Business Calculus Quiz: Net Change And Total Change
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Net Change And Total ChangeQuestion 1 of 17

The marginal cost for a company to produce qq units of a product is given by C(q)=0.03q21.2q+15C'(q) = 0.03q^2 - 1.2q + 15 dollars per unit. What is the total increase in cost if the company raises its production level from q=20q=20 to q=40q=40 units?

$15
$140
$280
$7
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Business Calculus Quiz

Business Calculus Quiz: Net Change And Total Change

Practice Net Change And Total Change in Business Calculus 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 Net Change And Total Change, giving you a quick way to practice the rules, question types, and explanations that matter most for Business Calculus.

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Question 1

The marginal cost for a company to produce qq units of a product is given by C(q)=0.03q21.2q+15C'(q) = 0.03q^2 - 1.2q + 15 dollars per unit. What is the total increase in cost if the company raises its production level from q=20q=20 to q=40q=40 units?

  1. $15
  2. $140 (correct answer)
  3. $280
  4. $7
Explanation: The total increase in cost is the net change in the cost function from q=20q=20 to q=40q=40. This is found by integrating the marginal cost function over this interval: 2040C(q)dq=2040(0.03q21.2q+15)dq\int_{20}^{40} C'(q) dq = \int_{20}^{40} (0.03q^2 - 1.2q + 15) dq. The antiderivative is C(q)=0.01q30.6q2+15qC(q) = 0.01q^3 - 0.6q^2 + 15q. Evaluating the definite integral: C(40)C(20)=(0.01(40)30.6(40)2+15(40))(0.01(20)30.6(20)2+15(20))=(640960+600)(80240+300)=280140=140C(40) - C(20) = (0.01(40)^3 - 0.6(40)^2 + 15(40)) - (0.01(20)^3 - 0.6(20)^2 + 15(20)) = (640 - 960 + 600) - (80 - 240 + 300) = 280 - 140 = 140. Thus, the cost increases by $140.

Question 2

A new technology startup has a continuous revenue stream with a rate of flow modeled by R(t)=150e0.1tR'(t) = 150e^{-0.1t} thousands of dollars per month, where tt is in months. The project had initial start-up costs, resulting in a total revenue of -5050 thousand (a loss) at time t=0t=0. What is the total accumulated revenue, in thousands of dollars, after the first 10 months?

  1. $948.2
  2. $898.2 (correct answer)
  3. $55.2
  4. $1003.2
Explanation: Total accumulated revenue at time t=10t=10 is given by R(10)=R(0)+010R(t)dtR(10) = R(0) + \int_0^{10} R'(t) dt. First, calculate the net change in revenue over the 10 months: 010150e0.1tdt=150[e0.1t0.1]010=1500[e1e0]=1500(e11)948.18\int_0^{10} 150e^{-0.1t} dt = 150 [\frac{e^{-0.1t}}{-0.1}]_0^{10} = -1500[e^{-1} - e^0] = -1500(e^{-1} - 1) \approx 948.18. Now, add the initial revenue: R(10)=50+948.18=898.18R(10) = -50 + 948.18 = 898.18. The total revenue is approximately $898.2 thousand.

Question 3

V(t)V(t) represents the value of a commercial property in thousands of dollars, tt years after its purchase. Financial analysts determine that 510V(t)dt=150\int_5^{10} V'(t) dt = -150. Which of the following statements is the correct interpretation of this result?

  1. At the end of year 10, the property was worth $150,000 less than it was at the time of purchase.
  2. The rate of change of the property's value was exactly -$150,000 per year at some point between year 5 and year 10.
  3. The property's value decreased by a total of $150,000 between the end of year 5 and the end of year 10. (correct answer)
  4. The average value of the property between year 5 and year 10 was negative.
Explanation: By the Net Change Theorem, the integral of a rate of change function V(t)V'(t) from t=at=a to t=bt=b gives the net change in the original function, V(b)V(a)V(b) - V(a). Therefore, 510V(t)dt=V(10)V(5)=150\int_5^{10} V'(t) dt = V(10) - V(5) = -150. This means the change in value from the end of year 5 to the end of year 10 is -150150 thousand, which is a decrease of $150,000. Distractor A refers to the wrong time interval. Distractor B confuses the total change with an instantaneous rate. Distractor D incorrectly interprets the meaning of the integral of the derivative.

