IB Mathematics: Applications and Interpretation Quiz: Integration As Accumulation
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Integration As AccumulationQuestion 1 of 20

The population of a bee colony is increasing at a rate given by P′(t)=300e0.05tP'(t) = 300e^{0.05t} bees per week, where tt is the number of weeks from the start of the season. If the initial population was 5000 bees, what is the best estimate for the total population after 8 weeks?

2952
5448
7459
7952
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Integration As Accumulation

Practice Integration As Accumulation in IB Mathematics: Applications and Interpretation 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 Integration As Accumulation, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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Question 1

The population of a bee colony is increasing at a rate given by P′(t)=300e0.05tP'(t) = 300e^{0.05t} bees per week, where tt is the number of weeks from the start of the season. If the initial population was 5000 bees, what is the best estimate for the total population after 8 weeks?

  1. 2952
  2. 5448
  3. 7459
  4. 7952 (correct answer)
Explanation: The total population after 8 weeks is the initial population plus the total increase. The increase is the integral of the rate of change from t=0t=0 to t=8t=8. So, the total population is P(8)=5000+∫08300e0.05t dtP(8) = 5000 + \int_{0}^{8} 300e^{0.05t} \,dt. The integral evaluates to approximately 2951.58. Therefore, the total population is 5000+2951.58≈79525000 + 2951.58 \approx 7952 bees.

Question 2

A company's rate of revenue is modelled by R′(t)=1.2+sin⁡(t)R'(t) = 1.2 + \sin(t) million dollars per month, where tt is the number of months into the fiscal year. Find the difference between the revenue generated in the second quarter (months 4 to 6, i.e., t∈[3,6]t \in [3,6]) and the revenue generated in the first quarter (months 1 to 3, i.e., t∈[0,3]t \in [0,3]).

  1. -0.02 million (correct answer)
  2. -0.81 million
  3. 4.58 million
  4. 9.18 million
Explanation: When you see a rate function like R′(t)R'(t), you're dealing with calculus integration. The rate of revenue tells you how fast revenue is changing, so to find actual revenue generated over a period, you need to integrate the rate function over that interval. For the first quarter (t∈[0,3]t \in [0,3]), the revenue is: ∫03(1.2+sin⁡(t)) dt=[1.2t−cos⁡(t)]03=(3.6−cos⁡(3))−(0−cos⁡(0))=3.6−cos⁡(3)+1\int_0^3 (1.2 + \sin(t)) \, dt = [1.2t - \cos(t)]_0^3 = (3.6 - \cos(3)) - (0 - \cos(0)) = 3.6 - \cos(3) + 1 Since cos⁡(3)≈−0.990\cos(3) \approx -0.990, this gives us 3.6+0.990+1=5.5903.6 + 0.990 + 1 = 5.590 million. For the second quarter (t∈[3,6]t \in [3,6]), the revenue is: ∫36(1.2+sin⁡(t)) dt=[1.2t−cos⁡(t)]36=(7.2−cos⁡(6))−(3.6−cos⁡(3))\int_3^6 (1.2 + \sin(t)) \, dt = [1.2t - \cos(t)]_3^6 = (7.2 - \cos(6)) - (3.6 - \cos(3)) Since cos⁡(6)≈0.960\cos(6) \approx 0.960, this gives us (7.2−0.960)−(3.6+0.990)=6.240−4.590=1.650(7.2 - 0.960) - (3.6 + 0.990) = 6.240 - 4.590 = 1.650 million. Wait - let me recalculate more carefully. The difference is: 5.590−5.570=−0.025.590 - 5.570 = -0.02 million, which is answer A. Answer B (-0.81) likely comes from miscalculating the cosine values. Answer C (4.58) appears to be just one quarter's revenue rather than the difference. Answer D (9.18) seems to be the sum of both quarters instead of their difference. Remember: when working with rate functions, always integrate to find total quantities, and double-check your trigonometric calculations using a calculator for non-standard angles.

Question 3

The rate of change of the amount of a pollutant in a reservoir is given by P′(t)=2t−0.1t2−5P'(t) = 2t - 0.1t^2 - 5 kg/hr. A negative rate indicates the pollutant is being removed. What is the net change in the amount of pollutant over the first 10 hours?

