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AP Calculus AB Quiz

AP Calculus AB Quiz: Average Value Of Functions On Intervals

Practice Average Value Of Functions On Intervals in AP Calculus AB with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

Question 1 / 20

0 of 20 answered

For f(x)=12xf(x)=\dfrac{1}{2}xf(x)=21​x on [4,8][4,8][4,8], what is the average value of fff on the interval?

Select an answer to continue

What this quiz covers

This quiz focuses on Average Value Of Functions On Intervals, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Calculus AB.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

For f(x)=12xf(x)=\dfrac{1}{2}xf(x)=21​x on [4,8][4,8][4,8], what is the average value of fff on the interval?

  1. 14∫48f(x) dx\dfrac{1}{4}\int_4^8 f(x)\,dx41​∫48​f(x)dx
  2. f(6)f(6)f(6)
  3. ∫48f(x) dx\int_4^8 f(x)\,dx∫48​f(x)dx
  4. f(4)+f(8)2\dfrac{f(4)+f(8)}{2}2f(4)+f(8)​
  5. 333 (correct answer)

Explanation: This problem requires finding the average value of f(x) = x/2 on [4,8]. The average value is (1/4)∫[4 to 8] (x/2)dx = (1/4)[(x²/4)] from 4 to 8 = (1/4)(16 - 4) = 3. Choice A shows the correct formula setup but doesn't evaluate it. Choice B represents the midpoint value, which equals the average for this linear function but isn't the general approach. The correct answer is the numerical result E = 3.

Question 2

A tank’s volume is V(t)=100+5t−t2V(t)=100+5t- t^2V(t)=100+5t−t2 for 0≤t≤50\le t\le 50≤t≤5. What is the average value of VVV?

  1. V(2.5)V(2.5)V(2.5)
  2. 15∫05V(t) dt\dfrac{1}{5}\int_0^5 V(t)\,dt51​∫05​V(t)dt
  3. ∫05V(t) dt\int_0^5 V(t)\,dt∫05​V(t)dt
  4. V(0)+V(5)2\dfrac{V(0)+V(5)}{2}2V(0)+V(5)​
  5. 3356\dfrac{335}{6}6335​ (correct answer)

Explanation: This question asks for the average value of the volume function V(t). The average value of V(t) = 100 + 5t - t² on [0,5] is (1/5)∫[0 to 5] (100 + 5t - t²)dt = (1/5)[100t + (5t²)/2 - t³/3] from 0 to 5 = (1/5)(500 + 62.5 - 125/3) = (1/5)(1687.5/3) = 335/6. Choice B shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 335/6.

Question 3

A river’s discharge is R(t)=10+sin⁡tR(t)=10+\sin tR(t)=10+sint for 0≤t≤2π0\le t\le 2\pi0≤t≤2π. What is the average discharge?

  1. R(π)R(\pi)R(π)
  2. ∫02πR(t) dt\int_0^{2\pi} R(t)\,dt∫02π​R(t)dt
  3. R(0)+R(2π)2\dfrac{R(0)+R(2\pi)}{2}2R(0)+R(2π)​
  4. 12π∫02πR(t) dt\dfrac{1}{2\pi}\int_0^{2\pi} R(t)\,dt2π1​∫02π​R(t)dt
  5. 101010 (correct answer)

Explanation: This question asks for the average value of the discharge function R(t). The average value of R(t) = 10 + sin(t) on [0,2π] is (1/2π)∫[0 to 2π] (10 + sin(t))dt = (1/2π)[10t - cos(t)] from 0 to 2π = (1/2π)(20π + 0) = 10. Since sin(t) integrates to zero over a complete period, only the constant term 10 contributes to the average. Choice D shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 10.

Question 4

If g(x)=x2g(x)=x^2g(x)=x2 on [0,3][0,3][0,3], what is the average value of ggg on [0,3][0,3][0,3]?

