IB Mathematics: Applications and Interpretation Quiz: Numerical Integration
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Numerical IntegrationQuestion 1 of 20

The rate of profit for two products, A and B, are modelled by PA(t)=60e−0.1tP_A(t) = 60e^{-0.1t} and PB(t)=20+2tP_B(t) = 20 + 2t respectively, in thousands of dollars per year. Find the total excess profit of product A over product B during the time interval t=0t=0 to t=5t=5 years.

$52,600
$81,400
$156,400
$386,400
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IB Mathematics: Applications and Interpretation Quiz

IB Mathematics: Applications and Interpretation Quiz: Numerical Integration

Practice Numerical Integration 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.

What this quiz covers

This quiz focuses on Numerical Integration, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Applications and Interpretation.

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

The rate of profit for two products, A and B, are modelled by PA(t)=60e−0.1tP_A(t) = 60e^{-0.1t} and PB(t)=20+2tP_B(t) = 20 + 2t respectively, in thousands of dollars per year. Find the total excess profit of product A over product B during the time interval t=0t=0 to t=5t=5 years.

  1. $52,600
  2. $81,400 (correct answer)
  3. $156,400
  4. $386,400
Explanation: The excess profit is the integral of the difference between the two profit rates. We need to calculate ∫05(PA(t)−PB(t)) dt=∫05(60e−0.1t−(20+2t)) dt\int_{0}^{5} (P_A(t) - P_B(t)) \,dt = \int_{0}^{5} (60e^{-0.1t} - (20 + 2t)) \,dt. Using a calculator, this integral is approximately 81.4. Since the units are in thousands of dollars, the total excess profit is $81,400.

Question 2

The marginal cost of producing xx computer chips is given by C′(x)=40−0.1x+0.0015x2C'(x) = 40 - 0.1x + 0.0015x^2 dollars per chip. Calculate the increase in cost when production is increased from 200 chips to 300 chips.

  1. $75.00
  2. $5500 (correct answer)
  3. $7500
  4. $9750
Explanation: The increase in cost is the definite integral of the marginal cost function from x=200x=200 to x=300x=300. The integral to calculate is ∫200300(40−0.1x+0.0015x2) dx\int_{200}^{300} (40 - 0.1x + 0.0015x^2) \,dx. Using a calculator's numerical integration feature, the result is $5500.

Question 3

The rate of change of a country's population is modelled by P′(t)P'(t), where PP is the population in millions and tt is the number of years since 2020. What is the correct interpretation of the statement ∫510P′(t) dt=2.5\int_{5}^{10} P'(t) \,dt = 2.5?

  1. The population in the year 2030 was 2.5 million.
  2. The total population was 2.5 million people on average between 2025 and 2030.
  3. The average rate of population growth between 2025 and 2030 was 2.5 million per year.
  4. The population increased by a total of 2.5 million between 2025 and 2030. (correct answer)
Explanation: When you encounter an integral of a derivative like ∫510P′(t) dt\int_{5}^{10} P'(t) \,dt, you're looking at one of the most fundamental relationships in calculus: the Fundamental Theorem of Calculus. This theorem tells us that integrating a rate of change function gives us the total change over that interval. Since P′(t)P'(t) represents the rate of population change, ∫510P′(t) dt=P(10)−P(5)\int_{5}^{10} P'(t) \,dt = P(10) - P(5). This equals 2.5, meaning the population at t=10t = 10 (year 2030) minus the population at t=5t = 5 (year 2025) is 2.5 million. In other words, the population increased by a total of 2.5 million between 2025 and 2030, making D correct. Option A confuses the change in population with the actual population value. The integral doesn't tell us what the population was in 2030, only how much it changed. Option B misinterprets the integral as an average population. To find average population, you'd need 15∫510P(t) dt\frac{1}{5}\int_{5}^{10} P(t) \,dt, not the integral of the derivative. Option C confuses total change with average rate of change. The average rate would be ∫510P′(t) dt10−5=2.55=0.5\frac{\int_{5}^{10} P'(t) \,dt}{10-5} = \frac{2.5}{5} = 0.5 million per year. Remember this key pattern: whenever you see ∫abf′(x) dx\int_{a}^{b} f'(x) \,dx, it always equals f(b)−f(a)f(b) - f(a)—the total change in the original function. This appears frequently on IB exams in contexts involving rates and accumulation.

