FREE AP CALCULUS AB PRACTICE

Free AP Calculus AB Practice Questions with Full Solutions

Work through AP Calculus AB questions that test limits, derivatives, integrals, accumulation, area, volume, and calculator setup. Each question includes the route, the trap, and the AP-style reason behind the answer.

Work through these multiple-choice problems and their explanations. Full practice tests add mixed topics, timing, and free-response work.

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The published bank contains multiple-choice problems with full solutions. Start with a topic below, or take the free AP Calculus AB diagnostic to identify where to focus.

Across limits, derivatives, integration, and differential equations, check the setup and interpretation as carefully as the calculation. These single questions help you isolate a method; a full practice test combines topics, multiple-choice and free-response work under exam conditions.

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Quick AP Calculus AB practice tips

AP Calculus AB questions often test whether you can choose the right method before calculating. Use these tips before working through the free questions below.

Read for the concept first. A question may look computational, but it might really be testing FTC, MVT, concavity, or total area.
Separate setup from calculation. Calculator questions are often hard because of the setup, not the arithmetic.
Watch notation carefully. Bounds, inverse functions, derivative notation, and composite functions can change the entire route.
Total area is not net area. AP questions often check whether you understand the difference between signed accumulation and geometric area.

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Definite Integrals

AP Calculus AB absolute value definite integral practice question

Definite Integrals Medium Multiple Choice Evaluate a definite integral involving absolute value
∫60 |x − 4| dx =
A. 6
B. 8
C. 10
D. 12

Correct answer: Choice C

10

Full solution The absolute value changes behavior at x = 4. From 0 to 4, |x − 4| = 4 − x. From 4 to 6, |x − 4| = x − 4. So the integral is ∫40(4 − x) dx + ∫64(x − 4) dx. Visually, these are two triangles. The first has area 8, and the second has area 2. Total: 10.
Common mistake The common trap is ignoring the absolute value and integrating x − 4 across the whole interval. That gives signed behavior, not the area created by the absolute value.
Trap to watch Absolute value integrals must be split at the sign-change point.
Method note When you see an absolute value inside an integral, first ask where the inside expression changes sign.
AP note AP note: This is a non-calculator question. The fastest route is not a long antiderivative; it is splitting at x = 4 and reading the geometry.
Calculator note Non-calculator.

AP Calculus AB u-substitution definite integral cos squared sine

Definite Integrals Medium Multiple Choice Evaluate a definite integral using substitution
∫π/20 cos2x sin x dx =
A. −1
B. −13
C. 13
D. 1

Correct answer: Choice C

13

Full solution Use u = cos x. Then du = −sin x dx. Change the bounds: when x = 0, u = 1; when x = π/2, u = 0. The integral becomes −∫01u2du, which is the same as ∫10u2du = 13.
Common mistake The sign is the trap. Many students choose −13 because they remember du = −sin x dx but forget to reverse the bounds.
Trap to watch The negative from du must be handled by reversing the bounds or carrying the sign carefully.
Method note For definite integrals, changing the bounds during substitution usually keeps the work cleaner.
AP note AP note: This is a clean non-calculator u-substitution question. The test is checking substitution control, not hard algebra.
Calculator note Non-calculator.

Limits and Derivatives

AP Calculus AB derivative limit definition cosine at pi over two

Limits and Derivatives Medium Multiple Choice Recognize a derivative from the limit definition
limh→0 cos(π/2 + h) − cos(π/2)h is
A. 1
B. nonexistent
D. −1

Correct answer: Choice D

−1

Full solution This is the derivative definition for cos x at x = π/2. In the form f(a+h) − f(a)h, here f(x)=cos x and a=π/2. Since ddxcos x = −sin x, the limit is −sin(π/2)=−1.
Common mistake The mistake is trying direct substitution and stopping at 0/0. That form is not failure here; it is the signal that the expression is a derivative limit.
Trap to watch This is a derivative-definition question, not a direct substitution question.
Method note Before doing algebra, check whether a limit is secretly a derivative definition.
AP note AP note: On AP Calculus, recognizing the structure can save more time than simplifying the trigonometric expression.
Calculator note Non-calculator.