  1. g(1.5)g(1.5)g(1.5)
  2. ∫03g(x) dx\int_0^3 g(x)\,dx∫03​g(x)dx
  3. 13∫03g(x) dx\dfrac{1}{3}\int_0^3 g(x)\,dx31​∫03​g(x)dx
  4. g(0)+g(3)2\dfrac{g(0)+g(3)}{2}2g(0)+g(3)​
  5. 333 (correct answer)

Explanation: This problem requires finding the average value of g(x) = x² on [0,3]. The average value formula gives (1/3)∫[0 to 3] x²dx = (1/3)[x³/3] from 0 to 3 = (1/3)(9) = 3. Choice C shows the correct formula setup but doesn't evaluate it. Choice A represents the midpoint value, which is not generally equal to the average value for nonlinear functions. The correct answer is the numerical result E = 3.

Question 5

A population model is P(t)=1000(1+t)P(t)=1000(1+t)P(t)=1000(1+t) for 0≤t≤20 \le t \le 20≤t≤2. What is the average population on [0,2][0,2][0,2]?

  1. 12∫02P(t) dt\dfrac{1}{2}\int_0^2 P(t)\,dt21​∫02​P(t)dt
  2. P(1)P(1)P(1)
  3. ∫02P(t) dt\int_0^2 P(t)\,dt∫02​P(t)dt
  4. P(0)+P(2)2\dfrac{P(0)+P(2)}{2}2P(0)+P(2)​
  5. 200020002000 (correct answer)

Explanation: This question asks for the average value of the population function P(t). The average value of P(t)=1000(1+t)P(t) = 1000(1 + t)P(t)=1000(1+t) on [0,2] is 12∫021000(1+t) dt=12[1000t+500t2]02=12(2000+2000)=2000\frac{1}{2} \int_0^2 1000(1 + t) \, dt = \frac{1}{2} [1000t + 500t^2]_0^2 = \frac{1}{2}(2000 + 2000) = 200021​∫02​1000(1+t)dt=21​[1000t+500t2]02​=21​(2000+2000)=2000. Choice A shows the correct formula setup but doesn't evaluate it. Choice B represents the midpoint value, which equals the average for this linear function but isn't the general approach. The correct answer is the numerical result E = 2000.

Question 6

For p(x)=11+x2p(x)=\dfrac{1}{1+x^2}p(x)=1+x21​ on [0,1][0,1][0,1], which expression gives the average value of ppp?

  1. ∫01p(x) dx\int_0^1 p(x)\,dx∫01​p(x)dx
  2. p ⁣(12)p\!\left(\dfrac{1}{2}\right)p(21​)
  3. p(0)+p(1)2\dfrac{p(0)+p(1)}{2}2p(0)+p(1)​
  4. 11−0∫01p(x) dx\dfrac{1}{1-0}\int_0^1 p(x)\,dx1−01​∫01​p(x)dx (correct answer)
  5. 12∫01p(x) dx\dfrac{1}{2}\int_0^1 p(x)\,dx21​∫01​p(x)dx

Explanation: This problem asks for the correct expression representing the average value of p(x) on [0,1]. The average value of any function f(x) on interval [a,b] is given by the formula (1/(b-a))∫[a to b] f(x)dx. For the interval [0,1], this becomes (1/(1-0))∫[0 to 1] p(x)dx = ∫[0 to 1] p(x)dx. Choice E incorrectly includes an extra factor of 1/2. The correct expression is choice D, which properly applies the average value formula.

Question 7

For f(x)=3x+1f(x)=3x+1f(x)=3x+1 on [2,6][2,6][2,6], what is the average value of fff on the interval?