Question 4

Two particles, A and B, start from the same position and move along the same line. Their velocities are given by vA(t)=10+2cos⁡(t)v_A(t) = 10 + 2\cos(t) and vB(t)=11v_B(t) = 11, for t≥0t \ge 0. What is the distance between the two particles at time t=4πt=4\pi?

  1. 0
  2. 2
  3. π\pi
  4. 4π4\pi (correct answer)
Explanation: The position of each particle is the integral of its velocity. The distance between them is the absolute difference in their positions. Position of A is xA(t)=∫04π(10+2cos⁡(t)) dt=[10t+2sin⁡(t)]04π=40πx_A(t) = \int_0^{4\pi} (10+2\cos(t)) \,dt = [10t+2\sin(t)]_0^{4\pi} = 40\pi. Position of B is xB(t)=∫04π11 dt=[11t]04π=44πx_B(t) = \int_0^{4\pi} 11 \,dt = [11t]_0^{4\pi} = 44\pi. The distance between them is ∣44π−40π∣=4π|44\pi - 40\pi| = 4\pi.

Question 5

The area of a healing wound is decreasing at a rate of A′(t)=−15(t+1)−2A'(t) = -15(t+1)^{-2} square millimetres per day. What is the total decrease in the area of the wound from the end of day 1 (t=1t=1) to the end of day 4 (t=4t=4)?

  1. 4.50 mm² (correct answer)
  2. 3.75 mm²
  3. 6.00 mm²
  4. 9.00 mm²
Explanation: When you encounter a rate of change function and need to find the total change over an interval, you're looking at a definite integral problem. The rate A′(t)=−15(t+1)−2A'(t) = -15(t+1)^{-2} tells you how fast the wound area is changing at any time tt, so integrating this rate gives you the total change. To find the total decrease from day 1 to day 4, you need to evaluate ∫14A′(t) dt=∫14−15(t+1)−2 dt\int_1^4 A'(t) \, dt = \int_1^4 -15(t+1)^{-2} \, dt. First, find the antiderivative: ∫−15(t+1)−2 dt=−15⋅(t+1)−1−1=15(t+1)−1=15t+1\int -15(t+1)^{-2} \, dt = -15 \cdot \frac{(t+1)^{-1}}{-1} = 15(t+1)^{-1} = \frac{15}{t+1} Now evaluate the definite integral: [15t+1]14=154+1−151+1=155−152=3−7.5=−4.5\left[\frac{15}{t+1}\right]_1^4 = \frac{15}{4+1} - \frac{15}{1+1} = \frac{15}{5} - \frac{15}{2} = 3 - 7.5 = -4.5 Since we want the decrease (positive value), the total decrease is 4.50 mm². Answer A (4.50 mm²) is correct. Answer B (3.75 mm²) likely comes from an integration error or incorrect bounds. Answer C (6.00 mm²) might result from forgetting the negative sign in the rate function. Answer D (9.00 mm²) could come from doubling the correct answer or making multiple calculation errors. Remember: when given a rate function, always integrate to find total change. Pay careful attention to signs—negative rates mean decreasing quantities, but the question asks for the amount of decrease, which should be positive.

Question 6

The rate at which water leaks from a storage tank is modelled by the function R(t)=15e−0.2tR(t) = 15e^{-0.2t} litres per hour, where tt is the number of hours since the leak started. Find the total amount of water, in litres, that has leaked from the tank between t=3t=3 and t=7t=7.

