Score More. Spend Less. Bundle 3+ College Prep products for 20% off.

AP® Calculus AB Practice Tests & Questions

From Practice to 5’s: AP Calc AB Made Easy
AP Calc AB Practice Question Bank shown on multiple devices

Ace AP® Calculus AB with practice test questions from our Question Bank (QBank) and earn college credit on your schedule and budget.

Choose Your Subscription
Dream Score Plan
360-Day Access
$69
Deep Learning Plan
180-Day Access
$59
Score Booster Plan
90-Day Access
$49
Introductory Plan
30-Day Access
$39
Buy Now! Try 7-Day Free Trial

Access Includes

  • 1400+ Exam-style Practice Questions
  • Customizable Quiz Generator
  • Realistic Timed Test Simulation
  • Colorful Visual Explanations
  • Step-by-Step Solutions
  • Adjustable Smart Study Planner
  • Progress Dashboard
  • Smart Flashcards

Try These AP Calc AB Practice Test Questions

Check out sample AP Calculus MCQs before you commit. Stop wasting hours on material you already know; our practice tests let you focus on the topics you actually need to study. Get exam-style questions with step-by-step answer explanations and visuals that actually make sense. Practice smarter, build confidence, and see how UWorld helps you nail that 5.
Limits and Continuity Practice Test

Question

The function f is defined for all real values of x. The graph above shows the portion of the graph of f on [−1, 3], and the table above gives selected values of f near x = 5. Which of the following limit statements is likely to be true?

A. I and II only
B. I and III only
C. II and III only
D. I, II, and III

Explanation

As x approaches 1 from the left, the graph of f approaches 3. Therefore, lim x → 1 - f x = 3 and Statement I is true.

As x approaches 5 from the right, the values of f in the given table get very close to 2. Therefore, lim x → 5 + f x = 2 and Statement II is true.

As x approaches 5 from the left, the values of f in the given table also get very close to 2. Both limits of f approach the same value as x approaches 5, so lim x → 5 f x exists. Therefore, Statement III is false.

Question

Let f, g, and h be functions such that g(x) ≤ f(x) ≤ h(x) for all values of x. Let g(x) = x2 + 1 and h(x) = −x2 + 1. At what value of x can the squeeze theorem be applied, and what is the limit of f as x approaches this value?

A. The squeeze theorem can be applied at x = 0 for the given functions, and lim x → 0 f (x) = 0 .
B. The squeeze theorem can be applied at x = 0 for the given functions, and lim x → 0 f (x) = 1 .
C. The squeeze theorem can be applied at x = 1 for the given functions, and lim x → 1 f (x) = 0 .
D. The squeeze theorem cannot be applied to the given functions.

Explanation

The functions g(x) = x2 + 1 and h(x) = −x2 + 1 only approach the same value as x approaches 0, so it is possible to apply the squeeze theorem at x = 0.

The limits of g and h as x approaches 0 are both 1, so the limit of f as x approaches 0 must equal 1.

Question

lim x → 2 x 2 - 4 x - 2 =
A. 0
B. 2
C. 4
D. The limit does not exist.

Explanation

To evaluate the given limit, first try to substitute x = 2 into the limit directly.

Direct substitution yields an indeterminate form, so factor the numerator to evaluate.

Therefore, the given limit equals 4.

Question

If lim x → 7 ln (x - 1) - ln (6) x - 7 =

A. 0
B. 1 7
C. 1 6
D. nonexistent

Hint :

A limit of the form lim x → a f (x)   -  f (a) x  -  a is the definition of f '(a), the derivative of a function f(x) at x = a.

Explanation

A limit of the form lim x → a f (x)   -  f (a) x  -  a is the definition of f '(a), the derivative of a function f(x) at x = a.

Usual limit evaluation techniques do not apply to a limit of this form, so identify the function f(x), the value of a, and then calculate f'(a) to evaluate the limit.

Compare the given limit to the definition of the derivative f ′ a = lim x → a f x   -  f a x  -  a to identify f(x) and a, which is the x-value the limit is approaching.

If a = 7, then ln(6) must equal f(7), ln(x - 1) must equal f(x), and lim x → 7 ln ⁡ ( x − 1 ) − ln ⁡ ( 6 ) x − 7 must equal f'(7). Note: To verify that f(7) = ln(6), plug 7 into f(x) = ln(x - 1).

To evaluate f'(7), differentiate f(x) and then plug in x = 7. The function f(x) = \ln(x - 1) is a composite function, so use the chain rule to find f'(x).

f ( x ) = ln ⁡ ( x − 1 ) Given function
f ′ ( x ) = 1 x − 1 ⋅ 1 Differentiate
f ′ ( x ) = 1 x − 1 Multiply
f ′ ( 7 ) = 1 7 − 1 Plug in x = 7
f ′ ( 7 ) = 1 6 Simplify

Therefore, lim x → 7 ln ⁡ ( x − 1 ) − ln ⁡ ( 6 ) x − 7 = 1 6 .

(Choice A) 0 may result from attempting to evaluate the limit by direct substitution and assuming that the indeterminate form 0 0 equals 0.

(Choice B) 1 7 may result from not using u-substitution to integrate ln(x - 1).

(Choice D) "Nonexistent" may result from attempting to evaluate the limit by direct substitution and assuming that the indeterminate form 0 0 does not exist.

Things to remember:

A limit of the form lim x → a f ( x ) − f ( a ) x − a is the definition of f'(a), the derivative of a function f(x) at x = a.

Alternate Method :

An alternate way to evaluate the limit is to plug in x = 7 and try direct substitution.

lim x → 7 ln ⁡ ( x − 1 ) − ln ⁡ ( 6 ) x − 7 Given limit
ln ⁡ ( 7 − 1 ) − ln ⁡ ( 6 ) 7 − 7 Plug in x = 7
0 0 Simplify

Direct substitution yields the indeterminate form 0 0 , so apply L'Hospital's Rule. Differentiate the numerator and denominator independently using derivative rules, and then try direct substitution on the resulting fraction.

The function ln(x-1) is a composite function, so use the chain rule to differentiate.

lim x → 7 ln ⁡ ( x − 1 ) − ln ⁡ ( 6 ) x − 7

Equation from given limit

lim x → 7 d d x [ ln ⁡ ( x − 1 ) − ln ⁡ ( 6 ) ] d d x [ x − 7 ]

Apply L'Hospital's Rule

lim x → 7 1 x − 1 − 0 1 − 0

Differentiate

1 7 − 1

Simplify and plug in x = 7

1 6

Subtract

Therefore, lim x → 7 ln ⁡ ( x − 1 ) − ln ⁡ ( 6 ) x − 7 = 1 6 .

Things to remember:

  • To evaluate a limit of the form lim x → c f ( x ) g ( x ) first try directly substituting x = c into the function. If direct substitution yields the indeterminate form 0 0 , simplify the expression or apply L'Hospital's Rule.
  • To differentiate a composite function, use the chain rule:
    d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) g ′ ( x )

Question

If f (x) = 7 cot x - 3 x 4 , then f ′(x) =

A. - 7 csc x cot x - 12 x 3
B. 7 csc x cot x + 12 x 5
C. 7 csc 2 x + 12 x 3
D. - 7 csc 2 x + 12 x 5

Hint :

The derivative of the difference of two or more functions is equal to the difference of their derivatives.

Explanation

The derivative of the sum or difference of two or more functions is equal to the sum or difference of their derivatives.

The given function f(x) = 7 cot x - frac{3}{x^4} is a difference of two functions, so apply the sum and difference of derivatives property to find f'(x).

To differentiate f(x), first rewrite the second term in the form x−n.  Then use differentiation rules to find f′(x).

f ′ ( x ) = d d x [ 7 cot ⁡ x ] − d d x [ 3 x 4 ] Result of sum and difference rule
f ′ ( x ) = d d x [ 7 cot ⁡ x ] − d d x [ 3 x − 4 ] Rewrite 1 x n = x - n
f ′ ( x ) = − 7 csc 2 x + 12 x − 5 Differentiate cot x and apply power rule
f ′ ( x ) = − 7 csc 2 x + 12 x 5 Rewrite x - n = 1 x n to match answer choices

Therefore, the derivative of the given function f is f'(x) = −7csc2x + 12x5 .