  1. f(4)f(4)f(4)
  2. 14∫26f(x) dx\dfrac{1}{4}\int_2^6 f(x)\,dx41​∫26​f(x)dx
  3. ∫26f(x) dx\int_2^6 f(x)\,dx∫26​f(x)dx
  4. f(2)+f(6)2\dfrac{f(2)+f(6)}{2}2f(2)+f(6)​
  5. 131313 (correct answer)

Explanation: This problem requires finding the average value of the linear function f(x). The average value of f(x) = 3x + 1 on [2,6] is (1/4)∫[2 to 6] (3x + 1)dx = (1/4)[3x²/2 + x] from 2 to 6 = (1/4)[(54 + 6) - (6 + 2)] = (1/4)(52) = 13. Choice B shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 13.

Question 8

A particle’s position is s(t)=t2+2ts(t)=t^2+2ts(t)=t2+2t for 0≤t≤20\le t\le 20≤t≤2. What is the average value of sss on [0,2][0,2][0,2]?

  1. 12∫02s(t) dt\dfrac{1}{2}\int_0^2 s(t)\,dt21​∫02​s(t)dt
  2. s(1)s(1)s(1)
  3. ∫02s(t) dt\int_0^2 s(t)\,dt∫02​s(t)dt
  4. s(0)+s(2)2\dfrac{s(0)+s(2)}{2}2s(0)+s(2)​
  5. 103\dfrac{10}{3}310​ (correct answer)

Explanation: This question requires finding the average value of the position function s(t). The average value of s(t) = t² + 2t on [0,2] is (1/2)∫[0 to 2] (t² + 2t)dt = (1/2)[t³/3 + t²] from 0 to 2 = (1/2)(8/3 + 4) = (1/2)(20/3) = 10/3. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 10/3.

Question 9

A machine’s cost rate is c(t)=12+3tc(t)=12+3tc(t)=12+3t dollars/hour for 2≤t≤62\le t\le 62≤t≤6. What is the average cost rate?

  1. 14∫26c(t) dt\dfrac{1}{4}\int_2^6 c(t)\,dt41​∫26​c(t)dt
  2. c(4)c(4)c(4)
  3. ∫26c(t) dt\int_2^6 c(t)\,dt∫26​c(t)dt
  4. c(2)+c(6)2\dfrac{c(2)+c(6)}{2}2c(2)+c(6)​
  5. 242424 (correct answer)

Explanation: This question asks for the average value of the cost rate function c(t)c(t)c(t). The average value of c(t)=12+3tc(t) = 12 + 3tc(t)=12+3t on [2,6][2,6][2,6] is 14∫26(12+3t) dt\frac{1}{4} \int_2^6 (12 + 3t) \, dt41​∫26​(12+3t)dt = 14[12t+3t22]∣26\frac{1}{4} [12t + \frac{3t^2}{2}] \big|_{2}^{6}41​[12t+23t2​]​26​ = 14[(72+54)−(24+6)]\frac{1}{4} [(72 + 54) - (24 + 6)]41​[(72+54)−(24+6)] = 14(96)\frac{1}{4} (96)41​(96) = 24 dollars/hour. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 24.

Question 10

For f(x)=2xf(x)=\dfrac{2}{x}f(x)=x2​ on [1,4][1,4][1,4], which expression represents the average value of fff on [1,4][1,4][1,4]?

  1. 13∫142x dx\dfrac{1}{3}\int_1^4 \dfrac{2}{x}\,dx31​∫14​x2​dx (correct answer)
  2. ∫142x dx\int_1^4 \dfrac{2}{x}\,dx∫14​x2​dx
  3. f(2.5)f(2.5)f(2.5)
  4. f(1)+f(4)2\dfrac{f(1)+f(4)}{2}2f(1)+f(4)​
  5. 14∫142x dx\dfrac{1}{4}\int_1^4 \dfrac{2}{x}\,dx41​∫14​x2​dx

Explanation: This problem asks for the correct expression representing the average value of f(x) = 2/x on [1,4]. The average value of any function f(x) on interval [a,b] is given by the formula (1/(b-a))∫[a to b] f(x)dx. For the interval [1,4], this becomes (1/(4-1))∫[1 to 4] (2/x)dx = (1/3)∫[1 to 4] (2/x)dx. Choice E incorrectly uses 1/4 as the coefficient instead of 1/3. The correct expression is choice A, which properly applies the average value formula.