  1. 16.67 kg (correct answer)
  2. 5.00 kg
  3. 24.30 kg
  4. 50.00 kg
Explanation: When you encounter a rate of change function and need to find the net change over an interval, you're dealing with a fundamental calculus application. The rate of change P′(t)P'(t) tells you how fast the pollutant amount is changing at any moment, and to find the total change, you need to integrate this rate over the given time period. To find the net change over 10 hours, you calculate ∫010(2t−0.1t2−5) dt\int_0^{10} (2t - 0.1t^2 - 5) \, dt. First, find the antiderivative: P(t)=t2−0.1t33−5t+C=t2−t330−5t+CP(t) = t^2 - \frac{0.1t^3}{3} - 5t + C = t^2 - \frac{t^3}{30} - 5t + C. Then evaluate from 0 to 10: P(10)−P(0)=[100−100030−50]−[0]=100−33.33−50=16.67 kgP(10) - P(0) = [100 - \frac{1000}{30} - 50] - [0] = 100 - 33.33 - 50 = 16.67 \text{ kg} Answer A (16.67 kg) is correct. Answer B (5.00 kg) likely comes from incorrectly handling the integration, perhaps making arithmetic errors with the cubic term. Answer C (24.30 kg) suggests a sign error or incorrect coefficient handling during integration. Answer D (50.00 kg) appears to come from only considering the linear term 2t2t and ignoring the other components entirely. Remember: net change always requires integration of the rate function over the interval. Don't be tempted to just plug values into the rate equation itself—that only gives you the instantaneous rate, not the accumulated change. Practice identifying when a problem gives you f′(x)f'(x) and asks for change in f(x)f(x).

Question 4

The rate of deforestation in a region is modelled by A′(t)=−750(1.03)tA'(t) = -750(1.03)^t hectares per year, where tt is the number of years after 2010. Calculate the total area of forest lost from the beginning of 2015 to the end of 2018.

  1. 108 hectares
  2. 2890 hectares
  3. 3988 hectares (correct answer)
  4. 4265 hectares
Explanation: The beginning of 2015 corresponds to t=5t=5. The end of 2018 corresponds to the beginning of 2019, which is t=9t=9. The total area lost is the absolute value of the integral of the rate of change over this interval: ∣∫59−750(1.03)t dt∣|\int_{5}^{9} -750(1.03)^t \,dt|. The integral is approximately -3988. The area lost is the positive value, 3988 hectares.

Question 5

An ice sculpture initially has a volume of 12,000 cm³. It melts at a rate given by V′(t)=−0.1(t2+5)V'(t) = -0.1(t^2 + 5) cm³ per minute, where tt is the time in minutes. What is the remaining volume of the sculpture after 10 minutes?

  1. 383 cm³
  2. 11617 cm³ (correct answer)
  3. 11989.5 cm³
  4. 11990 cm³
Explanation: The final volume is the initial volume plus the net change. The net change is the integral of the rate of change from t=0t=0 to t=10t=10. We calculate V(10)=12000+∫010−0.1(t2+5) dtV(10) = 12000 + \int_{0}^{10} -0.1(t^2 + 5) \,dt. The integral evaluates to −0.1[t33+5t]010=−0.1(10003+50)=−0.1(333.33+50)=−38.33-0.1[\frac{t^3}{3} + 5t]_{0}^{10} = -0.1(\frac{1000}{3} + 50) = -0.1(333.33 + 50) = -38.33 cm³. So the final volume is 12000−383.33=11616.67≈1161712000 - 383.33 = 11616.67 \approx 11617 cm³.

Question 6

The rate of change of the amount of sand on a beach is given by S′(t)=1500−30t2S'(t) = 1500 - 30t^2 kg per hour, where tt is the time in hours. At t=0t=0, there were 25,000 kg of sand. How much sand is on the beach after 6 hours?