Question 4

A company's cash reserves change based on its rate of income, I(t)=25e0.05tI'(t) = 25e^{-0.05t}, and its rate of expenditure, E(t)=10+0.2tE'(t) = 10 + 0.2t, both in thousands of dollars per month. What is the net change in the company's cash reserves over the first six months (from t=0t=0 to t=6t=6)?

  1. An increase of $193.2 thousand
  2. An increase of $7.3 thousand
  3. A decrease of $66.0 thousand
  4. An increase of $66.0 thousand (correct answer)
Explanation: The rate of change of cash reserves, S(t)S'(t), is the rate of income minus the rate of expenditure: S(t)=I(t)E(t)=25e0.05t(10+0.2t)S'(t) = I'(t) - E'(t) = 25e^{-0.05t} - (10 + 0.2t). The net change is the integral of this rate from t=0t=0 to t=6t=6: 06(25e0.05t100.2t)dt\int_0^6 (25e^{-0.05t} - 10 - 0.2t) dt. The antiderivative is [250.05e0.05t10t0.1t2]06=[500e0.05t10t0.1t2]06[\frac{25}{-0.05}e^{-0.05t} - 10t - 0.1t^2]_0^6 = [-500e^{-0.05t} - 10t - 0.1t^2]_0^6. Evaluating at the limits: (500e0.3603.6)(500e0)=500e0.363.6+500500(0.7408)63.6+500=370.463.6+500=66.0(-500e^{-0.3} - 60 - 3.6) - (-500e^0) = -500e^{-0.3} - 63.6 + 500 \approx -500(0.7408) - 63.6 + 500 = -370.4 - 63.6 + 500 = 66.0. The net change is an increase of $66.0 thousand.

Question 5

A company's marginal profit function is P(q)P'(q) dollars per unit, where qq is the number of units produced. Which of the following is the best interpretation of the statement 500600P(q)dq=4000\int_{500}^{600} P'(q) dq = 4000?

  1. The total profit earned from selling the first 600 units is $4000.
  2. At a production level of 600 units, the profit is increasing at a rate of $4000 per unit.
  3. The total profit increases by $4000 when production is increased from 500 to 600 units. (correct answer)
  4. The average profit for producing the first 600 units is $4000.
Explanation: The definite integral of a marginal function (a rate of change) over an interval gives the net change in the original function over that interval. Thus, 500600P(q)dq\int_{500}^{600} P'(q) dq represents the total change in profit, P(600)P(500)P(600) - P(500), when the production level goes from 500 to 600 units. The value of $4000 indicates that this change is an increase of $4000.

Question 6

The population of subscribers to a streaming service is growing at a rate of N(t)=1000t+1N'(t) = \frac{1000}{\sqrt{t+1}} subscribers per month, where tt is the number of months since the service launched. Three months after launch (t=3t=3), the service had 2500 subscribers. How many subscribers, to the nearest whole number, did the service have after one year (t=12t=12)?

  1. 3211
  2. 5211
  3. 5711 (correct answer)
  4. 7711
Explanation: We need to find N(12)N(12), given N(3)=2500N(3)=2500. The value is N(12)=N(3)+312N(t)dtN(12) = N(3) + \int_3^{12} N'(t) dt. First, calculate the integral representing the net change in subscribers from month 3 to month 12: 3121000(t+1)1/2dt=1000[2(t+1)1/2]312=2000[12+13+1]=2000(134)2000(3.605552)=2000(1.60555)3211.1\int_3^{12} 1000(t+1)^{-1/2} dt = 1000[2(t+1)^{1/2}]_3^{12} = 2000[\sqrt{12+1} - \sqrt{3+1}] = 2000(\sqrt{13} - \sqrt{4}) \approx 2000(3.60555 - 2) = 2000(1.60555) \approx 3211.1. This is the increase in subscribers. The total number is the initial amount at t=3t=3 plus this increase: N(12)=2500+3211.1=5711.1N(12) = 2500 + 3211.1 = 5711.1. To the nearest whole number, this is 5711.

