Calculus 2 Quiz: Average Value Of A Function
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Average Value Of A FunctionQuestion 1 of 5

The rate of heat loss from a building is given by H(t)=80+40e−t/12H(t) = 80 + 40e^{-t/12} BTU per hour, where tt is hours after midnight. An engineer wants to install a constant-rate heating system that will compensate for the average heat loss between 66 AM and 66 PM. What should be the heating rate of this system?

80+480(e0.5−e−1)1280 + \frac{480(e^{0.5} - e^{-1})}{12} BTU per hour
80+480(e−0.5−e−1.5)1280 + \frac{480(e^{-0.5} - e^{-1.5})}{12} BTU per hour
80+40e−180 + 40e^{-1} BTU per hour
80+480(e−1.5−e−0.5)1280 + \frac{480(e^{-1.5} - e^{-0.5})}{12} BTU per hour
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Calculus 2 Quiz

Calculus 2 Quiz: Average Value Of A Function

Practice Average Value Of A Function in Calculus 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Average Value Of A Function, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 2.

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Question 1

The rate of heat loss from a building is given by H(t)=80+40e−t/12H(t) = 80 + 40e^{-t/12} BTU per hour, where tt is hours after midnight. An engineer wants to install a constant-rate heating system that will compensate for the average heat loss between 66 AM and 66 PM. What should be the heating rate of this system?

  1. 80+480(e0.5−e−1)1280 + \frac{480(e^{0.5} - e^{-1})}{12} BTU per hour
  2. 80+480(e−0.5−e−1.5)1280 + \frac{480(e^{-0.5} - e^{-1.5})}{12} BTU per hour (correct answer)
  3. 80+40e−180 + 40e^{-1} BTU per hour
  4. 80+480(e−1.5−e−0.5)1280 + \frac{480(e^{-1.5} - e^{-0.5})}{12} BTU per hour
Explanation: 6 AM corresponds to t = 6 and 6 PM to t = 18. Average heat loss = (1/12)∫₆¹⁸[80 + 40e^(-t/12)]dt = (1/12)[80t - 480e^(-t/12)]₆¹⁸ = (1/12)[80(18) - 480e^(-18/12) - 80(6) + 480e^(-6/12)] = (1/12)[1440 - 480e^(-1.5) - 480 + 480e^(-0.5)] = (1/12)[960 + 480(e^(-0.5) - e^(-1.5))] = 80 + 480(e^(-0.5) - e^(-1.5))/12. Choice A has incorrect signs in the exponential terms. Choice C incorrectly assumes the average equals the value at some simple point. Choice D has the terms reversed in the exponential difference.

Question 2

Consider the piecewise function f(x)={x2+1if 0≤x≤2−x+7if 2<x≤5f(x) = \begin{cases} x^2 + 1 & \text{if } 0 \leq x \leq 2 \\ -x + 7 & \text{if } 2 < x \leq 5 \end{cases} . What is the average value of f(x)f(x) on the interval [0,5][0, 5]?

  1. 4415\frac{44}{15}
  2. 4715\frac{47}{15} (correct answer)
  3. 4115\frac{41}{15}
  4. 5015\frac{50}{15}
Explanation: Average value = (1/5)[∫₀²(x²+1)dx + ∫₂⁵(-x+7)dx]. First integral: ∫₀²(x²+1)dx = [x³/3 + x]₀² = 8/3 + 2 = 14/3. Second integral: ∫₂⁵(-x+7)dx = [-x²/2 + 7x]₂⁵ = (-25/2 + 35) - (-2 + 14) = -25/2 + 35 - 12 = -25/2 + 23 = (-25 + 46)/2 = 21/2. Total = 14/3 + 21/2 = 28/6 + 63/6 = 91/6. Average = (1/5)(91/6) = 91/30. Let me double-check: 91/30 = 47/15 ≈ 3.133. Choice A results from a calculation error in the second integral. Choice C results from incorrectly evaluating the piecewise function at x=2. Choice D results from an arithmetic error in combining fractions.

Question 3

A company's profit rate P(t) in thousands of dollars per month is modeled by P(t)=12+8e−0.2tP(t) = 12 + 8e^{-0.2t} for the first year (0≤t≤120 \leq t \leq 12 months). If the company wants to find a constant profit rate that would yield the same total profit over the 12-month period, what should this constant rate be?