(Choice A) −7cscxcotx −12x3 may result from incorrectly differentiating the first term using the derivative of csc x (instead of cot x) and not using the negative exponent rule when rewriting the second term.

(Choice B) 7cscxcotx + 12x5 may result from incorrectly differentiating the first term using the derivative of csc x and not including the negative sign.

(Choice C) 7csc2x + 12x3 may result from not including the negative sign in the derivative of cot x and mistakenly adding 1 (instead of subtracting 1) to the exponent when applying the power rule to the second term.

Things to remember:
The derivative of the difference of two or more functions is equal to the difference of their derivatives:

d dx [f (x) - g (x)] = d dx [f (x)] - d dx [g (x)]

Question

What is the slope of the line tangent to the graph of y = x sin x at x = π ?

A. - π
B. - 1
C. 0
D. π

Hint :

The slope of the line tangent to a function f at x=a is the value of the derivative of f at x=a.

Explanation

The slope of the line tangent to a function f at x=a is the value of the derivative of f at x=a.

To determine the slope of the line tangent to y=xsinx at x=π , first differentiate to find dydx and then plug in x=π. The given function is the product of two functions, so use the product rule to differentiate:

d dx [uv] = u·v' + v·u'

Let u=x and v=sinx , and then differentiate both equations to find u' and v'.

Now apply the product rule to find dydx .

dy dx = u·v' + v·u' Product rule
dy dx = (x) (cosx) + (sinx) (1) Substitute u=x, v=sinx, u'=1, and v'=cosx
dy dx = xcosx + sinx Simplify

To find the slope of the line tangent to y at x=π , plug x=π into the equation for dy dx and evaluate the resulting expression.

dy dx = xcosx + sinx Derivative of y
dy dx |x=π = πcosπ + sinπ Plug in x=π
dy dx |x=π = π(-1)+(0) Evaluate cosπ=-1 and sinπ=0
dy dx |x=π = -π Simplify

Therefore, the slope of the line tangent to the graph of y=xsinx at x=π is -π.

(Choice B) -1 may be the result of incorrectly calculating the values of the trigonometric functions at x=π: sinπ≠-1 and cosπ≠0.

(Choice C) 0 may be the result of mistakenly plugging x=π into y, rather than dy dx , to calculate the slope of the tangent line.

(Choice D) π may be the result of subtracting, rather than adding, when using the product rule: d dx [uv] ≠ u'v − v'u .

Things to remember:

  • The slope of the line tangent to a function fatx=a is the value of the derivative of fatx=a .
  • To find the derivative of the product of two functions, use the product rule:
    d dx [uv] = u·v' + v·u'

Question

If f (x) = cos ( x 2 - 3 π 2 ) , then f ′ ( 3 π ) =

A. - 2 3 π
B. - sin ( 2 3 π )
C. 1
D. 2 3 π

Hint :

Use the chain rule to differentiate a composite function.

Explanation

To find f ′(c), the value of the derivative of a function f at x = c, first differentiate f and then plug x = c into f ′(x).

The given function f (x) = cos ( x 2 - 3 π 2 ) is a composite function of the form cos u, where u = x 2 - 3 π 2 is the inside function.

To find the derivative of a function of the form cos u, use the chain rule.

First differentiate the outside function with respect to the inside function, leaving the inside function intact. Then multiply by the derivative of the inside function with respect to x.

f ( x ) = cos ( x 2 − 3 π 2 ) Given function
f ′ (x) = - sin ( x 2 - 3 π 2 ) ⋅ d dx ( x 2 - 3 π 2 ) Apply chain rule
f ′ ( x ) = − sin ⁡ ( x 2 − 3 π 2 ) ( 2 x ) Differentiate

To evaluate f ′ ( 3 π ) , plug x = 3 π into f ′ (x) and simplify the result.

f ′ ( x ) = − sin ⁡ ( x 2 − 3 π 2 ) ( 2 x ) Derivative of f
f ′ ( 3 π ) = − sin ⁡ ( ( 3 π ) 2 − 3 π 2 ) ( 2 3 π ) Plug in x = 3 π
f ′ ( 3 π ) = − sin ⁡ ( 3 π − 3 π 2 ) ( 2 3 π ) Evaluate: ( 3 π ) 2 = 3 π
f ′ ( 3 π ) = − sin ⁡ ( 3 π 2 ) ( 2 3 π ) Simplify: 3 π - 3 π 2 = 6 π 2 - 3 π 2
f ′ ( 3 π ) = 2 3 π Evaluate: sin 3 π 2 = - 1

Therefore, f ′ ( 3 π ) = 2 3 π .

(Choice A) − 2 3 π may result from differentiating cos ⁡ x incorrectly: d d x [ cos ⁡ x ] ≠ sin ⁡ x .

(Choice B) − sin ⁡ ( 2 3 π ) may result from not using the chain rule: d d x [ f ( g ( x ) ) ] ≠ f ′ ( g ′ ( x ) ) .

(Choice C) 1 may result from not using the chain rule: d d x [ f ( g ( x ) ) ] ≠ f ′ ( g ( x ) ) .

Things to remember:

To find the derivative of a composite function, use the chain rule:

d d x [ f ( u ( x ) ) ] = f ′ ( u ( x ) ) ⋅ u ′ ( x )

Question

If y = 2sinx − e3x , then d3y dx3 =

A. 2 cos x − 3e3x
B. −2 sin x − 9e3x
C. −2 cos x − e3x
D. −2 cos x − 27e3x

Hint :

The notation d 3 y d x 3 represents the third derivative of y with respect to x.

Explanation

The notation d 3 y d x 3 represents the third derivative of y with respect to x. To find the third derivative:

Differentiate each term of the given function y = 2sinx − e3x to find the first derivative.

Use trigonometric differentiation rules to differentiate the first term. The second term is a composite function of the form eu where u = 3x, so use the exponential chain rule to differentiate.

y = 2 sin ⁡ x − e 3 x Given function
d y d x = 2 cos ⁡ x − e 3 x Apply constant and trigonometric rules to differentiate first term
d y d x = 2 cos ⁡ x − 3 e 3 x Differentiate second term

To find the second derivative, differentiate the first derivative.

d y d x = 2 cos ⁡ x − 3 e 3 x First derivative
d 2 y d x 2 = − 2 sin ⁡ x − 3 e 3 x Apply constant and trigonometric rules to differentiate first term
d 2 y d x 2 = − 2 sin ⁡ x − 3 e 3 x ⋅ 3 Apply constant and exponential rules to differentiate second term
d 2 y d x 2 = − 2 sin ⁡ x − 9 e 3 x Simplify

Now differentiate the second derivative to find the third derivative.

d 2 y d x 2 = − 2 sin ⁡ x − 9 e 3 x Second derivative
d 3 y d x 3 = − 2 cos ⁡ x − 9 e 3 x Apply constant and trigonometric rules to differentiate first term
d 3 y d x 3 = − 2 cos ⁡ x − 9 e 3 x ⋅ 3 Apply constant and exponential rules to differentiate second term
d 3 y d x 3 = − 2 cos ⁡ x − 27 e 3 x Simplify

Therefore, the third derivative of y = 2 sin x − e3x is − 2  cos ⁡ x − 27 e 3 x .

(Choice A) 2 cosx − 3e3x is the first derivative of 2sinx − e3x, but the question asks for the third derivative.

(Choice B) −2sinx − 9e3x is the second derivative of 2sinx − e3x, but the question asks for the third derivative.

(Choice C) − 2 cos ⁡ x − e 3 x may result from not applying the chain rule to differentiate the second term.

Things to remember:

  • The notation d 3 y d x 3 represents the third derivative of y with respect to x.
  • To find the derivative of a composite function, use the chain rule.

Question

The table above gives selected values for a differentiable and increasing function g and its derivative. If g−1 is the inverse function of g, what is the value of (g −1)′(4) ?