Continuity and limits practice question

Limits and Derivatives Medium Multiple Choice Continuity and limits
For x ≠ 2, let f(x) = (x² − 4)/(x − 2). What value should f(2) be assigned to make f continuous at x = 2?
B. 2
C. 4
D. The function cannot be made continuous.

Correct answer: Choice C

4

Full solution For x ≠ 2, factor and cancel to get f(x) = x + 2. The limit as x approaches 2 is 4, so defining f(2) = 4 fills the removable discontinuity.
Common mistake Substituting x = 2 into the original fraction gives 00, which signals that the limit needs analysis; it is not the limit value.
Method note Find the limit first, then set the function value equal to it.

Applications of Derivatives

AP Calculus AB tangent line approximation underestimates concavity

Applications of Derivatives Easy Multiple Choice Connect tangent line approximation to concavity
If f is differentiable, we can use the line tangent to f at x = a to approximate values of f near x = a. Suppose that for a certain function f this method always underestimates the correct values. If so, then in an interval surrounding x = a, the graph of f must be
A. increasing
B. decreasing
C. concave upward
D. concave downward

Correct answer: Choice C

concave upward

Full solution If a tangent line approximation underestimates the actual function values, the tangent line is below the graph near the point of tangency. That happens when the graph is concave upward. Concavity controls whether the tangent line sits above or below the curve.
Common mistake A common mistake is choosing increasing or decreasing. But tangent-line underestimate is about curvature, not whether the function is going up or down.
Trap to watch Underestimate from a tangent line points to concavity, not monotonicity.
Method note Tangent line below the curve means concave upward. Tangent line above the curve means concave downward.
AP note AP note: This is a conceptual approximation question. No computation is needed; the key word is underestimates.
Calculator note Non-calculator.

Related rates practice question

Applications of Derivatives Medium Multiple Choice Related rates
A circle has radius r that increases at 2 centimeters per second. At the instant r = 3 centimeters, what is the rate of change of its area?
A. 6π cm²/s
B. 9π cm²/s
C. 12π cm²/s
D. 18π cm²/s

Correct answer: Choice C

12π cm²/s

Full solution Differentiate A = πr² with respect to time: dAdt = 2πr(drdt). Substitute r = 3 and drdt = 2 to get 12π square centimeters per second.
Common mistake Substituting r before differentiating can hide the changing quantity and lose the factor drdt.
Method note Write the relationship, differentiate with respect to time, then substitute the instant.

Derivative Rules and Tables

AP Calculus AB chain rule table values P of x equals g of x squared

Derivative Rules and Tables Medium Multiple Choice Apply the chain rule using table values
Use the table below.

xff′gg′
1212−35
23104
34223
464312

If P(x) = (g(x))2, then P′(3) equals
A. 4
B. 6
C. 9
D. 12

Correct answer: Choice D

12

Full solution First choose the rule, then read the table. Since P(x)=(g(x))2, the derivative is P′(x)=2g(x)g′(x). From the row x=3, the table gives g(3)=2 and g′(3)=3. Therefore, P′(3)=2(2)(3)=12.
Common mistake A common mistake is reading the table first and grabbing random values. The rule tells you exactly which values you need: g(3) and g′(3).
Trap to watch Use both g(3) and g′(3); do not square g′.
Method note For table problems, identify the derivative rule before pulling numbers from the table.
AP note AP note: This is a table-based chain rule question. It rewards rule recognition more than calculation.
Calculator note Non-calculator.