  1. 7.46
  2. 11.1
  3. 25.9 (correct answer)
  4. 57.3
Explanation: The total amount of water leaked between t=3t=3 and t=7t=7 is the definite integral of the rate function R(t)R(t) over this interval. The calculation is ∫3715e−0.2t dt\int_{3}^{7} 15e^{-0.2t} \,dt. Using a graphing calculator, this integral evaluates to approximately 25.9 litres.

Question 7

A company's daily profit is changing at a rate of P′(t)=50−3t2P'(t) = 50 - 3t^2 hundreds of dollars per day, where tt is the number of days from the start of a new campaign. What is the net change in profit over the first 5 days of the campaign?

  1. $12,500 (correct answer)
  2. -$2500
  3. $16,667
  4. $20,833
Explanation: When you see a rate of change function like P′(t)P'(t), you're dealing with a derivative that tells you how fast profit is changing at each moment. To find the total change over an interval, you need to integrate this rate function - this is a fundamental application of the Fundamental Theorem of Calculus. To find the net change in profit over the first 5 days, integrate P′(t)=50−3t2P'(t) = 50 - 3t^2 from t=0t = 0 to t=5t = 5: ∫05(50−3t2)dt=[50t−t3]05=(50⋅5−53)−(0)=250−125=125\int_0^5 (50 - 3t^2) dt = [50t - t^3]_0^5 = (50 \cdot 5 - 5^3) - (0) = 250 - 125 = 125 Since the rate is given in hundreds of dollars per day, the result is 125 hundreds of dollars, which equals $12,500. Looking at the wrong answers: Answer B (-2500)likelycomesfromasignerrororincorrectlycalculatingthedefiniteintegral.AnswerC(2500) likely comes from a sign error or incorrectly calculating the definite integral. Answer C (16,667) might result from using an incorrect antiderivative or evaluation error. Answer D ($20,833) could come from forgetting to subtract $t3t^3 whenfindingtheantiderivative,usingonlythewhen finding the antiderivative, using only the 50t50t $ term. The correct answer is A ($12,500). Remember this key principle: when given a rate function, always integrate to find total change over an interval. Also, pay careful attention to units - here the rate was in "hundreds of dollars," so don't forget to convert your final numerical answer to actual dollars.

Question 8

The rate at which water leaks from a storage tank is modelled by the function R(t)=15e−0.2tR(t) = 15e^{-0.2t} litres per hour, where tt is the number of hours since the leak started. Find the total amount of water, in litres, that has leaked from the tank between t=3t=3 and t=7t=7.

  1. 7.46
  2. 11.1
  3. 25.9 (correct answer)
  4. 57.3
Explanation: The total amount of water leaked between t=3t=3 and t=7t=7 is the definite integral of the rate function R(t)R(t) over this interval. The calculation is ∫3715e−0.2t dt\int_{3}^{7} 15e^{-0.2t} \,dt. Using a graphing calculator, this integral evaluates to approximately 25.9 litres.

Question 9

The rate of profit for two products, A and B, are modelled by PA(t)=60e−0.1tP_A(t) = 60e^{-0.1t} and PB(t)=20+2tP_B(t) = 20 + 2t respectively, in thousands of dollars per year. Find the total excess profit of product A over product B during the time interval t=0t=0 to t=5t=5 years.

  1. $52,600
  2. $81,400 (correct answer)
  3. $156,400
  4. $386,400
Explanation: The excess profit is the integral of the difference between the two profit rates. We need to calculate ∫05(PA(t)−PB(t)) dt=∫05(60e−0.1t−(20+2t)) dt\int_{0}^{5} (P_A(t) - P_B(t)) \,dt = \int_{0}^{5} (60e^{-0.1t} - (20 + 2t)) \,dt. Using a calculator, this integral is approximately 81.4. Since the units are in thousands of dollars, the total excess profit is $81,400.

Question 10

A wooden bowl is designed by rotating the curve y=0.2(x−1)2+3y = 0.2(x-1)^2 + 3 about the x-axis, for 1≤x≤61 \le x \le 6. The dimensions are in centimetres.

Find the volume of wood required to make the bowl, in cm³.

