ (at $x = 0$ and $x = 3$ ); the smallest is $-3$ (at $x = -1$ and $x = 2$ ).

The absolute maximum value is 1 and the absolute minimum value is -3. You never needed the first or second derivative test here: on a closed interval, comparing values at the candidates is faster, and that is exactly what the Candidates Test is for.

Frequently asked questions

Is AP Calculus AB hard?

It is manageable with steady practice. In 2025, 64.2% of students scored 3 or higher and 20.3% earned a 5, with a mean of 3.21. The students who struggle usually know the mechanics but freeze on choosing which method a problem needs, and that is a habit you can train.

How many hours should I study for AP Calculus AB?

There is no fixed number, but a useful rule is to weight your time by exam weight. Put the most into Units 5 and 6 (each 15-20% of the multiple-choice), a solid block into Units 1, 2, 4, and 8, and lighter targeted review into Units 3 and 7. Steady weekly practice beats a cram.

What is the hardest unit in AP Calculus AB?

By weight, Units 5 (analytical applications of derivatives) and 6 (integration) carry the most and are worth the most study. Unit 5 leans on justification and Unit 6 on choosing the right antidifferentiation technique, which is where a lot of points are won or lost.

Can I use a calculator on the AP Calculus AB exam?

Only on part of it. A graphing calculator is required for Section I Part B and Section II Part A, and not permitted for Section I Part A or Section II Part B. The two no-calculator parts are 66.7% of the score, so practice most of your work by hand.