A. - 5
B. - 3
C. 1 2
D. 1 5

Hint :

If g−1 is the inverse of the function g, then the derivative of g−1 at the point (a, b) is the reciprocal of the derivative of g at the point (b, a).

Explanation

If g−1 is the inverse of the function g, then the derivative of g−1 at the point (a, b) is the reciprocal of the derivative of g at the point (b, a):

To find (g−1)′(4), first use the table to find the value of x for which g(x) = 4. The table shows that g contains the ordered pair (−1, 4), so it's inverse g−1 must contain the ordered pair (4, −1).

Plug a = 4 an b = −1 into the equation above, and then identify and plug in the value of gʹ(−1) from the given table.

( g − 1 ) ′ ( a ) = 1 g ′ ( b ) Derivative of the inverse of g
( g − 1 ) ′ ( 4 ) = 1 g ′ ( − 1 ) Plug in a=4 and b=-1
( g − 1 ) ′ ( 4 ) = 1 2 Plug in g'(-1)=2

Therefore, the value of ( g − 1 ) ′ ( 4 ) is 1 2 .

(Choice A) − 5 is the value of g ′ ( 4 ) , but the question asks for the value of ( g − 1 ) ′ ( 4 ) .

(Choice B) − 3 is the value of g ′ ( 4 ) , but the question asks for the value of ( g − 1 ) ′ ( 4 ) .

(Choice D) 1 5 result from mistakenly taking the negative reciprocal of g ′ ( 4 ) , instead of the reciprocal of g ′ ( − 1 ) , to calculate ( g − 1 ) ′ ( 4 ) .

Things to remember:

If g-1 is the inverse of the function g, then the derivative of g-1 at the point (a, b) is the reciprocal of the derivative of g at the point (b, a).

( g − 1 ) ′ ( a ) = 1 g ′ ( b )

Question

The graph above shows the position x t at time t of a particle moving along a straight line for 0 < t < 7 The graph has horizontal tangents at t = 3 and t = 6 and a point of inflection at t = 5 . For what values of t is the velocity of the particle decreasing?

A. 0 < t < 5
B. 5 < t < 7 only
C. 0 < t < 3 and 6 < t < 7
D. 0 < t < 1 and 5 < t < 7

Hint :

The velocity v t of a particle moving along a straight line is decreasing when v ′ t < 0

Explanation

To determine the interval(s) on which the velocity of a particle is decreasing from the graph of the position function, first recall that the velocity function v(t) is the derivative of the position function x(t).

v ( t ) = x ′ ( t )

If the velocity is decreasing, then the derivative of velocity v'(t) must be negative and therefore the second derivative of the position function x''(t) must also be negative.

v ′ ( t ) < 0
⇓
x ″ ( t ) < 0

The second derivative of a function determines the concavity of its graph. Since x''(t) < 0, identify where the graph of x(t) is concave down.

The graph of x(t) has a point of inflection at t = 5, so the graph changes concavity at that point.

The graph of x(t) is concave down on the interval 5 < t < 7, so the velocity is decreasing on 5 < t < 7 only.

(Choice A) 0 < t < 5 is the interval on which the velocity is increasing, but the question asks for the interval on which the velocity is decreasing.

(Choice C) 0 < t < 3 and 6 < t < 7 may be a result of identifying the intervals on which the velocity is negative rather than the interval on which the velocity is decreasing.

(Choice D) 0 < t < 1 and 5 < t < 7 may be a result of identifying intervals on which x(t) > 0 rather than on which the velocity is decreasing.

Things to remember:

To determine when the velocity of a function is decreasing from the graph of its position function, identify when the graph of x(t) is concave down.

Question

The volume of a cube with edges of length x centimeters is given by V = x 3 cubic centimeters. If the volume of the cube is increasing at a constant rate of 45 cubic centimeters per minute, at what rate is x changing, in centimeters per minute, with respect to time at the instant when x = 3 ?

A. 5 3
B. 5 2
C. 15
D. 5

Hint :

To solve a related rates problem, first identify the rate to be found, the given rate, and any other given values.

Explanation

The volume of a cube with edges of length x centimeters is given by V = x 3 cubic centimeters. If the volume of the cube is increasing at a constant rate of 45 cubic centimeters per minute, at what rate is x changing, in centimeters per minute, with respect to time at the instant when x = 3 ?

To find the rate at which x is changing with respect to time dx dt when x = 3 , differentiate both sides of the volume equation with respect to time t. Then plug in the given values for dV dt and x and solve for dx dt .

V=x3 Volume of a cube with edge length x
dV dt = 3x2 dx dt Differentiate both sides with respect to t: d dt [un] =nun-1 dudt
45=3(3)2 dxdt Plug in dVdt=45 and x=3
45=27 dxdt Simplify right side: 3(3)2=3(9)=27
45 27 = dxdt Divide both sides by 27
5 3 = dxdt Reduce: 45 27 = 5·9 3·9 = 53

Therefore, x is changing at the rate of 5 3 centimeters per minute at the instant when x = 3 .

(Choice B) 52 may be a result of incorrectly differentiating the given expression for volume: ddt [x3] ≠ 3·2 dxdt .

(Choice C) 15 may be a result of improperly differentiating the volume equation to get dVdt =3x2 and solving for x rather than differentiating both sides of the volume equation with respect to t to get dVdt =3x2 dxdt and solving for dxdt .

(Choice D) 5 may be a result of neglecting to multiply by 3 when differentiating the expression for volume: ddt [x3] ≠ x2 dxdt .

Things to remember:

  • To solve a related rates problem in which one rate is given and another rate is to be found, first identify the rate to be found, when it is to be found, and the given rate.

Question

lim x → 1 2 [ sin(2πx) ln(2x) ] is

A. - π
B. 0
C. π
D. nonexistent

Hint :

To evaluate a limit of the form lim x→c f(x) g(x) , first directly substitute x=c into the function.

Explanation

To evaluate a limit of the form lim x→c f(x) g(x) , first directly substitute x=c into the function and simplify. If the result yields an indeterminate form, apply L'Hospital's Rule.

lim x→12 [ sin(2πx) ln(2x) ] Given limit
sin(2π·12) ln(2·12) Plug in x=12
0 0 Evaluate: sinπ=0 and ln1=0

Direct substitution results in the indeterminate form 0 0 , so apply L'Hospital's Rule by differentiating the numerator and denominator independently, and then evaluate the limit again.

The numerator and denominator are composite functions of the form sinu and lnu, so use the chain rule to differentiate.

lim x → 1 2 d dx [sin(2πx)] d dx [ln(2x)] Apply L'Hospital's Rule
lim x → 12 2πcos(2πx) 1x Differentiate numerator and differentiate denominator independently
-2π 2 Evaluate limits of numerator and denominator
-π Simplify

Therefore, lim x → 12 [ sin(2πx) ln(2x) ] is −π.

(Choice B) 0 may result from assuming the indeterminate form 00 = 0.

(Choice C) π may result from using an incorrect derivative: d dx [sinx] ≠ −cosx.

(Choice D) "Nonexistent" may result from assuming that the indeterminate form 00 means that the limit does not exist.

Things to remember:

  • If a limit yields an indeterminate form, apply L'Hospital's Rule: Differentiate the numerator and denominator independently and then evaluate the limit.

Question

The derivative of a continuous function f is given by f ′ x = 4 x 2 - 12 x 2 , what value of x in the open interval (1, 4) satisfies the conclusion of the Mean Value Theorem for f on the closed interval [1, 4] ?

A. 1
B. 3
C. 2 3
D. 2

Hint :

The Mean Value Theorem guarantees that the instantaneous rate of change of a continuous and differentiable function f equals the average rate of change of f on [a, b] for at least one x-value in the interval (a, b).

Explanation

The Mean Value Theorem (MVT) guarantees that the instantaneous rate of change of a continuous and differentiable function f equals the average rate of change (AROC) of f on [a, b] for at least one x-value c in the interval (a, b).