  1. 157
  2. 218
  3. 686 (correct answer)
  4. 2154
Explanation: The volume of a solid of revolution about the x-axis is given by the formula V=π∫aby2 dxV = \pi \int_a^b y^2 \,dx. In this case, we must calculate V=π∫16(0.2(x−1)2+3)2 dxV = \pi \int_{1}^{6} (0.2(x-1)^2 + 3)^2 \,dx. Using a calculator's numerical integration function, the integral evaluates to approximately 218.3. Therefore, V≈π×218.3≈685.8V \approx \pi \times 218.3 \approx 685.8 cm³.

Question 11

A company's initial investment of $50,000 is expected to generate a continuous income stream at a rate of R(t)=15000e0.04tR(t) = 15000e^{0.04t} dollars per year, where tt is in years. Find the total income generated during the first five years.

  1. $75,000
  2. $83,214 (correct answer)
  3. $91,578
  4. $125,000
Explanation: The total income generated over a period is the definite integral of the income rate function. We need to calculate ∫0515000e0.04t dt\int_{0}^{5} 15000e^{0.04t} \,dt. Using a calculator, this integral is approximately $83,214. The initial investment is extra information not needed for this calculation.

Question 12

The rate at which a social media post gains likes is modelled by L′(t)=50t2e−0.5tL'(t) = 50t^2 e^{-0.5t} likes per hour, for t≥0t \ge 0. To the nearest hour, how long does it take for the post to accumulate its first 500 likes?

  1. 3 hours
  2. 5 hours
  3. 8 hours (correct answer)
  4. 10 hours
Explanation: We need to find the value of TT for which the total number of likes, ∫0TL′(t) dt\int_{0}^{T} L'(t) \,dt, is equal to 500. We need to solve the equation ∫0T50t2e−0.5t dt=500\int_{0}^{T} 50t^2 e^{-0.5t} \,dt = 500. Using a graphing calculator's solver or by finding the intersection of y=∫0x50t2e−0.5t dty = \int_{0}^{x} 50t^2 e^{-0.5t} \,dt and y=500y = 500, we find T≈8.16T \approx 8.16 hours. To the nearest hour, this is 8 hours.

Question 13

A wooden bowl is designed by rotating the curve y=0.2(x−1)2+3y = 0.2(x-1)^2 + 3 about the x-axis, for 1≤x≤61 \le x \le 6. The dimensions are in centimetres.

Find the volume of wood required to make the bowl, in cm³.

  1. 157
  2. 218
  3. 686 (correct answer)
  4. 2154
Explanation: The volume of a solid of revolution about the x-axis is given by the formula V=π∫aby2 dxV = \pi \int_a^b y^2 \,dx. In this case, we must calculate V=π∫16(0.2(x−1)2+3)2 dxV = \pi \int_{1}^{6} (0.2(x-1)^2 + 3)^2 \,dx. Using a calculator's numerical integration function, the integral evaluates to approximately 218.3. Therefore, V≈π×218.3≈685.8V \approx \pi \times 218.3 \approx 685.8 cm³.

Question 14

A company's initial investment of $50,000 is expected to generate a continuous income stream at a rate of R(t)=15000e0.04tR(t) = 15000e^{0.04t} dollars per year, where tt is in years. Find the total income generated during the first five years.

  1. $75,000
  2. $83,214 (correct answer)
  3. $91,578
  4. $125,000
Explanation: The total income generated over a period is the definite integral of the income rate function. We need to calculate ∫0515000e0.04t dt\int_{0}^{5} 15000e^{0.04t} \,dt. Using a calculator, this integral is approximately $83,214. The initial investment is extra information not needed for this calculation.

Question 15

The rate of ice melting on a glacier is given by M(t)=0.5+sin⁡2(πt12)M(t) = 0.5 + \sin^2(\frac{\pi t}{12}) billion tonnes per month, where tt is the number of months from the start of the year. Approximately how much ice melts in total during the first 6 months of the year?