Implicit differentiation practice question

Derivative Rules and Tables Medium Multiple Choice Implicit differentiation
If x² + xy + y² = 7, what is dydx at the point (1, 2)?
A. −45
B. −54
C. 45
D. 54

Correct answer: Choice A

−45

Full solution Differentiate implicitly: 2x + y + x(dydx) + 2y(dydx) = 0. Thus dydx = −(2x + y)/(x + 2y). At (1, 2), this is −45.
Common mistake The derivative of xy uses the product rule: y + x(dydx).
Method note Differentiate every y term with the chain rule and evaluate only after isolating dydx.

Inverse Functions

AP Calculus AB inverse derivative table values

Inverse Functions Medium Multiple Choice Use the derivative of an inverse function with table values
Use the table below.

xff′gg′
1212−35
23104
34223
464312

If H(x) = f−1(x), then H′(3) equals
A. −116
B. −18
C. 12
D. 1

Correct answer: Choice D

1

Full solution For inverse derivatives, do not start by looking at x=3. Since H(x)=f−1(x), H(3) means the input where f equals 3. From the table, f(2)=3, so H(3)=2. Then H′(3)=1f′(2)=11=1.
Common mistake The common trap is using f′(3). For inverse derivative questions, the input shifts: first find where the original function output is 3.
Trap to watch Find the original input first; do not use the derivative row at x = 3 automatically.
Method note For (f−1)′(a), find the x-value where f(x)=a, then use the reciprocal of f′ there.
AP note AP note: This is one of the most common inverse-function table patterns on AP Calculus.
Calculator note Non-calculator.

Applications of Integration

AP Calculus AB total area bounded by x root one minus x squared

Applications of Integration Hard Multiple Choice Find total area using symmetry and integration
The total area of the region bounded by the graph of y = x√(1 − x2) and the x-axis is
A. 13
B. 12
C. 23
D. 1

Correct answer: Choice C

23

Full solution The function y=x√(1−x2) is defined from x=−1 to x=1 and is odd. The signed integral over the symmetric interval would be 0, but the question asks for total area. Use symmetry and double the positive half: 2∫10x√(1−x2)dx. With u=1−x2, the integral from 0 to 1 is 13. Doubling gives 23.
Common mistake The trap is confusing total area with net area. Net area cancels positive and negative regions; total area adds their sizes.
Trap to watch Total area means use absolute value or symmetry, not cancellation.
Method note When a function is odd and the question says total area, be careful: the answer is usually not zero.
AP note AP note: AP Calculus frequently separates net signed area from total geometric area.
Calculator note Non-calculator.

AP Calculus AB accumulation from rate arctan leaking tank

Applications of Integration Medium Multiple Choice Use a definite integral to accumulate a rate over time
Water is leaking from a tank at the rate of R(t) = 5 arctan(t5) gallons per hour, where t is the number of hours since the leak began. To the nearest gallon, how much water will leak out during the first day?
A. 7
B. 12
C. 24
D. 124

Correct answer: Choice D

124

Full solution The function gives a rate in gallons per hour. The total amount leaked during the first day is the accumulation of that rate from 0 to 24 hours: ∫2405 arctan(t5)dt. Calculator evaluation gives approximately 123.58, so to the nearest gallon the amount is 124.
Common mistake A common mistake is evaluating the rate at t=24. A rate at one time is not the total leaked amount.
Trap to watch Rate must be accumulated with an integral to get total amount.
Method note When the question asks for total amount from a rate, integrate the rate over the interval.
AP note AP note: This is a calculator-allowed accumulation question. The setup matters more than the arithmetic.
Calculator note Calculator allowed.

Average value of a function practice question

Applications of Integration Medium Multiple Choice Average value of a function
What is the average value of f(x) = x² on the interval [0, 3]?
A. 3
B. 92
C. 6
D. 9

Correct answer: Choice A

3

Full solution The average value is (1/(3 − 0))∫ from 0 to 3 of x² dx = (13)[x³/3] from 0 to 3 = (13)(9) = 3.
Common mistake The definite integral is 9; divide by the interval length 3 to obtain average value.
Method note Average value is total signed area divided by interval length.