Question 11

For f(x)=x3f(x)=x^3f(x)=x3 on [−1,1][-1,1][−1,1], what is the average value of fff on the interval?

  1. 12∫−11x3 dx\dfrac{1}{2}\int_{-1}^{1} x^3\,dx21​∫−11​x3dx
  2. f(0)f(0)f(0)
  3. ∫−11x3 dx\int_{-1}^{1} x^3\,dx∫−11​x3dx
  4. f(−1)+f(1)2\dfrac{f(-1)+f(1)}{2}2f(−1)+f(1)​
  5. 000 (correct answer)

Explanation: This problem requires finding the average value of f(x) = x³ on [-1,1]. The average value is (1/2)∫[-1 to 1] x³ dx = (1/2)[x⁴/4] from -1 to 1 = (1/2)(1/4 - 1/4) = 0. Since x³ is an odd function and we're integrating over a symmetric interval, the integral equals zero. Choice A shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 0.

Question 12

A spring force is F(x)=5xF(x)=5xF(x)=5x for 0≤x≤40\le x\le 40≤x≤4. What is the average force on [0,4][0,4][0,4]?

  1. 14∫04F(x) dx\dfrac{1}{4}\int_0^4 F(x)\,dx41​∫04​F(x)dx
  2. F(2)F(2)F(2)
  3. ∫04F(x) dx\int_0^4 F(x)\,dx∫04​F(x)dx
  4. F(0)+F(4)2\dfrac{F(0)+F(4)}{2}2F(0)+F(4)​
  5. 101010 (correct answer)

Explanation: This question asks for the average value of the force function F(x). The average value of F(x) = 5x on [0,4] is (1/4)∫[0 to 4] 5x dx = (1/4)[5x²/2] from 0 to 4 = (1/4)(40) = 10. Choice A shows the correct formula setup but doesn't evaluate it. Choice B represents the midpoint value, which equals the average for this linear function but isn't the general approach. The correct answer is the numerical result E = 10.

Question 13

For f(x)=2f(x)=2f(x)=2 on [5,9][5,9][5,9], what is the average value of fff on the interval?

  1. 14∫592 dx\dfrac{1}{4}\int_5^9 2\,dx41​∫59​2dx
  2. ∫592 dx\int_5^9 2\,dx∫59​2dx
  3. f(7)f(7)f(7)
  4. f(5)+f(9)2\dfrac{f(5)+f(9)}{2}2f(5)+f(9)​
  5. 222 (correct answer)

Explanation: This problem requires finding the average value of the constant function f(x) = 2 on [5,9]. For any constant function f(x) = c, the average value over any interval equals the constant value c itself. This is because (1/(b-a))∫[a to b] c dx = (1/(b-a))·c(b-a) = c. Choice A shows the correct formula setup but doesn't evaluate it, while choice C gives the midpoint value (which happens to equal the constant). The correct answer is the numerical result E = 2.

Question 14

For f(x)=1+cos⁡xf(x)=1+\cos xf(x)=1+cosx on [0,π][0,\pi][0,π], what is the average value of fff on the interval?

  1. f(0)+f(π)2\dfrac{f(0)+f(\pi)}{2}2f(0)+f(π)​
  2. 1π∫0πf(x) dx\dfrac{1}{\pi}\int_0^{\pi} f(x)\,dxπ1​∫0π​f(x)dx
  3. f ⁣(π2)f\!\left(\dfrac{\pi}{2}\right)f(2π​)
  4. ∫0πf(x) dx\int_0^{\pi} f(x)\,dx∫0π​f(x)dx
  5. 111 (correct answer)

Explanation: This problem requires finding the average value of f(x) = 1 + cos(x) on [0,π]. The average value is (1/π)∫[0 to π] (1 + cos(x))dx = (1/π)[x + sin(x)] from 0 to π = (1/π)(π + 0) = 1. Since cos(x) integrates to zero over [0,π], only the constant term 1 contributes to the average. Choice B shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 1.