To determine the value of x = c that satisfies the MVT for f on the closed interval [1, 4], first use the given function values to calculate the (AROC) of f(x) on [1, 4].

f 4 - f 1 4 - 1 AROC formula for a = 1 and b = 4
19 - 16 4 - 1 Plug in f(4) = 19 and f(1) = 16
1 Simplify

The average rate of change of f(x) on [1, 4] is equal to 1, so set the given instantaneous rate of change  f ′ x = 4 x 2 - 12 x 2 equal to the average rate of change 1 and solve for x.

4 x 2 - 12 x 2 = 1 Set instantaneous rate of change equal to AROC
4 x 2 - 12 = x 2 Multiply both sides by x2
3 x 2 = 12 Add 12 and subtract x2 on both sides
x 2 = 4 Divide both sides by 3
x = ± 2 Take the square root of both sides

The solution x = −2 is not in the interval (1, 4), so the only value that satisfies the MVT for f on the closed interval [1, 4] is x = 2.

(Choice A) 1 is the value of the average rate of change of f on [1, 4], but the question asks for the value of x that satisfies the MVT on [1, 4].

(Choice B) 3  may result from mistakenly setting f ′ equal to zero instead of setting f ′ equal to the average rate of change.

(Choice C) 2 3  may result from incorrectly calculating the average rate of change as 3.

Things to remember:
The Mean Value Theorem guarantees that the instantaneous rate of change of a continuous and differentiable function f equals the average rate of change of f on [a, b] for at least one x-value in the interval (a, b).

Question

The function g is defined by g x = - 5 + 6 + x 2 - 7   sin   x . If the function f is an antiderivative of g, which of the following is the x-coordinate of the location of a local minimum for the graph of y = f(x) ?

A. −5.058
B. −3.391
C. 1.216
D. 3.821

Hint :

An antiderivative f of a function g is any function whose derivative is g.

Explanation

An antiderivative f of a function g is any function whose derivative is g. It is given that f is an antiderivative of g, so the derivative of f is g.

A local (relative) minimum on the graph of f(x) occurs at any value of x where f′(x) = g(x) is zero or undefined (a critical point on the graph of f) and changes from negative to positive.

To find the x-value of the relative minimum of f, graph its derivative  g x = - 5 + 6 + x 2 - 7   sin   x  and locate any zeros where the graph changes from negative to positive.

The graph of f′ = g changes from negative to positive at only one point, so the graph of y = f(x) has a relative minimum at x = 3.821.

(Choice A) −5.058 is the x-coordinate of a local maximum on the graph of f, but the question asks for the x-value of a local minimum.

(Choices B and C) −3.391 and 1.216 are x-coordinates of relative minimums on the graph of g(x), but the question asks for a relative minimum on the graph of f(x) = g′(x).

Things to remember:

  • An antiderivative of a function f is any function F whose derivative is f.

  • A local minimum of a continuous function f(x) occurs at any value of x at which f′(x) is zero or undefined and f′ changes from negative to positive.

Question

The function f is continuous on the closed interval [−1, 5]. If f(−1) = −3 and f(5) = 3, which of the following conditions would guarantee that there is a number c in the open interval (−1, 5) where f′(c) = 1 ?

A.

f is differentiable on the open interval (−1, 5).
B. f is not differentiable on the open interval (−1, 5).
C. lim x → k f x = f k for all k on the open interval (−1, 5).
D. f′(k) = 0 for at least one value of k in the open interval (−1, 5).

Hint :

The Mean Value Theorem guarantees that the instantaneous rate of change of a continuous and differentiable function f equals the average rate of change of f on [a, b] for at least one x-value c in the interval (a, b).

Explanation

The Mean Value Theorem (MVT) guarantees that the instantaneous rate of change of a continuous and differentiable function f equals the average rate of change of f on [a, b] for at least one x-value c in the interval (a, b).

To guarantee there is a number c in the open interval (−1, 5) where f′(c) = 1, first verify the average rate of change is 1 between the given points f(−1) = −3 and f(5) = 3 (calculation).

The MVT guarantees that f′(c) = 1 if the MVT conditions are met for f on the interval [−1, 5]. It is given that f is continuous on [−1, 5], but no information is given about the differentiability of f on (1, 5).

The MVT applies to the continuous function f on [−1, 5] and guarantees that f′(c) = 1 for at least one value of c in (−1, 5) only if f is differentiable on the open interval (−1, 5).

(Choice B) "f is not differentiable on the open interval (−1, 5)" indicates that the MVT does not apply to the function on the interval, but the question implies that the MVT does apply.

(Choice C) " lim x → k f x = f k for all k on the open interval (−1, 5)" indicates that the function is continuous on the open interval, but the MVT also requires differentiability.

(Choice D) "f′(k) = 0 for at least one value of k in the open interval (−1, 5)" is the conclusion of Rolle's Theorem, but the question requires justifying the conditions of the MVT.

Things to remember:
The Mean Value Theorem guarantees that the instantaneous rate of change of a continuous and differentiable function f equals the average rate of change of f on [a, b] for at least one x-value c in the interval (a, b).

Question

The graph of a differentiable function f is shown above. g (x) = ∫ 2 x f (t) dt , which of the following is true?

A. g(5) < g′(5) < g″(5)
B. g(5) < g″(5) < g′(5)
C. g′(5) < g(5) < g″(5)
D. g″(5) < g′(5) < g(5)

Hint :

Determine whether each quantity is negative, zero, or positive and then compare their relative values.

Explanation

To compare the values of g(5), g′(5), and g″(5), evaluate g(5) and find the first and second derivatives of g(x) at x = 5.

Evaluate g(5)

Plug x = 5 into g(x) to evaluate g(5): g (5) = ∫ 2 5 f (t) dt

This definite integral represents the area bounded by the graph of f and the x-axis between x = 2 and x = 5. The exact value of g(5) cannot be determined, but this region is above the x-axis, so g(5) > 0.

Evaluate g′(5)

The given function is of the form ∫ a x f (t) dt , so use the Second Fundamental Theorem of Calculus (Second FTC) to differentiate g(x) and then plug in x = 5.

g (x) = ∫ 2 x f (t) dt Given function
g ′ (x) = d dx [ ∫ 2 x f (t) dt ] Differentiate both sides
g ′ (x) = f (x) Apply Second FTC
g ′ (5) = f (5) Plug in x = 5

Examine the graph of f at x = 5 to see that g′(5) = 0.

Evaluate g″(5)

Differentiate g′(x) = f(x) and then plug in x = 5 to evaluate g″(5).

g ′ (x) = f (x) First derivative of g x
d dx [ g ′ (x) ] = d dx [ f (x) ] Differentiate both sides
g ″ (x) = f ′ (x) Simplify
g ″ (5) = f ′ (5) Plug in x = 5

The value of f ′ 5 is the slope of the line tangent to f at x = 5. The exact value of g″(5) cannot be found, but the slope of the tangent line is negative at x = 5, so g″(5) < 0.

Therefore, the statement g″(5) < g′(5) < g(5) is true because g″(5) < 0, g′(5) = 0, and g(5) > 0.

(Choice A)g(5) < g′(5) < g″(5) may result from mistakenly reversing the correct inequality.

(Choice B)g(5) < g″(5) < g′(5) may result from incorrectly calculating the relative values of the three quantities.

(Choice C)g′(5) < g(5) < g″(5) may result from assuming the graph represents g(x) rather than f(x).

Things to remember:
The derivative of an integral of the form ∫ a x f (t) dt , where f is a continuous function, is f(x):

d dx ∫ a x f (t) dt = f (x)

Question

The rate at which gravel is deposited at a landfill on a certain day is modeled by the function R, where R(t) is measured in tons per hour and t is the number of hours since the landfill opened. Using a trapezoidal sum with three subintervals as indicated by the data in the table above, approximate the number of tons of gravel deposited during the first 8 hours since the landfill opened.

A. 75
B. 81
C. 128
D. 150

Hint :

The net change in a quantity over an interval of time is equal to the definite integral of the rate of change in the quantity over that interval.