  1. 0.75 billion tonnes
  2. 3.00 billion tonnes
  3. 4.50 billion tonnes (correct answer)
  4. 6.00 billion tonnes
Explanation: The total amount of ice melted is the definite integral of the melting rate from t=0t=0 to t=6t=6. We need to calculate ∫06(0.5+sin⁡2(πt12)) dt\int_0^6 (0.5 + \sin^2(\frac{\pi t}{12})) \,dt. Using a calculator's numerical integration feature, this integral evaluates to approximately 4.50.

Question 16

The rate of ice melting on a glacier is given by M(t)=0.5+sin⁡2(πt12)M(t) = 0.5 + \sin^2(\frac{\pi t}{12}) billion tonnes per month, where tt is the number of months from the start of the year. Approximately how much ice melts in total during the first 6 months of the year?

  1. 0.75 billion tonnes
  2. 3.00 billion tonnes
  3. 4.50 billion tonnes (correct answer)
  4. 6.00 billion tonnes
Explanation: The total amount of ice melted is the definite integral of the melting rate from t=0t=0 to t=6t=6. We need to calculate ∫06(0.5+sin⁡2(πt12)) dt\int_0^6 (0.5 + \sin^2(\frac{\pi t}{12})) \,dt. Using a calculator's numerical integration feature, this integral evaluates to approximately 4.50.

Question 17

The marginal cost of producing xx computer chips is given by C′(x)=40−0.1x+0.0015x2C'(x) = 40 - 0.1x + 0.0015x^2 dollars per chip. Calculate the increase in cost when production is increased from 200 chips to 300 chips.

  1. $75.00
  2. $5500 (correct answer)
  3. $7500
  4. $9750
Explanation: The increase in cost is the definite integral of the marginal cost function from x=200x=200 to x=300x=300. The integral to calculate is ∫200300(40−0.1x+0.0015x2) dx\int_{200}^{300} (40 - 0.1x + 0.0015x^2) \,dx. Using a calculator's numerical integration feature, the result is $5500.

Question 18

The rate of change of a country's population is modelled by P′(t)P'(t), where PP is the population in millions and tt is the number of years since 2020. What is the correct interpretation of the statement ∫510P′(t) dt=2.5\int_{5}^{10} P'(t) \,dt = 2.5?

  1. The population in the year 2030 was 2.5 million.
  2. The total population was 2.5 million people on average between 2025 and 2030.
  3. The average rate of population growth between 2025 and 2030 was 2.5 million per year.
  4. The population increased by a total of 2.5 million between 2025 and 2030. (correct answer)
Explanation: When you encounter an integral of a derivative like ∫510P′(t) dt\int_{5}^{10} P'(t) \,dt, you're looking at one of the most fundamental relationships in calculus: the Fundamental Theorem of Calculus. This theorem tells us that integrating a rate of change function gives us the total change over that interval. Since P′(t)P'(t) represents the rate of population change, ∫510P′(t) dt=P(10)−P(5)\int_{5}^{10} P'(t) \,dt = P(10) - P(5). This equals 2.5, meaning the population at t=10t = 10 (year 2030) minus the population at t=5t = 5 (year 2025) is 2.5 million. In other words, the population increased by a total of 2.5 million between 2025 and 2030, making D correct. Option A confuses the change in population with the actual population value. The integral doesn't tell us what the population was in 2030, only how much it changed. Option B misinterprets the integral as an average population. To find average population, you'd need 15∫510P(t) dt\frac{1}{5}\int_{5}^{10} P(t) \,dt, not the integral of the derivative. Option C confuses total change with average rate of change. The average rate would be ∫510P′(t) dt10−5=2.55=0.5\frac{\int_{5}^{10} P'(t) \,dt}{10-5} = \frac{2.5}{5} = 0.5 million per year. Remember this key pattern: whenever you see ∫abf′(x) dx\int_{a}^{b} f'(x) \,dx, it always equals f(b)−f(a)f(b) - f(a)—the total change in the original function. This appears frequently on IB exams in contexts involving rates and accumulation.