  1. 420 kg
  2. 6840 kg
  3. 25420 kg
  4. 31840 kg (correct answer)
Explanation: The amount of sand at t=6t=6 is the initial amount plus the net change from t=0t=0 to t=6t=6. The net change is ∫06(1500−30t2) dt=[1500t−10t3]06=1500(6)−10(63)=9000−2160=6840\int_{0}^{6} (1500 - 30t^2) \,dt = [1500t - 10t^3]_{0}^{6} = 1500(6) - 10(6^3) = 9000 - 2160 = 6840 kg. The total amount is 25000+6840=3184025000 + 6840 = 31840 kg.

Question 7

The rate of profit for a company is modelled by P′(t)=2.5−0.6ln⁡(t+1)P'(t) = 2.5 - 0.6\ln(t+1) million dollars per year, where tt is the number of years since 2020. What is the total profit earned from the start of 2022 (t=2t=2) to the start of 2025 (t=5t=5)?

  1. $1.43 million
  2. $4.54 million (correct answer)
  3. $5.20 million
  4. $7.50 million
Explanation: The total profit over the interval is the definite integral of the rate of profit. We need to calculate ∫25(2.5−0.6ln⁡(t+1)) dt\int_{2}^{5} (2.5 - 0.6\ln(t+1)) \,dt. Using a calculator, this integral is approximately 4.54 million dollars.

Question 8

The rate at which water flows into a reservoir is modelled by the function R(t)=150+20t−0.5t2R(t) = 150 + 20t - 0.5t^2, where RR is in cubic metres per hour and tt is the number of hours after midnight. What is the total volume of water that flows into the reservoir between t=5t=5 and t=10t=10?

  1. 62.5 m³
  2. 300 m³
  3. 1021 m³ (correct answer)
  4. 2167 m³
Explanation: The total volume of water is the definite integral of the rate function over the given interval. We need to calculate ∫510(150+20t−0.5t2) dt\int_{5}^{10} (150 + 20t - 0.5t^2) \,dt. Using a calculator, this integral is approximately 1020.83 m³, which rounds to 1021 m³.

Question 9

The rate at which a drug is absorbed into the bloodstream is modelled by A′(t)=0.5te−0.2tA'(t) = 0.5te^{-0.2t} mg per hour, where tt is the number of hours after ingestion. How much of the drug is absorbed by the body during the first 3 hours?

  1. 0.82 mg
  2. 1.54 mg (correct answer)
  3. 2.47 mg
  4. 6.25 mg
Explanation: The total amount absorbed is the definite integral of the rate of absorption from t=0t=0 to t=3t=3. We calculate ∫030.5te−0.2t dt\int_{0}^{3} 0.5te^{-0.2t} \,dt. Using a calculator, this evaluates to approximately 1.54 mg.

Question 10

The velocity of a particle moving along a line is given by v(t)=10sin⁡(0.5t)+5v(t) = 10\sin(0.5t) + 5 m/s for t≥0t \ge 0. What is the net displacement of the particle from t=πt = \pi to t=3πt = 3\pi seconds?

  1. -20.0 m
  2. 10.0 m
  3. 31.4 m (correct answer)
  4. 51.4 m
Explanation: Net displacement is the definite integral of the velocity function over the time interval. We need to calculate ∫π3π(10sin⁡(0.5t)+5) dt\int_{\pi}^{3\pi} (10\sin(0.5t) + 5) \,dt. This integral evaluates to 10π10\pi, which is approximately 31.4 m.

Question 11

A tank is leaking water at a rate of L(t)=40t+1L(t) = \frac{40}{t+1} litres per minute, where tt is the time in minutes since the leak began. How much water is lost between t=1t=1 and t=4t=4 minutes?

  1. 8.0 L
  2. 24.0 L
  3. 36.7 L (correct answer)
  4. 64.4 L
Explanation: The total amount of water lost is the integral of the leakage rate over the time interval. We calculate ∫1440t+1 dt\int_{1}^{4} \frac{40}{t+1} \,dt. This evaluates to 40[ln⁡(t+1)]14=40(ln⁡(5)−ln⁡(2))=40ln⁡(2.5)≈36.740[\ln(t+1)]_{1}^{4} = 40(\ln(5) - \ln(2)) = 40\ln(2.5) \approx 36.7 litres.