Explanation

The number of tons of gravel deposited during the first 8 hours since the landfill opened is the net change in tons of gravel deposited at the rate R(t) from t1 = 0 to t2 = 8.

The net change in a quantity over an interval of time [t1, t2] is equal to the definite integral of the rate of change r(t) in the quantity over that interval.

Therefore, the number of tons of gravel deposited during the first 8 hours since the landfill opened is equal to ∫ 0 8 R t dt .

To approximate the net change integral using a trapezoidal sum with three subintervals, calculate the areas of three trapezoids, where the trapezoid on subinterval n has height hn and base lengths bn and bn+1.

The height hn of each trapezoid is the length of the subinterval (distance between the t-values). The base lengths bn and bn+1 of each trapezoid are the values of R(t) at the endpoints of the subinterval.

The three subintervals from the given table are [0, 2], [2, 5], and [5, 8], so the heights of the trapezoids are 2, 3, and 3, respectively. The base lengths are the given values of R at the endpoints.

Plug values of bn and h for the three trapezoids into the trapezoidal sum expression and simplify. Factor the greatest common factor (GCF)  1 2  out of the trapezoidal sum expression for a more efficient calculation.

1 2 h 1 (b 1 + b 2 ) + 1 2 h 2 (b 2 + b 3 ) + 1 2 h 3 (b 3 + b 4 ) Trapezoidal sum
1 2 [h 1 (b 1 + b 2 ) + h 2 (b 2 + b 3 ) + h 3 (b 3 + b 4 ) ] . Factor out GCF 12
1 2 [(2) (4 + 8) + (3) (8 + 10) + (3) (10 + 14)] Plug in values of bn and hn
1 2 [24 + 54 + 72] Simplify inside brackets
75 Simplify

Therefore, the approximate number of tons of gravel deposited during the first 8 hours since the landfill opened is 75.

(Choice B) 81 may result from mistakenly assuming that each of the three trapezoids has the same height (3).

(Choice C) 128 may result from approximating the number of tons of gravel using a left Riemann sum, but the question asks for the approximation using a trapezoidal sum.

(Choice D) 150 may result from neglecting the factors of  1 2  in the trapezoidal sum equation.

Things to remember:

  • The net change in a quantity over an interval of time [t1, t2] is given by the definite integral of the rate of change r(t) in the quantity from t1 to t2.
  • A trapezoidal sum approximates a definite integral  ∫ a b f (x) dx  using trapezoids with height hn and base lengths bn and bn+1 to estimate the area under f in each subinterval.

∫ a b f (x) dx ≈ 1 2 h 1 (b 1 + b 2) + 1 2 h 2 (b 2 + b 3) + …

Question

∫ 2 x 3 x 4 - 8 5 d x =

A. 1 2 x 4 1 5 x 5 - 8 x 5 + C
B. 1 12 x 4 - 8 6 + C
C. 1 3 x 4 - 8 6 + C
D. 4 3 x 4 - 8 6 + C

Hint :

If one factor of an integrand is a constant multiple of the derivative of the inside of a composite function, use  u-substitution to integrate.

Explanation

If one term of the integrand is a constant multiple of the derivative of the inside of a composite function, use u -substitution to integrate.

.

The given integrand contains the composite function  x 4 - 8 5 , where  u = x 4 - 8  is the inside function.

Check to see if the factor 2x3 from the integrand is a constant multiple of the derivative of x4−8 .

Since  2 x 3  is a constant multiple of  x 4 - 3 , use u-substitution, where u=x4−8 , First differentiate u with respect to x, and then solve for dx.

u = x 4 - 8 Inside function
d u d x = 4 x 3 Differentiate both sides with respect to  x
d u = 4 x 3 d x Multiply both sides by  d x
d u 4 x 3 = d x Divide both sides by  4 x 3  to solve for  d x

Now substitute  x 4 - 8  and d u 4 x 3  into the given integral, and then simplify and integrate.

∫ 2 x 3 x 4 - 8 5 d x Given integral
∫ 2 x 3 u 5 d u 4 x 3 Substitute  u = x 4 - 8  and  d u 4 x 3 = d x
2 4 ∫ x 3 u 5 d u x 3 Reduce:  x 3 x 3 = 1  and apply:  ∫ c f x d x = c ∫ f x d x
1 2 ∫ u 5 d u Reduce:  2 4 = 1 2
1 2 ⋅ u 5 + 1 6 + C Apply:  ∫ x n   d x = x n + 1 n + 1 + C
u 6 12 + C Simplify

Substitute x 4 - 8 back in for u to write the answer in terms of x .

u 6 12 + C Answer in terms of  u
x 4 - 8 6 12 + C Substitute  x 4 - 8  back in for  u
1 12 x 4 - 8 6 + C Rewrite to match answer choice

Therefore,  ∫ 2 x 3 x 4 - 8 5 d x = 1 12 x 4 - 8 6 + C .

(Choice A) 1 2 x 4 1 5 x 5 - 8 x 5 + C  is a result of integrating each factor independently instead of using  u -substitution.

(Choice C) 1 3 x 4 - 8 6 + C  may be a result of omitting  1 4  when substituting in  d u 4 x 3  for  d x .

(Choice D) 4 3 x 4 - 8 6 + C  may be a result of mistakenly rewriting  d u 4 x 3  as  4 d u x 3 .

Things to remember:
If one term of the integrand is a constant multiple of the derivative of the inside function of a composite function, use u-substitution to integrate.

Question

If the population P(t) of microorganisms in an experiment changes linearly, where t is the time in hours, which of the following differential equations could model the rate of change in the number of microorganisms?

A. dP dt = - 362
B. dP dt = - 362 P
C. dP dt = - 418 t
D. dP dt = - 418 P 2

Hint :

If the growth of a quantity is linear, then the rate of growth of the quantity is constant.

Explanation

The rate of change of a linear function y is a differential equation of the form  dy dt = k , where k is the slope of the line (a constant) and  dy dt  is the derivative of y with respect to time t.

It is given that the population of microorganisms P(t) at time t in the experiment changes linearly, so the rate of change of the population  dP dt  is constant. Eliminate all choices that include a variable.

The only choice that could model the rate of change in the number of microorganisms is  dP dt = - 362 .

(Choices B and C) dP dt = - 362 P  and  dP dt = - 418 t  model linear rates of change. However, the rate of change of the population must be a constant value for the population to change linearly.

(Choice D) dP dt = - 418 P 2  contains a quadratic expression, but the rate of change of a linear function must equal a constant.

Things to remember:
The rate of change of a linear function y is a differential equation of the form  dy dt = k , where k is the slope of the line (a constant) and  dy dt  is the derivative of a function y with respect to time t.

Question

Which of the following is the solution and its domain for the differential equation dy dx = 16 x y with the initial condition y 2 = - 2 ?

A. y = 16 x 2 - 60
B. y = - 16 x 2 - 60
C. y = 16 x 2 - 60
D. y = - 16 x 2 - 60

Hint :

To find a particular solution to a separable differential equation, first separate the variables.

Explanation

To find a particular solution to a separable differential equation, separate the variables, integrate both sides of the equation, and then use an initial condition to find the expression for y that includes the constant of integration.

First multiply the given differential equation by dx and y to separate the variables on opposite sides of the equation.

dy dx = 16 x y Given differential equation
y dy = 16 x dx Multiply both sides by dx and y

Now integrate both sides of the equation to find the general solution to the differential equation.

∫ 0 0 y dy = ∫ 0 0 16 x dx Integrals of resulting equation
y 2 2 + C 1 = 8 x 2 + C 2 Integrate
y 2 2 = 8 x 2 + C Combine constants of integration

To find the value of C, plug in the initial condition y(2) = −2 and solve for C.

y 2 2 = 8 x 2 + C General solution to differential equation
(- 2) (2) 2 = 8 2 2 + C Plug in y(2) = −2
2 = 32 + C Simplify
- 30 = C Subtract 32 from both sides

Substitute C = −30 into the general solution to the differential equation, and then solve for y to find the particular solution to the differential equation.

y 2 2 = 8 x 2 + C General solution to differential equation
y 2 2 = 8 x 2 - 30 Substitute C = −30
y 2 = 16 x 2 - 60 Multiply both sides by 2
y = ± 16 x 2 - 60 Take square root of both sides

The equation  y = ± 16 x 2 - 60  represents two different solutions,  y = - 16 x 2 - 60  and  y = 16 x 2 - 60 . Use the given initial condition to determine which case is the particular solution to the differential equation.

