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AP Calculus AB

College Board · 8 units · exam-weighted · high school

AP Calculus AB: every unit with its share of the exam, as College Board publishes it, with practice questions per unit and a countdown to the exam.

Set up AP Calculus AB → One set per unit, 8 of them pre-written and answer-checked. Free, no account needed.

Units and exam weighting

Unit 1: Limits and Continuity 10–12% of the exam

Limits from graphs, tables and algebra; limit laws; squeeze theorem; continuity; removable and jump discontinuities; infinite limits and asymptotes; Intermediate Value Theorem.

3 of 14 questions in this unit, with the worked answer

Evaluate $\lim_{x\to 2} \dfrac{x^2-5x+6}{x-2}$.

Answer. -1

Direct substitution gives 0/0. Factor the numerator as (x-2)(x-3), cancel the common factor, leaving x-3. Substituting x=2 gives -1.

A table shows f(0.9)=2.71, f(0.99)=2.97, f(0.999)=2.997, f(1.001)=3.003, f(1.01)=3.03, f(1.1)=3.31. Based on this table, what is $\lim_{x\to 1} f(x)$?

Answer. 3

As x approaches 1 from both the left and right, f(x) gets arbitrarily close to 3, so the two-sided limit is 3, even though we cannot confirm f(1) itself from a table.

Use the squeeze theorem to evaluate $\lim_{x\to 0} x^2\sin(1/x)$.

Answer. 0

Since $-1 \le \sin(1/x) \le 1$, multiplying by $x^2\ge 0$ gives $-x^2 \le x^2\sin(1/x) \le x^2$. Both bounding functions approach 0 as $x\to 0$, so by the squeeze theorem the limit is 0.

All 14 questions for Unit 1, with the working →

What people get wrong

  1. Plugging in the target value and stopping when you get 0/0 instead of recognizing it as a signal to factor, rationalize, or combine fractions before trying again.
  2. Treating the limit value and the function value f(c) as automatically the same thing — at a removable discontinuity they differ, and the limit exists independently of what f(c) is defined to be.
  3. Checking only that one-sided limits are equal and forgetting the third continuity condition, that this common value must also equal f(c), which is exactly what makes removable discontinuities easy to miss.
  4. Getting the sign of an infinite limit wrong by not separately checking the sign of the expression as x approaches from the left versus the right of a vertical asymptote.

Unit 2: Differentiation: Definition and Fundamental Properties 10–12% of the exam

Average versus instantaneous rate of change, the limit definition of the derivative, differentiability and continuity, power, product and quotient rules, derivatives of trig, exponential and log functions.

3 of 13 questions in this unit, with the worked answer

Find the average rate of change of $f(x)=x^3-2x$ on the interval $[1,3]$.

Answer. 11

Average rate of change is $\frac{f(3)-f(1)}{3-1}$. Compute $f(3)=27-6=21$ and $f(1)=1-2=-1$, so the quotient is $\frac{21-(-1)}{2}=\frac{22}{2}=11$.

A limit is given as $\lim_{h\to 0}\frac{\sin(\pi/4+h)-\sin(\pi/4)}{h}$. What does this limit represent, and what is its value?

Answer. It represents f'(π/4) for f(x)=sin x, and equals $\frac{\sqrt{2}}{2}$

This is exactly the limit definition of the derivative applied to f(x)=sin x at a=π/4. Since $\frac{d}{dx}\sin x=\cos x$, the value is $\cos(\pi/4)=\frac{\sqrt{2}}{2}$.

Use the limit definition to find f'(x) for $f(x)=3x^2+1$.

Answer. $f'(x)=6x$

Form the difference quotient: $\frac{3(x+h)^2+1-(3x^2+1)}{h}=\frac{6xh+3h^2}{h}=6x+3h$. Taking the limit as h→0 gives $6x$.

All 13 questions for Unit 2, with the working →

What people get wrong

  1. Plugging h=0 directly into the difference quotient before simplifying, which produces 0/0 and stalls the problem. Always algebraically simplify (expand, factor, or rationalize) to cancel the h in the denominator first.
  2. Assuming continuity guarantees differentiability. Continuity is necessary but not sufficient — functions like $|x|$ or $x^{1/3}$ are continuous but not differentiable at x=0. Always separately check for corners, cusps, or vertical tangents.
  3. Reversing the order of terms in the quotient rule numerator, giving $fg'-f'g$ instead of $f'g-fg'$. Memorize it as 'low d-high minus high d-low' (using the denominator as 'low') to keep the sign correct.
  4. Treating the derivative of $a^x$ the same as $e^x$ and forgetting the $\ln a$ factor, or forgetting the extra $\ln a$ in the denominator for $\log_a x$. Only $e^x$ and $\ln x$ are 'clean'; other bases pick up a $\ln a$ factor.