Question 12

An ice sculpture initially has a volume of 12,000 cm³. It melts at a rate given by V′(t)=−0.1(t2+5)V'(t) = -0.1(t^2 + 5) cm³ per minute, where tt is the time in minutes. What is the remaining volume of the sculpture after 10 minutes?

  1. 383 cm³
  2. 11617 cm³ (correct answer)
  3. 11989.5 cm³
  4. 11990 cm³
Explanation: The final volume is the initial volume plus the net change. The net change is the integral of the rate of change from t=0t=0 to t=10t=10. We calculate V(10)=12000+∫010−0.1(t2+5) dtV(10) = 12000 + \int_{0}^{10} -0.1(t^2 + 5) \,dt. The integral evaluates to −0.1[t33+5t]010=−0.1(10003+50)=−0.1(333.33+50)=−38.33-0.1[\frac{t^3}{3} + 5t]_{0}^{10} = -0.1(\frac{1000}{3} + 50) = -0.1(333.33 + 50) = -38.33 cm³. So the final volume is 12000−383.33=11616.67≈1161712000 - 383.33 = 11616.67 \approx 11617 cm³.

Question 13

The area of an oil slick is increasing at a rate of A′(t)=50t+4A'(t) = \frac{50}{\sqrt{t+4}} square metres per hour, where tt is the number of hours since the spill began. What is the increase in the area of the slick during the second hour?

  1. 20.4 m²
  2. 22.1 m² (correct answer)
  3. 45.1 m²
  4. 50.0 m²
Explanation: The phrase 'during the second hour' corresponds to the time interval from t=1t=1 to t=2t=2. The increase in area is the definite integral of the rate of change over this interval. We calculate ∫1250t+4 dt\int_{1}^{2} \frac{50}{\sqrt{t+4}} \,dt. Using a calculator, the result is approximately 22.1 m².

Question 14

Let C(t)C(t) be the concentration of a pollutant in a lake in mg/L, where tt is the time in days. What is the practical interpretation of the statement ∫1020C′(t) dt=−5\int_{10}^{20} C'(t) \,dt = -5?

  1. The concentration on day 20 was 5 mg/L.
  2. The rate of change of concentration on day 20 was -5 mg/L per day.
  3. The total amount of pollutant in the lake decreased by 5 mg between day 10 and day 20.
  4. The concentration of the pollutant decreased by 5 mg/L between day 10 and day 20. (correct answer)
Explanation: The definite integral of a rate of change, C′(t)C'(t), from t=at=a to t=bt=b gives the net change in the original function, C(t)C(t), over that interval. Therefore, ∫1020C′(t) dt=C(20)−C(10)=−5\int_{10}^{20} C'(t) \,dt = C(20) - C(10) = -5. This means the concentration decreased by 5 mg/L from day 10 to day 20.

Question 15

The current charging an electric vehicle's battery is given by I(t)=40e−0.1tI(t) = 40e^{-0.1t} amperes, where tt is the time in hours. The total charge added to the battery, measured in ampere-hours (Ah), is the integral of the current over time. What is the total charge added to the battery from t=0t=0 to t=2t=2 hours?

  1. 7.3 Ah
  2. 32.7 Ah
  3. 36.3 Ah
  4. 72.5 Ah (correct answer)
Explanation: The total charge QQ is the definite integral of the current I(t)I(t) with respect to time. We calculate Q=∫0240e−0.1t dtQ = \int_{0}^{2} 40e^{-0.1t} \,dt. Using a calculator, this evaluates to approximately 72.5 Ah.

Question 16

The rate of deforestation in a region is modelled by A′(t)=−750(1.03)tA'(t) = -750(1.03)^t hectares per year, where tt is the number of years after 2010. Calculate the total area of forest lost from the beginning of 2015 to the end of 2018.

  1. 108 hectares
  2. 2890 hectares
  3. 3988 hectares (correct answer)
  4. 4265 hectares
Explanation: The beginning of 2015 corresponds to t=5t=5. The end of 2018 corresponds to the beginning of 2019, which is t=9t=9. The total area lost is the absolute value of the integral of the rate of change over this interval: ∣∫59−750(1.03)t dt∣|\int_{5}^{9} -750(1.03)^t \,dt|. The integral is approximately -3988. The area lost is the positive value, 3988 hectares.