The y-value of the initial condition y(2) = −2 is negative, so the particular solution to the differential equation  dy dx = 16 x y  is the negative case y = - 16 x 2 - 60 . Eliminate Choices A and C.

There are only two possible domains given in the answer choices,  x ≥ 15 2  and  x ≤ - 15 2 . Use the x-value of the initial condition (x = 2) to see that the domain must be  x ≥ 15 2  because  2 ≥ 15 2 , so it is not necessary to derive the domain.

Therefore, the solution to the differential equation is  y = - 16 x 2 - 60  and its domain is  x ≥ 15 2 .

(Choice A) y = 16 x 2 - 60  for  x ≥ 15 2  may result from mistakenly using the x-value of the initial condition to choose the positive solution to the differential equation instead of the value of f(2).

(Choice C) y = 16 x 2 - 60  for  x ≤ - 15 2  may result from the combination of errors described in Choices A and D.

(Choice D) y = - 16 x 2 - 60  for  x ≤ - 15 2  may result from mistakenly using the negative value of f(2), instead of the positive x-value, to determine the domain from the given condition.

Things to remember:
To determine a particular solution to a separable differential equation, separate the variables, integrate both sides of the equation, and use the given initial condition to find the expression for f(x) that includes the constant of integration.

Question

If  dy dx = 3 - y , and if y = 2 when x = 4, then y =

A. 3 + e−x
B. 3 − e−x
C. 3 − e4 − x
D. 3 − ex − 4

Hint :

To find a particular solution to a separable differential equation, separate the variables and integrate each side of the equation independently.

Explanation

To determine a particular solution to a separable differential equation, first separate the variables, integrate both sides of the equation, then add the constant of integration.

Multiply the differential equation by dx and divide by 3 − y to separate the variables.

Integrate both sides of the resulting equation independently. The left side is a composite function of the form  1 u , where u = 3 − y, so use u-substitution and the natural logarithm rule. Apply the power rule to the right side.

∫ dy 3 - y = ∫ dx Integral of both sides
- ∫ du u = ∫ dx Substitute 3 − y = u and dy = −du
- ln u = x + C Integrate
- ln 3 - y = x + C Substitute u = 3 − y

Plug in the given initial condition of x = 4 and y = 2 and solve for C to find the particular value of C.

−ln|3 − y| = x + C General solution to differential equation
−ln|3 − 2| = 4 + C Plug in x = 4 and y = 2
−ln|1| = 4 + C Subtract
0 = 4 + C Simplify ln1 = 0
−4 = C Subtract 4 from both sides

Substitute C = −4 into the general solution to the differential equation and solve for y.

−ln|3 − y| = x + C General solution to differential equation
−ln|3 − y| = x + (−4) Substitute C = −4
ln|3 − y| = 4 − x Multiply both sides by −1
|3 − y| = e4 − x Apply logarithmic exponential relationship
3 − y = ±e4 − x Remove absolute value bars
3 ± e4 − x = y Solve for y

The expression e4 − x is positive for all values of x and the given value of y = 2 is less than 3, so the exponential term must be negative.

Therefore, the solution to the given differential equation if y = 2 when x = 4 is y = 3 − e4 − x.

(Choices A and B) 3 + e−x and 3 − e−x may result from not adding a constant of integration when integrating both sides of the differential equation.

(Choice D) 3 − ex − 4 may result from not using u-substitution to integrate the y-term, leading to the incorrect integral  ∫ dy 3 - y = ln 3 - y + C .

Things to remember:
To determine a particular solution to a separable differential equation, separate the variables, integrate both sides of the equation, and use the given initial condition to find the constant of integration.

It is also possible to find the solution to the given differential equation by using the given information to determine which choice satisfies the initial condition and differential equation.

First, plug the given value of x = 4 into each choice to verify which choices yield the given value of y = 2.

Eliminate Choices A and B because they do not yield values of y = 2 when x = 4.

Substitute each of the remaining choices in for y in the given differential equation to determine which results in a true statement.

Eliminate Choice D because it does not result in a true statement.

Therefore, the solution to the given differential equation if y = 2 when x = 4 is y = 3 − e4 − x.

Things to remember:
To find the particular solution to a differential equation given a finite number of choices, it may be possible to test each choice with given information to determine which satisfies the initial condition and differential equation.

Question

Rain is falling at the rate of r t = 3 e - 0.2 t inches per hour, where t is the time in hours since the rain began to fall. Which of the following gives the amount of rain, in inches, that falls from time t = 0 to time t = 6 hours?
A. 3 e - 1.2 - 3
B. 0.6 - 0.6 e - 1.2
C. 3 - 3 e - 1.2
D. 15 - 15 e - 1.2

Hint :

The definite integral of a rate of change of a quantity over an interval gives the amount of change in the quantity over the interval.

Explanation

The definite integral of a rate of change r(t) of a quantity over an interval of time from t1 to t2 gives the amount of change in the quantity over the interval.

∫ t 1 t 2 r t   d t = amount of change (net change) in quantity from t1 to t2

Since r t = 3 e - 0.2 t is the rate of change of the rainfall, the amount of rain that falls between times t1=0 and t2=6 is given by

∫ 0 6 3 e - 0.2 t d t

The integrand above is a composite function of the form e u where u=−0.2t is the inside function, so use u -substitution to integrate.

First differentiate u=−0.2t with respect to t to find dt in terms of u .

Definite integral expression derived from a rate of change problem, ready for evaluation using the Fundamental Theorem of Calculus

Calculate the limits of integration in terms of u and then substitute u = - 0.2 t , d u - 0.2 = d t , and the new limits of integration into the integral.

Definite integral expression derived from a rate of change problem, ready for evaluation using the Fundamental Theorem of Calculus

Now use the Fundamental Theorem of Calculus (FTC) to evaluate the resulting integral.

Definite integral expression derived from a rate of change problem, ready for evaluation using the Fundamental Theorem of Calculus

Therefore, the amount of rain in inches that falls from time t=0 to time t=6 hours is given by 15 - 15 e - 1.2

(Choice A) 3 e - 1.2 - 3 may be a result of neglecting to integrate the function or incorrectly integrating before applying the Fundamental Theorem of Calculus. This value is negative, which is not a valid option for the context of the question.

(Choice A) 0.6 - 0.6 e - 1.2 may be a result of differentiating rather than integrating the function before applying the Fundamental Theorem of Calculus.

(Choice C) 3 - 3 e - 1.2 may be a result of neglecting to integrate the function or incorrectly integrating before applying the Fundamental Theorem of Calculus and then incorrectly applying the Fundamental Theorem of Calculus: ∫ a b f ′ x d x ≠ f a - f b

Things to remember:
The definite integral of a rate of change of a quantity over an interval gives the amount of change in the quantity over the interval.

Question

The base of a solid is the region in the first quadrant bounded by the graph of y = ex, the vertical line x = 2, and the x- and y-axes. For this solid, each cross section perpendicular to the x-axis is a square. What is the volume of the solid?

A. e 2 - 1
B. 1 2 e 4 - 1 2
C. e 4 - 1
D. 2 e 4 - 2

Hint :

The volume of a solid with cross sections of area A(x) perpendicular to the x-axis is given by ∫ x 1 x 2 A x dx

Explanation

The volume of a solid with cross sections of area A(x) perpendicular to the x-axis is given by a definite integral of A(x) in terms of x.

Diagram of a solid with square cross sections perpendicular to the x-axis, showing base region in the xy-plane bounded by y = e^x, x = 2, and the axes

Sketch the given boundaries of the region defined by the function y = ex, the line x = 2, and the x- and y-axes that form the base of the solid and determine the limits of integration.