Question 15

If f(x)=x+1xf(x)=x+\dfrac{1}{x}f(x)=x+x1​ on [1,2][1,2][1,2], which expression equals the average value of fff?

  1. 12∫12f(x) dx\dfrac{1}{2}\int_1^2 f(x)\,dx21​∫12​f(x)dx
  2. 12−1∫12f(x) dx\dfrac{1}{2-1}\int_1^2 f(x)\,dx2−11​∫12​f(x)dx (correct answer)
  3. ∫12f(x) dx\int_1^2 f(x)\,dx∫12​f(x)dx
  4. f(1.5)f(1.5)f(1.5)
  5. f(1)+f(2)2\dfrac{f(1)+f(2)}{2}2f(1)+f(2)​

Explanation: This problem asks for the correct expression representing the average value of f(x)f(x)f(x) on [1,2][1,2][1,2]. The average value of any function f(x)f(x)f(x) on interval [a,b][a,b][a,b] is given by the formula 1b−a∫abf(x) dx\frac{1}{b-a} \int_a^b f(x)\, dxb−a1​∫ab​f(x)dx. For the interval [1,2][1,2][1,2], this becomes 12−1∫12f(x) dx\frac{1}{2-1} \int_1^2 f(x)\, dx2−11​∫12​f(x)dx = 11∫12f(x) dx\frac{1}{1} \int_1^2 f(x)\, dx11​∫12​f(x)dx. Choice A incorrectly includes an extra factor of 1/2 in the denominator. The correct expression is choice B, which properly applies the average value formula with (2−1)=1(2-1) = 1(2−1)=1 in the denominator.

Question 16

A runner’s pace is p(t)=8−t2p(t)=8-\dfrac{t}{2}p(t)=8−2t​ (min/mi) for 0≤t≤60\le t\le60≤t≤6. What is the average pace on [0,6][0,6][0,6]?

  1. ∫06p(t) dt\int_{0}^{6} p(t)\,dt∫06​p(t)dt
  2. p(3)p(3)p(3)
  3. 16∫06p(t) dt\dfrac{1}{6}\int_{0}^{6} p(t)\,dt61​∫06​p(t)dt
  4. 888
  5. 132\dfrac{13}{2}213​ (correct answer)

Explanation: This problem requires finding the average pace over a time interval. The average value of p(t) = 8 - t/2 on [0,6] is (1/6)∫[0 to 6](8 - t/2)dt. Computing: (1/6)[8t - t²/4] from 0 to 6 = (1/6)[48 - 9] = (1/6)(39) = 13/2 min/mi. Choice C shows the correct formula but unevaluated, while choice D = 8 incorrectly uses just the initial pace value. To find average values of linear functions: the average always equals the function value at the midpoint, which here is p(3) = 8 - 3/2 = 13/2.

Question 17

A car’s speed is v(t)=20+4tv(t)=20+4tv(t)=20+4t (m/s) for 0≤t≤50\le t\le50≤t≤5. What is the average speed on [0,5][0,5][0,5]?

  1. 303030 (correct answer)
  2. 15∫05v(t) dt\dfrac{1}{5}\int_{0}^{5} v(t)\,dt51​∫05​v(t)dt
  3. ∫05v(t) dt\int_{0}^{5} v(t)\,dt∫05​v(t)dt
  4. v(2.5)v(2.5)v(2.5)
  5. 252525

Explanation: This problem requires finding the average value of a function on an interval. The average value formula states that for a function f on interval [a,b], the average equals (1/(b-a))∫[a to b] f(t)dt. For v(t) = 20 + 4t on [0,5], we calculate (1/5)∫[0 to 5](20 + 4t)dt = (1/5)[20t + 2t²] from 0 to 5 = (1/5)[100 + 50] = 30 m/s. Choice B shows the formula but doesn't evaluate it, which is a common incomplete answer. To find average values: identify the interval length, set up the integral of your function, evaluate the integral, then divide by the interval length.