Unit 3: Differentiation: Composite, Implicit, and Inverse Functions 9–13% of the exam

Chain rule, implicit differentiation, derivatives of inverse functions and inverse trig functions, higher-order derivatives.

3 of 14 questions in this unit, with the worked answer

Find $f'(x)$ if $f(x)=(3x^2-5x)^4$.

Answer. $f'(x)=4(3x^2-5x)^3(6x-5)$

This is a composition with outer function (something)^4 and inner function $3x^2-5x$. Differentiate the outer power, keeping the inner function unchanged, then multiply by the derivative of the inner function, $6x-5$.

Find the derivative of $y=\sin(x^3)$.

Answer. $\frac{dy}{dx}=3x^2\cos(x^3)$

The outer function is sine and the inner function is $x^3$. Differentiate sine to get cosine of the same inner function, then multiply by the derivative of $x^3$, which is $3x^2$.

For the circle $x^2+y^2=25$, find $\frac{dy}{dx}$ at the point (3,4).

Answer. $\frac{dy}{dx}=-\frac34$

Differentiating implicitly: $2x+2y\frac{dy}{dx}=0$, so $\frac{dy}{dx}=-\frac{x}{y}$. Substituting x=3, y=4 gives $-\frac{3}{4}$.

All 14 questions for Unit 3, with the working →

What people get wrong

  1. Forgetting to multiply by the derivative of the inner function in a composition (e.g., writing $\frac{d}{dx}[\sin(x^3)]=\cos(x^3)$ instead of $3x^2\cos(x^3)$). Always ask 'what is the inner function, and what is its derivative?' before finishing.
  2. Differentiating a mixed term like $xy^2$ implicitly by treating it as a single variable and forgetting the product rule. Write it out as $\frac{d}{dx}[x]\cdot y^2+x\cdot\frac{d}{dx}[y^2]$ first, then apply the chain rule to the $y^2$ piece.
  3. Mixing up the derivatives of arcsin and arccos, or arctan and arccot — they differ only by a sign. Memorize arcsin/arctan/arcsec as positive and their co-function partners as the negatives of those same expressions.
  4. Leaving $\frac{dy}{dx}$ unsimplified inside a second-derivative expression instead of substituting the first-derivative formula back in. The AP rubric expects the final answer in terms of x and y (or a number), not in terms of $y'$.

Unit 4: Contextual Applications of Differentiation 10–15% of the exam

Motion along a line, rates of change in context, related rates, linearisation and approximation, L'Hospital's rule.

3 of 14 questions in this unit, with the worked answer

A particle moves along a line with position $s(t)=t^3-6t^2+9t$ meters, $t\ge0$ in seconds. At what time(s) is the particle at rest?

Answer. t = 1 and t = 3 seconds

Velocity is $v(t)=s'(t)=3t^2-12t+9=3(t-1)(t-3)$. The particle is at rest when $v(t)=0$, which happens at t=1 and t=3.

For the same particle, $v(t)=3t^2-12t+9$ and $a(t)=6t-12$, is the particle's speed increasing or decreasing at t = 1.5 seconds?

Answer. Speed is increasing

At t=1.5, $v(1.5)=3(2.25)-18+9=-2.25$ and $a(1.5)=6(1.5)-12=-3$. Both are negative, so velocity and acceleration point the same direction, meaning speed is increasing.

Using $s(t)=t^3-6t^2+9t$ and the fact that v(t)=0 at t=1 and t=3, find the total distance traveled by the particle from t=0 to t=3.

Answer. 8 meters

Compute s(0)=0, s(1)=1-6+9=4, s(3)=27-54+27=0. Distance = |s(1)-s(0)| + |s(3)-s(1)| = |4-0| + |0-4| = 4+4 = 8 meters, not the net displacement of 0.

All 14 questions for Unit 4, with the working →

What people get wrong

  1. Substituting the numerical values into the related-rates equation before differentiating. Differentiate the general equation with respect to t first, and only plug in the specific instant's numbers afterward.
  2. Treating speed and velocity as the same thing on a motion problem. Speed is $|v(t)|$; a particle can have decreasing speed while its velocity is becoming more negative. Always check the sign of both v and a.
  3. Computing total distance as $\int v(t)\,dt$ (which gives net displacement) instead of $\int|v(t)|\,dt$. Find where v(t)=0 first, then add the absolute values of displacement on each piece.
  4. Applying L'Hospital's rule to a limit that isn't actually 0/0 or ∞/∞. Plug in the limiting value first to confirm the indeterminate form; if it isn't indeterminate, just evaluate directly.