The enclosed region spans the interval 0 ≤ x ≤ 2, so the limits of integration are x1 = 0 and x2 = 2.

Shaded region in the first quadrant bounded by y = e^x from x = 0 to x = 2, the x-axis, and the line x = 2

The cross sections are squares, so A(x) = s2 is the integrand.

The side length s of each cross section is the vertical distance between the top and bottom functions, so s = top − bottom. Substitute y = ex for the top and y = 0 for the bottom and simplify.

Step-by-step derivation showing side length s = e^x − 0 = e^x, then area A(x) = (e^x)^2 = e^{2x}

To find an expression for the area function A ( x ) , plug s = e x into the formula for the area of a square to get A ( x ) = ( e x ) ² = e 2 x . Substitute A ( x ) = e 2 x , x 1 = 0 , and x 2 = 2 into the volume integral to get ∫ 0 2 e 2 x d x .

The integrand e 2 x is a composite function of the form e u , where u = 2 x , so rewrite the integral in terms of u. The derivative of u is d u d x = 2 , so solve for d x = d u 2 . Plug the limits of integration for x in u = 2 x to get 0 and 4.

Final definite integral setup: V = ∫ from 0 to 2 of e^{2x} dx, ready for evaluation

Apply the constant multiple rule to rewrite the resulting integral as  1 2 ∫ 0 4 e u du , then integrate and apply the Fundamental Theorem of Calculus (FTC).

V = 1/2 ∫04 eu du Resulting integral
V = 1/2 [ eu ]04 Integrate
V = 1/2 ( e4 − e0 ) Apply FTC
V = 1/2 e4 − 1/2 Simplify e0 = 1 and distribute 1/2

Therefore, the volume of the solid is 1 2 e 4 − 1 2 .

(Choice A) e 2 − 1 may result from not squaring the side length in the formula for the area of a square.

(Choice C) e 4 − 1 may result from not integrating with u-substitution.

(Choice D) 2 e 4 − 2 may result from finding an incorrect expression of d x ≠ 2 d u (instead of d x = d u 2 ) when applying u-substitution.

Things to remember:
The volume of a solid with cross sections of area A ( x ) perpendicular to the x-axis is given by ∫ x₁ x₂ A ( x ) d x .

Question

What is the area of the region R bounded by the graphs of x = y2 − y + 1 and x = y + 1 ?

A. 4 3
B. 8 3
C. 4
D. 20 3

Hint :

If one factor of an integrand is a constant multiple of the derivative of the inside of a composite function, use  u -substitution to integrate.

Explanation

The area between the given functions is equal to the integral of the right function minus the left function between the y-values of their intersection points, so it is necessary to determine which are the right and left functions and to find the intersection points.

The quadratic function has a positive leading coefficient, so the parabola opens to the right. All the choices are finite values, so the linear function must cross to the right of the quadratic (graph). Set the expressions equal and solve for y to find where they intersect.

y2 − y + 1 = y + 1 Set functions equal
y2 − 2y = 0 Subtract y and 1 from both sides
y(y − 2) = 0 Factor out GCF
y = 0 and y = 2 Apply zero product property

Therefore, the right function is f(y) = y + 1 and the left function is g(y) = y2 − y + 1, and they intersect at y = 0 and y = 2.

Integrate the difference between the functions on the interval of their intersection points to find the area of region R.

Definite integral expression derived from a rate of change problem, ready for evaluation using the Fundamental Theorem of Calculus

Therefore, the area of region R is  4 3 .

Make the AP Calculus AB Exam Feel Like Practice

Our AP Calculus AB practice tests and quizzes are just like the exam and make you think critically. They’ll teach you to spot trick answers, handle time pressure and boost your confidence for test day! Practice with realistic exam simulations so test day feels like just another practice run.

One Click AP Calc AB Quizzes

Generate tailored practice tests to target improvement in the calculus topics that you need to work on most. Build your knowledge and boost your critical thinking skills while saving prep time.

Practice Under Real Exam Conditions

Take timed AP Calc AB practice exams that look just like the actual BluebookTM test platform. Experience realistic exam pressure now so you stay calm and confident when it counts.

Study with Exam-Aligned Questions

Our AP Calc AB practice questions are written by in-house educators and aligned with official standards. Get detailed mathematical explanations with visuals that make complex concepts clear.

Achieve Higher AP Calc AB Exam Scores with Smart Study Tools

Exceptional Explanations

Understand the “why” with our simplified breakdowns of how to approach each question and topic. Our clear question explanations and vivid visuals help you spot and avoid trick answers so you’ll ace the AP Calculus AB exam. Our exclusive technique, backed by cognitive learning principles, maximizes learning and retention.

AP Calc AB Illustrated Explanation 1
AP Calc AB Illustrated Explanation 2
AP Calc AB Illustrated Explanation 3
Sample AP Calc AB Smart Flashcard
Sample AP Calc AB Digital Notebook

Flashcards & Notes That Help You Actually Remember

Create digital flashcards to practice the concepts you struggle with most, like how to apply the chain rule or the difference between the product rule and the quotient rule, helping you remember information months later, not just for the next unit quiz. Use My Notebook to organize key concepts from AP Calc practice questions in the format that works best for you, whether that’s bullet points for related rates problems, diagrams showing the relationship between a function and its derivative, or detailed notes on how to find the volume of revolution using different methods. One-click transfer from explanations to your flashcards or notebook means everything you need stays in one organized place.

Track Your Progress and Target Weak Spots

See exactly where you stand with detailed performance tracking across all your AP Calculus AB practice tests and practice questions. Our analytics dashboard shows which topics you're crushing and which ones need more work. Focus your study time on actual weak areas, stop wasting time on stuff you already know, and watch your scores improve.

Sample AP Calc AB Performance Dashboard 1
Sample AP Calc AB Performance Dashboard of Test Results
Sample AP Calc AB Performance Dashboard of Total Analysis
UWorld Dark Blue logo

Study on the go with our mobile-friendly practice questions. Get instant feedback on every AP Calc AB practice question (MCQ) and test your speed with built-in timers. Whether at home or between classes, you’ll always be ready to review and improve.

App Store
Google Play

Pick Your AP Calc AB Prep Package

From targeted AP Calculus AB practice questions to a complete prep course with video lessons, choose what fits your study style and budget
 
Best Value
Everything you need to pass the AP Calc AB exam

AP Calc AB Practice Tests

Starting at $39

Buy Now

AP Calc AB Review Course

Starting at $99

Buy Now
1400+ AP Calculus AB Practice Questions (MCQs)
Practice with exam-style multiple-choice questions that match what you'll see on test day. Every question includes detailed explanations with visuals.
Custom AP Calc AB Practice Tests & Quizzes
Build practice tests focused on YOUR weak topics. Create timed mock exams or quick topic quizzes. Study your way, not someone else's way.
Progress Tracking Dashboard
See exactly which topics you're nailing and which need more work. Track your progress across all AP Calculus AB practice tests and quizzes so you know where to focus.
Pick Your Study Topics
Stop wasting time on content you already know. Choose specific calculus units to practice and build a study plan that actually makes sense for you.
Smart Study Planner
Tell us your schedule and we'll create a personalized study plan that fits your life. Stay on track without the stress of figuring out what to study next.
Realistic FRQ Practice with Scoring Guide*
Practice free-response questions with the same scoring guides AP graders use. Learn exactly what earns points so you know how to write better answers.
Score-Predicting Full-Length AP Calculus AB Mock Exam*
Take a complete practice exam under real testing conditions and get an estimated AP score. See where you stand before test day.
Print and Digital Study Guide
Focused content review that works alongside video lessons and practice questions. Everything you need in one organized place.
150+ Check-for-Understanding Questions
Make sure you understand the basics before jumping into harder AP-level practice questions. Build a solid foundation first.
Expert-led Video Lessons
Former AP Calc AB teachers break down tough concepts with clear animations and step-by-step explanations. Finally understand the stuff that confuses you.

