AP Calculus practice problems: practice problems and exam questions

Practice for AP Calculus AB and BC topics: limits, derivatives and their applications, optimisation, integrals, differential equations, and the BC series topics with Taylor polynomials.

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Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

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Topics in this course

  1. Limitsdirect substitution and factoring · limits at infinity · L'Hôpital's rulecalculations
  2. Derivativespower rule · chain and product rules · tangents and extremacalculations
  3. Optimizationmaximum area or volume · minimum cost or material · profit and revenuecalculations
  4. Integralspolynomials · exponentials, logs and substitution · areas between curvescalculations
  5. Differential equationsexponential growth and decay · Newton's law of cooling · first-order linearcalculations
  6. Sequences & seriesarithmetic and geometric series · convergence and power series · Taylor and Maclaurin polynomialscalculations

Exam tips for AP Calculus practice problems

  • Check limits by direct substitution first.
  • For optimisation, check the endpoints as well as the critical points.
  • Separate variables before you integrate a differential equation.

Worked examples

LimitsEasy
xyx = 1

Find the limit L = limx→1 (x4 − 1)/(x − 1).

f(x) = (x4 − 1)/(x − 1)x → 1

Find L

Show hints
  1. Substitution gives 0/0. Factor the numerator.
  2. xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹).
  3. x4 − 1 = (x − 1)(x3 + x2 + x + 1).
Show the answer and the step-by-step solution

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Try direct substitution

numerator → 0 and denominator → 0, so the form is 0/0.

Factor and cancel

(x4 − 1)/(x − 1) = x3 + x2 + x + 1

Substitute

L = 1·13 + 1·12 + 1·1 + 1·1 = 4

Answer

L = 4

This is the derivative of xⁿ at x = a: n·aⁿ⁻¹.

DerivativesMedium
xyx = 1

Let f(x) = x3·e−x. Find f′(1).

f(x) = x3·e−xx = 1

Find f′(1)

Show hints
  1. Product rule with u = x3 and v = e−x.
  2. u′ = 3x2, v′ = −1·e−x.
  3. f′(x) = e−x·(3x2 + (−1)·x3).
Show the answer and the step-by-step solution

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Product rule

f′(x) = u′v + uv′ = e−x·(3x2 + (−1)·x3)

Substitute

f′(1) = 0.3679·(3 + (−1)) = 0.7358

Answer

f′(1) = 0.7358

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Half-life and exponential decay calculatorDerivative calculator with stepsIntegral calculator (definite and indefinite)

Related courses

Calculus 1Engineering MathematicsDifferential EquationsEngineering Mathematics 2Calculus 2Numerical MethodsIntroductory StatisticsStatistics 2: Statistical Inference
Practise this course for freePractise this courseTake a mock exam5 free problems every day. No card needed.