Fundamental Theorem of Calculus

AP Calculus AB Fundamental Theorem of Calculus upper bound 2x

Fundamental Theorem of Calculus Medium Multiple Choice Apply the Fundamental Theorem of Calculus with chain rule
If F(x) = ∫2x011 − t3 dt, then F′(x) =
A. 11 − x3
B. 21 − 2x3
C. 11 − 8x3
D. 21 − 8x3

Correct answer: Choice D

21 − 8x3

Full solution Use the Fundamental Theorem of Calculus, then apply the chain rule to the upper limit. Substitute 2x into the integrand and multiply by the derivative of 2x: F′(x)=11−(2x)3·2. Since (2x)3=8x3, F′(x)=21−8x3.
Common mistake The most common mistake is forgetting the outside factor 2. Another trap is writing 1−2x3 instead of 1−(2x)3.
Trap to watch Do not forget to multiply by the derivative of the upper limit.
Method note For ∫g(x)af(t)dt, the derivative is f(g(x))g′(x).
AP note AP note: This is a classic FTC Part 1 plus chain rule question.
Calculator note Non-calculator.

Derivative of an accumulation function practice question

Fundamental Theorem of Calculus Medium Multiple Choice Derivative of an accumulation function
Let F(x) = ∫ from 1 to x² of (t³ + 1) dt. What is F′(x)?
A. x⁶ + 1
B. 2x(x⁶ + 1)
C. 2x³ + 1
D. (x⁶ + 1)/2x

Correct answer: Choice B

2x(x⁶ + 1)

Full solution By the Fundamental Theorem of Calculus and the chain rule, F′(x) = ((x²)³ + 1)·(2x) = 2x(x⁶ + 1).
Common mistake Evaluating the integrand at x² without multiplying by the derivative of x² misses the chain rule.
Method note For an upper limit g(x), evaluate the integrand at g(x) and multiply by g′(x).

Volumes of Revolution

AP Calculus AB volume of revolution bounded by y axis y equals e to x and y equals 2

Volumes of Revolution Hard Multiple Choice Set up and evaluate a volume of revolution using washers
Find the volume of the solid generated when the region bounded by the y-axis, y = ex, and y = 2 is rotated around the y-axis.
A. 0.386
B. 0.592
C. 1.216
D. 3.998

Correct answer: Choice B

0.592

Full solution The region is bounded by x=0, y=ex, and y=2. Since the rotation is around the y-axis, rewrite the curve as x=ln y. The washers have radius ln y, and y runs from 1 to 2. So V=π∫21(ln y)2dy. Evaluating this integral with a calculator gives approximately 0.592.
Common mistake A common mistake is setting up the right idea with the wrong variable. Because the axis is the y-axis, washers are cleanest in terms of y, not x.
Trap to watch For y-axis rotation, washers often require writing x as a function of y.
Method note Before choosing washers or shells, look at the axis of rotation and decide which setup is less messy.
AP note AP note: The calculator helps after the setup. The main AP skill is recognizing the correct radius and bounds.
Calculator note Calculator allowed.

Numerical Integration

AP Calculus AB trapezoidal approximation unequal intervals hit rate table

Numerical Integration Hard Multiple Choice Use trapezoidal approximation with unequal time intervals
The table below shows the “hit rate” for an Internet site, measured at various intervals during a day. Use a trapezoid approximation with 6 subintervals to estimate the total number of people who visited that site.

TimeMidnight6 A.M.8 A.M.Noon5 P.M.8 P.M.Midnight
People per minute523810165
A. 5,280
B. 10,080
C. 10,440
D. 10,560

Correct answer: Choice C

10,440

Full solution The rates are in people per minute, so the time widths must be in minutes. The intervals are 360, 120, 240, 300, 180, and 240 minutes. Use one trapezoid for each interval: 3605+22 + 1202+32 + 2403+82 + 3008+102 + 18010+162 + 24016+52. This totals 10,440.
Common mistake The trap is assuming the subintervals are equal. They are not. The table jumps from 6 A.M. to 8 A.M., then to Noon, then to 5 P.M., so each width must be handled separately.
Trap to watch Unequal time intervals mean unequal trapezoid widths.
Method note For rate-table accumulation, always check the time units before multiplying.
AP note AP note: This is calculator-allowed, but the hard part is not computation. The hard part is respecting the unequal intervals.
Calculator note Calculator allowed.