Unit 5: Analytical Applications of Differentiation 15–18% of the exam

Mean Value Theorem, extreme values, increasing and decreasing intervals, first and second derivative tests, concavity, curve sketching, optimisation, implicit relations.

3 of 14 questions in this unit, with the worked answer

For $f(x)=x^3-x$ on $[0,2]$, find the value of c guaranteed by the Mean Value Theorem.

Answer. c = 2/√3 (≈1.155)

The average rate of change is (f(2)-f(0))/2 = (6-0)/2 = 3. Setting f'(x)=3x²-1 equal to 3 gives x²=4/3, so x=2/√3, which lies in (0,2).

Explain whether the Mean Value Theorem can be applied to $f(x)=|x-1|$ on $[0,2]$.

Answer. No — MVT cannot be applied.

f is continuous on [0,2] but not differentiable at x=1, which lies inside the open interval (0,2). Since the differentiability hypothesis fails, the theorem's conclusion is not guaranteed.

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

Answer. Absolute max is 2 (at x=0 and x=3); absolute min is -2 (at x=-1 and x=2).

f'(x)=3x²-6x=3x(x-2), giving critical numbers x=0,2. Evaluating f at critical numbers and endpoints: f(-1)=-2, f(0)=2, f(2)=-2, f(3)=2. The largest value is 2 and the smallest is -2, by the Extreme Value Theorem and candidates test.

All 14 questions for Unit 5, with the working →

What people get wrong

  1. Applying the Mean Value Theorem or Rolle's Theorem without checking both hypotheses — instead, explicitly verify continuity on the closed interval and differentiability on the open interval (watch for corners or vertical tangents) before invoking the conclusion.
  2. Treating every critical number as a local extremum — instead, always confirm with a sign change in f' (first derivative test) or the sign of f'' (second derivative test); some critical numbers are neither, like a horizontal inflection point.
  3. Making sign errors when building a sign chart for f' or f'' — instead, factor the derivative completely, test one value in each subinterval, and track the sign of each factor separately before multiplying them.
  4. Assuming the second derivative test always works — instead, when $f''(c)=0$, recognize the test is inconclusive and fall back on the first derivative test.

Unit 6: Integration and Accumulation of Change 17–20% of the exam

Riemann sums and the definite integral, accumulation functions, the Fundamental Theorem of Calculus, antiderivatives, integration by substitution, integrating with long division and completing the square.

3 of 14 questions in this unit, with the worked answer

A table gives f(0)=3, f(2)=5, f(4)=8, f(6)=12, f(8)=15. Use a left Riemann sum with 4 equal subintervals to approximate ∫ from 0 to 8 of f(x) dx.

Answer. 56

Each subinterval has width Δx=2, and a left sum uses f at x=0,2,4,6: 2(3+5+8+12)=2(28)=56.

Using the same table (f(0)=3, f(2)=5, f(4)=8, f(6)=12, f(8)=15), use a right Riemann sum with 4 equal subintervals to approximate the integral.

Answer. 80

A right sum uses f at x=2,4,6,8 with Δx=2: 2(5+8+12+15)=2(40)=80.

A function has values f(0)=2, f(1)=3, f(2)=5, f(3)=7, f(4)=10, f(5)=13, f(6)=16, f(7)=19, f(8)=22. Approximate ∫ from 0 to 8 of f(x) dx using a midpoint sum with 4 subintervals of width 2.

Answer. 84

The midpoints of the four subintervals [0,2],[2,4],[4,6],[6,8] are x=1,3,5,7. The sum is 2(f(1)+f(3)+f(5)+f(7))=2(3+7+13+19)=2(42)=84.

All 14 questions for Unit 6, with the working →

What people get wrong

  1. Forgetting the chain-rule factor in FTC Part 1 when the bound is a function of x other than x itself — always multiply by the derivative of that bound, not just plug it in.
  2. Forgetting to flip the sign when the variable is the lower limit of an accumulation function — rewrite ∫ from x to a as −∫ from a to x before differentiating.
  3. Dropping the +C on indefinite integral answers — it costs points on free-response even though it cancels out in definite integrals.
  4. Using u-substitution on a definite integral but evaluating the u-antiderivative at the original x-bounds — convert the limits to u-values, or convert the antiderivative back to x before plugging in the original bounds.

Unit 7: Differential Equations 6–12% of the exam

Modelling with differential equations, verifying solutions, slope fields, separation of variables, exponential models.