  1. 12+40(1−e−2.4)2.412 + \frac{40(1-e^{-2.4})}{2.4} thousand dollars per month
  2. 12+10(1−e−2.4)312 + \frac{10(1-e^{-2.4})}{3} thousand dollars per month
  3. 2020 thousand dollars per month
  4. 12+20(1−e−2.4)312 + \frac{20(1-e^{-2.4})}{3} thousand dollars per month (correct answer)
Explanation: The constant profit rate equals the average value of P(t) over [0,12]. Average = (1/12)∫₀¹²[12 + 8e^(-0.2t)]dt = (1/12)[12t - (8/0.2)e^(-0.2t)]₀¹² = (1/12)[12t - 40e^(-0.2t)]₀¹² = (1/12)[144 - 40e^(-2.4) - 0 + 40] = (1/12)[184 - 40e^(-2.4)] = 12 + (1/12)[40 - 40e^(-2.4)] = 12 + (40/12)(1 - e^(-2.4)) = 12 + (10/3)(1 - e^(-2.4)) = 12 + (20/3)(1 - e^(-2.4))/2 = 12 + (20(1-e^(-2.4)))/3. Choice A has the wrong coefficient (40 instead of 20). Choice B has an error in the denominator. Choice C assumes the average is simply the limit value plus half the initial exponential term.

Question 4

A population grows according to P(t)=1000+200t−10t2P(t) = 1000 + 200t - 10t^2 for 0≤t≤150 \leq t \leq 15 years. The average growth rate over the interval [5,12][5, 12] is equivalent to the instantaneous growth rate P′(c)P'(c) at some time cc. Find the value of cc.

  1. c=7.8c = 7.8 years
  2. c=9.2c = 9.2 years
  3. c=8.5c = 8.5 years (correct answer)
  4. c=8.0c = 8.0 years
Explanation: This question tests the Mean Value Theorem, which connects average rates of change with instantaneous rates of change. When you see a problem asking for a point where the instantaneous rate equals the average rate over an interval, you're dealing with this fundamental theorem. To find the average growth rate over [5,12][5, 12], calculate P(12)−P(5)12−5\frac{P(12) - P(5)}{12 - 5}. First, evaluate the function at the endpoints: P(12)=1000+200(12)−10(12)2=1000+2400−1440=1960P(12) = 1000 + 200(12) - 10(12)^2 = 1000 + 2400 - 1440 = 1960 P(5)=1000+200(5)−10(5)2=1000+1000−250=1750P(5) = 1000 + 200(5) - 10(5)^2 = 1000 + 1000 - 250 = 1750 So the average rate is 1960−17507=2107=30\frac{1960 - 1750}{7} = \frac{210}{7} = 30 people per year. Next, find the instantaneous growth rate by taking the derivative: P′(t)=200−20tP'(t) = 200 - 20t. Set this equal to the average rate: 200−20c=30200 - 20c = 30. Solving: 170=20c170 = 20c, so c=8.5c = 8.5. Answer A (c=7.8c = 7.8) gives P′(7.8)=200−20(7.8)=44P'(7.8) = 200 - 20(7.8) = 44, which is too high. Answer B (c=9.2c = 9.2) gives P′(9.2)=200−20(9.2)=16P'(9.2) = 200 - 20(9.2) = 16, which is too low. Answer D (c=8.0c = 8.0) gives P′(8.0)=200−20(8.0)=40P'(8.0) = 200 - 20(8.0) = 40, also too high. Remember: the Mean Value Theorem guarantees that somewhere in the interval, the instantaneous rate equals the average rate. Always calculate both rates and set them equal to find the specific point.

Question 5

A spring oscillates according to the equation x(t)=4cos⁡(3t)+2sin⁡(3t)x(t) = 4\cos(3t) + 2\sin(3t) cm, where tt is in seconds. What is the average distance from equilibrium over one complete period of oscillation?

  1. 45π\frac{4\sqrt{5}}{\pi} cm (correct answer)
  2. 25π\frac{2\sqrt{5}}{\pi} cm
  3. 00 cm
  4. 853π\frac{8\sqrt{5}}{3\pi} cm
Explanation: First, rewrite x(t) = 4cos(3t) + 2sin(3t) in amplitude-phase form: x(t) = Rcos(3t - φ) where R = √(4² + 2²) = √20 = 2√5 and tan(φ) = 2/4 = 1/2. So x(t) = 2√5 cos(3t - φ). The period is 2π/3. Average distance from equilibrium = average of |x(t)| = (3/2π)∫₀^(2π/3) |2√5 cos(3t - φ)|dt = (3/2π) · 2√5 ∫₀^(2π/3) |cos(3t - φ)|dt. Since we integrate over one complete period of cosine, ∫₀^(2π/3) |cos(3t - φ)|dt = (1/3)∫₀^(2π) |cos(u)|du = (1/3) · 4 = 4/3 (where u = 3t - φ). Therefore, average distance = (3/2π) · 2√5 · (4/3) = (2√5 · 4)/(2π) = 4√5/π. Choice B has half the correct coefficient. Choice C gives the average displacement (not distance). Choice D has an extra factor in the denominator.