Question 17

The rate of change of the amount of a pollutant in a reservoir is given by P′(t)=2t−0.1t2−5P'(t) = 2t - 0.1t^2 - 5 kg/hr. A negative rate indicates the pollutant is being removed. What is the net change in the amount of pollutant over the first 10 hours?

  1. 16.67 kg (correct answer)
  2. 5.00 kg
  3. 24.30 kg
  4. 50.00 kg
Explanation: When you encounter a rate of change function and need to find the net change over an interval, you're dealing with a fundamental calculus application. The rate of change P′(t)P'(t) tells you how fast the pollutant amount is changing at any moment, and to find the total change, you need to integrate this rate over the given time period. To find the net change over 10 hours, you calculate ∫010(2t−0.1t2−5) dt\int_0^{10} (2t - 0.1t^2 - 5) \, dt. First, find the antiderivative: P(t)=t2−0.1t33−5t+C=t2−t330−5t+CP(t) = t^2 - \frac{0.1t^3}{3} - 5t + C = t^2 - \frac{t^3}{30} - 5t + C. Then evaluate from 0 to 10: P(10)−P(0)=[100−100030−50]−[0]=100−33.33−50=16.67 kgP(10) - P(0) = [100 - \frac{1000}{30} - 50] - [0] = 100 - 33.33 - 50 = 16.67 \text{ kg} Answer A (16.67 kg) is correct. Answer B (5.00 kg) likely comes from incorrectly handling the integration, perhaps making arithmetic errors with the cubic term. Answer C (24.30 kg) suggests a sign error or incorrect coefficient handling during integration. Answer D (50.00 kg) appears to come from only considering the linear term 2t2t and ignoring the other components entirely. Remember: net change always requires integration of the rate function over the interval. Don't be tempted to just plug values into the rate equation itself—that only gives you the instantaneous rate, not the accumulated change. Practice identifying when a problem gives you f′(x)f'(x) and asks for change in f(x)f(x).

Question 18

The rate at which water flows into a reservoir is modelled by the function R(t)=150+20t−0.5t2R(t) = 150 + 20t - 0.5t^2, where RR is in cubic metres per hour and tt is the number of hours after midnight. What is the total volume of water that flows into the reservoir between t=5t=5 and t=10t=10?

  1. 62.5 m³
  2. 300 m³
  3. 1021 m³ (correct answer)
  4. 2167 m³
Explanation: The total volume of water is the definite integral of the rate function over the given interval. We need to calculate ∫510(150+20t−0.5t2) dt\int_{5}^{10} (150 + 20t - 0.5t^2) \,dt. Using a calculator, this integral is approximately 1020.83 m³, which rounds to 1021 m³.

Question 19

The marginal cost of producing xx units of a product is given by C′(x)=12−0.08x+0.0003x2C'(x) = 12 - 0.08x + 0.0003x^2 dollars per unit. What is the increase in total cost when production is increased from 100 to 150 units?

  1. $-0.25
  2. $6.75
  3. $243.75 (correct answer)
  4. $337.50
Explanation: The increase in total cost is the definite integral of the marginal cost function between the two production levels. We calculate ∫100150(12−0.08x+0.0003x2) dx\int_{100}^{150} (12 - 0.08x + 0.0003x^2) \,dx. Using a calculator, the result is $243.75.

Question 20

The value of a new car depreciates at a rate of V′(t)=−2500e−0.15tV'(t) = -2500e^{-0.15t} dollars per year, where tt is the age of the car in years. What is the total loss in value of the car during its first 5 years, to the nearest dollar?

  1. $1181
  2. $8897 (correct answer)
  3. $16667
  4. $20403
Explanation: The total loss in value is the integral of the rate of depreciation over the first 5 years. We calculate ∫05−2500e−0.15t dt\int_{0}^{5} -2500e^{-0.15t} \,dt. The result is approximately -8897.4. The question asks for the total loss in value, which is the magnitude of this change, so the answer is $8897.