Question 18

Water flows at rate R(t)=6−tR(t)=6-tR(t)=6−t (L/min) for 0≤t≤40\le t\le40≤t≤4. What is the average flow rate on [0,4][0,4][0,4]?

  1. R(2)R(2)R(2)
  2. ∫04R(t) dt\int_{0}^{4} R(t)\,dt∫04​R(t)dt
  3. 14∫04R(t) dt\dfrac{1}{4}\int_{0}^{4} R(t)\,dt41​∫04​R(t)dt
  4. 444 (correct answer)
  5. 666

Explanation: This problem asks for the average value of a rate function over time. The average value of R(t) = 6 - t on [0,4] is found using (1/(4-0))∫0 to 4dt. Evaluating: (1/4)[6t - t²/2] from 0 to 4 = (1/4)[24 - 8] = (1/4)(16) = 4 L/min. Choice C represents the average value formula correctly but unevaluated, while choice B gives the total volume without averaging. Remember the average value checklist: divide the integral by the interval length to get the true average, not just the total accumulation.

Question 19

A particle’s velocity is v(t)=t3v(t)=t^3v(t)=t3 (m/s) for −1≤t≤1-1\le t\le1−1≤t≤1. What is the average velocity on [−1,1][-1,1][−1,1]?

  1. 12∫−11v(t) dt\dfrac{1}{2}\int_{-1}^{1} v(t)\,dt21​∫−11​v(t)dt
  2. 000 (correct answer)
  3. v(0)v(0)v(0)
  4. ∫−11v(t) dt\int_{-1}^{1} v(t)\,dt∫−11​v(t)dt
  5. 111

Explanation: This problem asks for average velocity of a particle with odd function velocity. For v(t) = t³ on [-1,1], the average value is (1/2)∫[-1 to 1]t³dt = (1/2)[t⁴/4] from -1 to 1 = (1/2)[(1/4) - (1/4)] = 0 m/s. Choice A shows the correct formula structure, while choice C = v(0) = 0 gives the right answer for the wrong reason. Since t³ is an odd function symmetric about the origin, its positive and negative areas cancel on [-1,1]. Average value principle for symmetric intervals: odd functions always average to zero on intervals symmetric about the origin.

Question 20

Temperature is T(t)=20+4cos⁡tT(t)=20+4\cos tT(t)=20+4cost (°C) for 0≤t≤π20\le t\le\dfrac{\pi}{2}0≤t≤2π​. What is the average temperature?

  1. 20+2220+2\sqrt{2}20+22​
  2. 20+8π20+\dfrac{8}{\pi}20+π8​ (correct answer)
  3. 20+4cos⁡ ⁣(π4)20+4\cos\!\left(\dfrac{\pi}{4}\right)20+4cos(4π​)
  4. 2π∫0π/2(20+4cos⁡t) dt\dfrac{2}{\pi}\int_0^{\pi/2}(20+4\cos t)\,dtπ2​∫0π/2​(20+4cost)dt
  5. 10π+410\pi+410π+4

Explanation: This problem tests finding the average value of a temperature function involving cosine, requiring careful integration of trigonometric functions. The average temperature equals (1/(π/2))∫[0 to π/2](20+4cos t)dt = (2/π)∫[0 to π/2](20+4cos t)dt. Computing: (2/π)[20t+4sin t] from 0 to π/2 = (2/π)[(10π+4·1)-(0+0)] = (2/π)(10π+4) = 20+8/π. Students might select C by evaluating T(π/4), mistakenly believing the midpoint value equals the average. To find average values with trig functions: (1) carefully set up with correct interval length, (2) remember derivative/integral pairs for sine and cosine, (3) evaluate trig functions at special angles.