AP Calculus AB Practice Quiz: Finding the Average Value of a Function on an Interval

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 7 questions to check your progress.

Question 1 of 7

What is the average value of the function f(x) = x² on the interval [0, 3]?

A)3B)9C)27D)1/3

All Questions (7)

What is the average value of the function f(x) = x² on the interval [0, 3]?

A) 3

B) 9

C) 27

D) 1/3

Correct Answer: A

The average value of a function f on an interval [a, b] is given by the formula (1/(b-a)) ∫[a,b] f(x) dx. For f(x) = x² on [0, 3], the average value is (1/(3-0)) ∫[0,3] x² dx = (1/3) [x³/3] from 0 to 3 = (1/3) * (3³/3 - 0³/3) = (1/3) * (27/3) = (1/3) * 9 = 3.

Find the average value of the function g(x) = sin(x) over the interval [0, π].

A) 0

B) 2/π

C) 2

D) π/2

Correct Answer: B

Using the average value formula, we calculate (1/(π-0)) ∫[0,π] sin(x) dx. The integral of sin(x) is -cos(x). Evaluating from 0 to π gives (1/π) * [-cos(x)] from 0 to π = (1/π) * (-cos(π) - (-cos(0))) = (1/π) * (-(-1) - (-1)) = (1/π) * (1 + 1) = 2/π.

Which of the following expressions represents the average value of the function f(x) = e²ˣ on the interval [1, 5]?

A) ∫[1,5] e²ˣ dx

B) (1/5) ∫[1,5] e²ˣ dx

C) 4 ∫[1,5] e²ˣ dx

D) (1/4) ∫[1,5] e²ˣ dx

Correct Answer: D

The formula for the average value of a function f over an interval [a, b] is (1/(b-a)) ∫[a,b] f(x) dx. In this case, a=1 and b=5, so b-a = 4. The function is f(x) = e²ˣ. Therefore, the correct expression is (1/4) ∫[1,5] e²ˣ dx.

The average value of the function f(x) = 3x² on the interval [0, k] is 9. What is the value of k, where k > 0?

A) 1

B) √3

C) 3

D) 9

Correct Answer: C

We set up the average value equation: 9 = (1/(k-0)) ∫[0,k] 3x² dx. This simplifies to 9 = (1/k) * [x³] from 0 to k. Evaluating the integral gives 9 = (1/k) * (k³ - 0³) = (1/k) * k³ = k². Solving k² = 9 for k > 0 gives k = 3.

The temperature in a room over a 24-hour period is modeled by the function T(t) = 70 + 5cos(πt/12), where t is in hours. What is the average temperature, in degrees, over the 24-hour period from t=0 to t=24?

A) 65

B) 70

C) 72.5

D) 75

Correct Answer: B

To find the average temperature, we calculate the average value of T(t) on [0, 24]: (1/24) ∫[0,24] (70 + 5cos(πt/12)) dt. We can evaluate this as (1/24) ∫[0,24] 70 dt + (1/24) ∫[0,24] 5cos(πt/12) dt. The average value of a constant C is C, so the first part is 70. The function cos(πt/12) has a period of 2π/(π/12) = 24. Since the integral is over one full period of the cosine function, the value of ∫[0,24] 5cos(πt/12) dt is 0. Thus, the average value of the cosine term is 0. The total average temperature is 70 + 0 = 70.

What is the average value of the function f(x) = |x - 2| on the interval [0, 4]?

A) 0

B) 1/2

C) 1

D) 2

Correct Answer: C

The average value is (1/(4-0)) ∫[0,4] |x - 2| dx. We must split the integral at x=2, where the expression inside the absolute value changes sign. The integral becomes (1/4) * [∫[0,2] -(x - 2) dx + ∫[2,4] (x - 2) dx]. This evaluates to (1/4) * [∫[0,2] (2 - x) dx + ∫[2,4] (x - 2) dx] = (1/4) * [[2x - x²/2] from 0 to 2 + [x²/2 - 2x] from 2 to 4] = (1/4) * [((4-2)-0) + ((8-8)-(2-4))] = (1/4) * [2 + 2] = 1.

The average value of a continuous function f(x) on the interval [2, 8] is 10. What is the value of the definite integral ∫[2,8] f(x) dx?

A) 10/6

B) 10

C) 60

D) 100

Correct Answer: C

The formula for the average value is V = (1/(b-a)) ∫[a,b] f(x) dx. We are given V=10, a=2, and b=8. Plugging these values in gives 10 = (1/(8-2)) ∫[2,8] f(x) dx. This simplifies to 10 = (1/6) ∫[2,8] f(x) dx. To solve for the integral, we multiply both sides by 6, which gives ∫[2,8] f(x) dx = 10 * 6 = 60.