3 of 13 questions in this unit, with the worked answer

Verify that $y=3e^{2x}$ is a solution of the differential equation $\frac{dy}{dx}-2y=0$.

Answer. Yes, it is a solution.

Differentiate: $\frac{dy}{dx}=6e^{2x}$. Substitute: $6e^{2x}-2(3e^{2x})=6e^{2x}-6e^{2x}=0$, which matches the right side of the equation for all x.

Which differential equation is satisfied by every function of the form $y=Ce^{-5x}$?

Answer. $\frac{dy}{dx}=-5y$

Differentiating gives $\frac{dy}{dx}=-5Ce^{-5x}=-5y$, so the family satisfies $\frac{dy}{dx}=-5y$ for any constant C.

A rumor spreads through a school of 800 students. The rate at which the rumor spreads is proportional to the product of the number of students who have heard it, y, and the number who have not, (800-y). Write a differential equation modeling dy/dt.

Answer. $\frac{dy}{dt}=ky(800-y)$

The phrase 'proportional to the product of' two quantities becomes k times the product of those two quantities, giving the stated equation with proportionality constant k.

All 13 questions for Unit 7, with the working →

What people get wrong

  1. Forgetting to add the constant of integration when separating variables, or forgetting to solve for it using the initial condition — always write '+C' immediately after integrating, then substitute the given point before doing anything else.
  2. Dropping the absolute value when integrating $\int \frac{1}{y}dy = \ln|y|+C$, then mishandling the sign when exponentiating — carry the absolute value through and let the constant $A=\pm e^{C}$ absorb the sign, determined later by the initial condition.
  3. Treating $\frac{dy}{dx}=ky$ as if it produces a linear function $y=kx+C$ instead of the exponential solution $y=Ce^{kt}$ — memorize that this specific equation always gives exponential, not linear, growth.
  4. Verifying a solution by only checking one point instead of confirming the equation holds identically for all x — differentiate symbolically and simplify both sides completely.

Unit 8: Applications of Integration 10–15% of the exam

Average value of a function, motion problems, accumulation in context, area between curves, volumes by cross-sections and by discs and washers.

3 of 14 questions in this unit, with the worked answer

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

Answer. 3

Average value is $\frac{1}{3-0}\int_0^3 x^2\,dx = \frac{1}{3}\left[\frac{x^3}{3}\right]_0^3=\frac{1}{3}(9)=3$. Divide by the length of the interval, not just compute the integral.

Find the average value of $f(x)=\sin x$ on $[0,\pi]$.

Answer. $\dfrac{2}{\pi}$

$f_{avg}=\frac{1}{\pi}\int_0^\pi \sin x\,dx=\frac{1}{\pi}[-\cos x]_0^\pi=\frac{1}{\pi}(1-(-1))=\frac{2}{\pi}$.

A particle moves with velocity $v(t)=t^2-4t+3$ (in meters/sec) for $0\le t\le 4$. Find both the displacement and the total distance traveled.

Answer. Displacement = 4/3 m; total distance = 4 m

v(t)=(t-1)(t-3) changes sign at t=1,3. On [0,1] and [3,4] v>0 with integral 4/3 each; on [1,3] v<0 with integral −4/3. Displacement sums these signed values to 4/3; total distance sums their absolute values to 4/3+4/3+4/3=4.

All 14 questions for Unit 8, with the working →

What people get wrong

  1. Computing $\int_a^b f(x)\,dx$ and calling it the average value — you must also divide by $(b-a)$; write the formula down every time before plugging in.
  2. Finding total distance by just taking $\left|\int_a^b v(t)\,dt\right|$ — this only gives displacement's magnitude. Instead find every zero of $v(t)$ in $[a,b]$, integrate on each subinterval, and add the absolute values.
  3. Setting up a washer as $\pi\int (R-r)^2\,dx$ instead of $\pi\int (R^2-r^2)\,dx$ — these are algebraically different. Always square the two radii separately, then subtract.
  4. Forgetting to shift the radius when rotating about a line other than the x- or y-axis (e.g. using $R(x)=\sqrt{x}$ instead of $\sqrt{x}+1$ when the axis is $y=-1$). Always measure radius as distance from the axis of rotation to the curve.

How it works

Choosing this curriculum creates a course in Chalkset with one study set per unit and the exam date set to the start of the next testing window. Readiness is weighted the way the exam is, so a weak unit that carries 18% of the score pulls harder than one that carries 6%. Pre-written units come with worked explanations and an independently checked answer key; the rest are written by the agent when you open them.

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