Question 19

A company's daily profit is changing at a rate of P′(t)=50−3t2P'(t) = 50 - 3t^2 hundreds of dollars per day, where tt is the number of days from the start of a new campaign. What is the net change in profit over the first 5 days of the campaign?

  1. $12,500 (correct answer)
  2. -$2500
  3. $16,667
  4. $20,833
Explanation: When you see a rate of change function like P′(t)P'(t), you're dealing with a derivative that tells you how fast profit is changing at each moment. To find the total change over an interval, you need to integrate this rate function - this is a fundamental application of the Fundamental Theorem of Calculus. To find the net change in profit over the first 5 days, integrate P′(t)=50−3t2P'(t) = 50 - 3t^2 from t=0t = 0 to t=5t = 5: ∫05(50−3t2)dt=[50t−t3]05=(50⋅5−53)−(0)=250−125=125\int_0^5 (50 - 3t^2) dt = [50t - t^3]_0^5 = (50 \cdot 5 - 5^3) - (0) = 250 - 125 = 125 Since the rate is given in hundreds of dollars per day, the result is 125 hundreds of dollars, which equals $12,500. Looking at the wrong answers: Answer B (-2500)likelycomesfromasignerrororincorrectlycalculatingthedefiniteintegral.AnswerC(2500) likely comes from a sign error or incorrectly calculating the definite integral. Answer C (16,667) might result from using an incorrect antiderivative or evaluation error. Answer D ($20,833) could come from forgetting to subtract $t3t^3 whenfindingtheantiderivative,usingonlythewhen finding the antiderivative, using only the 50t50t $ term. The correct answer is A ($12,500). Remember this key principle: when given a rate function, always integrate to find total change over an interval. Also, pay careful attention to units - here the rate was in "hundreds of dollars," so don't forget to convert your final numerical answer to actual dollars.

Question 20

The area of a healing wound is decreasing at a rate of A′(t)=−15(t+1)−2A'(t) = -15(t+1)^{-2} square millimetres per day. What is the total decrease in the area of the wound from the end of day 1 (t=1t=1) to the end of day 4 (t=4t=4)?

  1. 4.50 mm² (correct answer)
  2. 3.75 mm²
  3. 6.00 mm²
  4. 9.00 mm²
Explanation: When you encounter a rate of change function and need to find the total change over an interval, you're looking at a definite integral problem. The rate A′(t)=−15(t+1)−2A'(t) = -15(t+1)^{-2} tells you how fast the wound area is changing at any time tt, so integrating this rate gives you the total change. To find the total decrease from day 1 to day 4, you need to evaluate ∫14A′(t) dt=∫14−15(t+1)−2 dt\int_1^4 A'(t) \, dt = \int_1^4 -15(t+1)^{-2} \, dt. First, find the antiderivative: ∫−15(t+1)−2 dt=−15⋅(t+1)−1−1=15(t+1)−1=15t+1\int -15(t+1)^{-2} \, dt = -15 \cdot \frac{(t+1)^{-1}}{-1} = 15(t+1)^{-1} = \frac{15}{t+1} Now evaluate the definite integral: [15t+1]14=154+1−151+1=155−152=3−7.5=−4.5\left[\frac{15}{t+1}\right]_1^4 = \frac{15}{4+1} - \frac{15}{1+1} = \frac{15}{5} - \frac{15}{2} = 3 - 7.5 = -4.5 Since we want the decrease (positive value), the total decrease is 4.50 mm². Answer A (4.50 mm²) is correct. Answer B (3.75 mm²) likely comes from an integration error or incorrect bounds. Answer C (6.00 mm²) might result from forgetting the negative sign in the rate function. Answer D (9.00 mm²) could come from doubling the correct answer or making multiple calculation errors. Remember: when given a rate function, always integrate to find total change. Pay careful attention to signs—negative rates mean decreasing quantities, but the question asks for the amount of decrease, which should be positive.