Pick Your AP Calc AB Prep Package

From targeted AP Calculus AB practice questions to a complete prep course with video lessons, choose what fits your study style and budget
Everything you need to pass the AP Calc AB exam

AP Calc AB Practice Tests

Starting at $39

Buy Now
1400+ AP Calculus AB Practice Questions (MCQs)
Practice with exam-style multiple-choice questions that match what you'll see on test day. Every question includes detailed explanations with visuals.
Build practice tests focused on YOUR weak topics. Create timed mock exams or quick topic quizzes. Study your way, not someone else's way.
See exactly which topics you're nailing and which need more work. Track your progress across all AP Calculus AB practice tests and quizzes so you know where to focus.
Stop wasting time on content you already know. Choose specific calculus units to practice and build a study plan that actually makes sense for you.
Tell us your schedule and we'll create a personalized study plan that fits your life. Stay on track without the stress of figuring out what to study next.

Best Value!

AP Calc AB Review Course

Starting at $99

Buy Now
See What’s Included
1400+ AP Calculus AB Practice Questions (MCQs)
Practice with exam-style multiple-choice questions that match what you'll see on test day. Every question includes detailed explanations with visuals.
Build practice tests focused on YOUR weak topics. Create timed mock exams or quick topic quizzes. Study your way, not someone else's way.
See exactly which topics you're nailing and which need more work. Track your progress across all AP Calculus AB practice tests and quizzes so you know where to focus.
Stop wasting time on content you already know. Choose specific calculus units to practice and build a study plan that actually makes sense for you.
Tell us your schedule and we'll create a personalized study plan that fits your life. Stay on track without the stress of figuring out what to study next.
Practice free-response questions with the same scoring guides AP graders use. Learn exactly what earns points so you know how to write better answers.
Take a complete practice exam under real testing conditions and get an estimated AP score. See where you stand before test day.
Focused content review that works alongside video lessons and practice questions. Everything you need in one organized place.
Make sure you understand the basics before jumping into harder AP-level practice questions. Build a solid foundation first.
Former AP Calc AB teachers break down tough concepts with clear animations and step-by-step explanations. Finally understand the stuff that confuses you.

* * Coming Winter 2025/Spring 2026

AP Calculus AB Practice Test Reviews

Here’s what our students have to say after using our practice tests to study for their AP Calc AB exam:

UWorlds multiple choice questions are similar to the ones on the official AP exam and allowed me to time myself for each question. This was very helpful for me as I was able to answer questions faster and could finish the questions on the actual exam. The explanations for each question went in-depth and gave important details pertaining to events in the timeline. Through this, I was able to gain important skills for the exam and get a 5.

See More
—Sanjana S.

Before, I had a hard time studying and staying focused because it was just boring, but now with UWorld, not only can I focus, but I actually feel motivated to learn!

—Arva P.

The explanations were clear and I could practice the question based on units. I got a 5 in the end!! So, I think it’s very helpful and I’ll be using it to study for my future exams 🙂 You guys provide so many different functions to help students like me, and I really appreciate it, it’s really worth the money.

See More
—Sophie Z.
Read More Reviews

AP Calc AB Practice Tests: Frequently Asked Questions (FAQs)

Our AP Calculus AB practice tests are meticulously designed to replicate the format, difficulty, and content of the real exam. Written by in-house experts, our multiple-choice and free-response questions align with what you’ll encounter on test day including the specific problem types, multi-step reasoning, and justification skills you’ll need to master. Every AP Calculus AB practice problem features a detailed, step-by-step solution so you understand the logic, methodology and mathematics behind every answer, not just the final number.

There are free AP Calc AB practice test resources available to help you study but they tend to focus on teaching the general concepts with limited solution and explanation sets. Our AP Calculus AB practice questions are authored by experienced educators and closely reflect the current exam.

With UWorld you get a comprehensive AP Calculus AB practice exam system that includes:

  • Exam-level AP Calc AB questions with detailed, step-by-step solutions.
  • Customizable quizzes to target specific units, like derivatives or integration.
  • Performance tracking to pinpoint your conceptual weak spots.
  • Timed modes to simulate the pressure of the AP Calculus AB practice exam.

You’re not just practicing; you’re getting a complete platform designed to help you master the material and improve your score.

Yes. Our AP Calculus AB question bank provides comprehensive coverage of all 8 units outlined in the official College Board course and exam description:

  • Unit 1: Limits and Continuity
  • Unit 2: Differentiation: Definition and Fundamental Properties
  • Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
  • Unit 4: Contextual Applications of Differentiation
  • Unit 5: Analytical Applications of Differentiation
  • Unit 6: Integration and Accumulation of Change
  • Unit 7: Differential Equations
  • Unit 8: Applications of Integration

You can create AP Calc AB practice tests from all topics or build focused quizzes on specific units where you need the most review.

Start by using topic-specific quizzes to solidify your understanding as you learn new units in class. Before the next unit, create and take at least one AP Calculus AB practice test to then review the step-by-step solutions for every single problem, even the ones you got right. This helps you master the underlying concepts, learn efficient problem-solving methods, and recognize common pitfalls. Use your performance dashboard to identify your weakest units, then create additional AP Calc AB practice tests and quizzes to drill those specific areas.

Ideally, you should start using AP Calculus AB practice questions as soon as you begin learning the material in class. Practicing topics and solution techniques like the chain rule or the fundamental theorem of calculus while they are fresh helps reinforce your learning and identifies misunderstandings early. If you’re starting later in the year, don’t worry. Focus on high-yield topics and use our customizable AP Calc AB practice exam generator to target your weakest areas. Consistent, focused practice even 4–6 weeks before the exam can still lead to significant score improvements.

Absolutely. Our platform allows you to create customizable AP Calc AB practice tests tailored to your exact study needs. You can focus on broad units, like “Applications of Integration,” or drill down into specific sub-topics. You can set the number of questions to fit your study schedule, allowing you to efficiently target the areas where you need the most practice. After every AP Calc AB practice exam or quiz you can review the integrated performance tracker to track your progress on specific concepts over time.

While there’s no single magic number, a good benchmark is consistently scoring 70% or higher on UWorld’s full-length AP Calculus AB practice tests (across both MCQ and FRQ sections). This is a strong indicator that you are on track for a 4 or 5. If you’re not at 70% yet, don’t panic. Use the performance dashboard data to see exactly which units (e.g., “Differential Equations”) are holding you back, then create targeted quizzes to improve in those areas.

Yes. UWorld’s AP Calculus AB preparation includes hundreds of multiple-choice questions and a robust set of free-response questions (FRQs). You can practice realistic AP Calculus AB exam questions that model the format and style of the real FRQs and our detailed solutions and scoring guidelines teach you how to structure your answers and show your work to earn maximum points, just as an AP grader would expect.

We recommend that every student take at least one full-length AP Calculus AB mock exam before test day. Practicing under realistic timed conditions is the best way to build stamina, manage your pacing across the different sections and reduce test-day anxiety. UWorld’s AP Calculus AB mock exams are built directly from our deep AB calculus question bank and will give you the realistic interface and style of questions you’ll see on exam day. Furthermore, coming in Spring 2026, we will also offer a pre-built, full-length AP Calculus AB mock exam as part of our AP Calculus AB Online Review Course.

You can try free AP Calculus AB practice questions by signing up for UWorld’s free 7-day trial. This will give you access to a sample of our high-quality MCQs and FRQs, complete with the detailed, step-by-step solutions that set our product apart. The free trial also provides a preview of our complete AP Calc AB Course, which includes video lessons and digital study guides to help you master the content.

Scroll to Top
AP Calculus AB Practice Tests
Select Your Plan
Fast Track
30 Day Access
$39
Buy
Power Boost
90 Day Access
$49
Buy
Recommended
Competitive Edge
180 Day Access
$59
Buy
Peak Performance
360 Day Access
$69
Buy
AP Calculus AB Courses
Select Your Plan
Introductory Plan
30 Day Access
(No print book, only online)
$99
Buy
Score Booster Plan
90 Day Access
(No print book, only online)
$129
Buy
Most Popular
Deep Learning Plan
180 Day Access
$149
Buy
Dream Score Plan
360 Day Access
$169
Buy