Trapezoidal approximation practice question

Numerical Integration Medium Multiple Choice Trapezoidal approximation
The values of f at x = 0, 1, and 2 are 2, 4, and 8. Using two trapezoids of width 1, what approximates ∫ from 0 to 2 of f(x) dx?
A. 8
B. 9
C. 10
D. 12

Correct answer: Choice B

9

Full solution The trapezoidal estimate is (12)(2 + 4) + (12)(4 + 8) = 3 + 6 = 9.
Common mistake Adding the function values without the trapezoid weights overcounts the endpoints.
Method note For each subinterval, multiply its width by the average of its endpoint heights.

Chain Rule

AP Calculus AB chain rule y equals f of x squared

Chain Rule Medium Multiple Choice Apply the chain rule to a composite function
If y = f(x2) and f′(x) = √(5x − 1), then dydx is equal to
A. 2x√(5x2 − 1)
B. √(5x − 1)
C. 2x√(5x − 1)
D. √(5x − 1)2x

Correct answer: Choice A

2x√(5x2 − 1)

Full solution Since y=f(x2), the outside function is f and the inside function is x2. By the chain rule, dydx=f′(x2)·2x. Because f′(x)=√(5x−1), replacing x with x2 gives f′(x2)=√(5x2−1). Therefore, dydx=2x√(5x2−1).
Common mistake A common mistake is using √(5x−1) without replacing the input by x2. Another common miss is forgetting the outside factor 2x.
Trap to watch Substitute x² into f′ and multiply by 2x.
Method note For f(g(x)), write f′(g(x))g′(x) before substituting.
AP note AP note: This is a direct chain rule question. The point is to substitute the inside expression into f′, not to find f itself.
Calculator note Calculator allowed, but not needed.

Mean Value Theorem

AP Calculus AB Mean Value Theorem e to x minus x squared calculator question

Mean Value Theorem Hard Multiple Choice Use the Mean Value Theorem and solve numerically
For what value of c on 0 < x < 1 is the tangent to the graph of f(x) = ex − x2 parallel to the secant line on the interval (0,1)?
A. 0.351
B. 0.500
C. 0.693
D. 0.718

Correct answer: Choice C

0.693

Full solution A tangent line parallel to the secant line means the derivative equals the secant slope. First find the secant slope from 0 to 1: f(1)−f(0)1−0. Since f(1)=e−1 and f(0)=1, the secant slope is e−2. Next, f′(x)=ex−2x. Set ec−2c=e−2. Solving this numerically gives c≈0.693.
Common mistake A common mistake is setting f(c) equal to the secant slope. The Mean Value Theorem uses the derivative: f′(c) equals the secant slope.
Trap to watch Parallel tangent and secant means equal slopes, so use f′(c).
Method note For Mean Value Theorem questions, compute the average rate of change first, then match it with the derivative.
AP note AP note: This is a calculator-allowed MVT question because the final equation is not meant to be solved by hand.
Calculator note Calculator allowed.

Differential Equations

Separable differential equations practice question

Differential Equations Medium Multiple Choice Separable differential equations
A function satisfies, dydx = 2xy and y(0) = 3. Which expression gives y(x)?
A. 3e2x
B. 3ex²
C. ex²+3
D. 3x²

Correct answer: Choice B

3ex²

Full solution Separate variables: (1y)dy = 2x dx. Integration gives ln|y| = x² + C, so y = Cex². The initial condition y(0) = 3 gives C = 3.
Common mistake Integrating 2x as 2x instead of x² produces the wrong exponent.
Method note Separate, integrate, and use the initial condition to find the constant.

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