Question 7

The rate of change of items in a warehouse is given by I(t)=3t218t+15I'(t) = 3t^2 - 18t + 15 items per hour for an 8-hour workday, 0t80 \le t \le 8. A positive rate indicates items are being added, and a negative rate indicates items are being removed. What is the total number of items moved (either in or out of the warehouse) during the first 4 hours?

  1. -20 items
  2. 34 items (correct answer)
  3. 7 items
  4. 27 items
Explanation: Total items moved is the integral of the absolute value of the rate, 04I(t)dt\int_0^4 |I'(t)| dt. First, find where I(t)I'(t) is positive or negative by finding its roots: 3t218t+15=3(t1)(t5)=03t^2 - 18t + 15 = 3(t-1)(t-5) = 0, so roots are at t=1t=1 and t=5t=5. On the interval [0,4][0, 4], I(t)>0I'(t) > 0 for [0,1)[0, 1) and I(t)<0I'(t) < 0 for (1,4](1, 4]. The integral becomes 01(3t218t+15)dt+14(3t218t+15)dt\int_0^1 (3t^2 - 18t + 15) dt + \int_1^4 -(3t^2 - 18t + 15) dt. Let F(t)=t39t2+15tF(t) = t^3 - 9t^2 + 15t. The first integral is F(1)F(0)=70=7F(1) - F(0) = 7-0=7. The second integral is (F(4)F(1))=(207)=(27)=27-(F(4)-F(1)) = -(-20 - 7) = -(-27) = 27. The total number of items moved is the sum of the magnitudes of these changes: 7+27=347 + 27 = 34.

Question 8

A company's rate of profit is modeled by P(t)=1.20.6tP'(t) = 1.2 - 0.6t million dollars per year, where tt is the number of years from the company's launch. The profit rate is positive for the first two years and then becomes negative. What is the net change in the company's profit during the first 4 years of operation?

  1. $0 million (correct answer)
  2. $1.2 million
  3. $2.4 million
  4. $-1.2 million
Explanation: The net change in profit is the definite integral of the rate of profit from t=0t=0 to t=4t=4. We calculate 04(1.20.6t)dt\int_0^4 (1.2 - 0.6t) dt. The antiderivative is P(t)=1.2t0.3t2P(t) = 1.2t - 0.3t^2. Evaluating the integral: [1.2t0.3t2]04=(1.2(4)0.3(42))(0)=4.80.3(16)=4.84.8=0[1.2t - 0.3t^2]_0^4 = (1.2(4) - 0.3(4^2)) - (0) = 4.8 - 0.3(16) = 4.8 - 4.8 = 0. The net change in profit over the first 4 years is $0 million, meaning the profit gained in the first two years was completely lost in the following two years.

Question 9

A manufacturer's inventory of a specific smartphone model changes based on a constant production rate of 1000 units per week and a sales rate of S(t)=20t2120t+1180S'(t) = 20t^2 - 120t + 1180 units per week, for 0t60 \le t \le 6. If the initial inventory at t=0t=0 was 500 units, what is the inventory level at the end of week 6?

  1. 140 units (correct answer)
  2. 860 units
  3. -360 units
  4. 6500 units
Explanation: The rate of change of inventory is I(t)=(rate in)(rate out)=1000(20t2120t+1180)=20t2+120t180I'(t) = (\text{rate in}) - (\text{rate out}) = 1000 - (20t^2 - 120t + 1180) = -20t^2 + 120t - 180. The net change in inventory over 6 weeks is 06I(t)dt=[203t3+60t2180t]06=203(63)+60(62)180(6)=1440+21601080=360\int_0^6 I'(t) dt = [-\frac{20}{3}t^3 + 60t^2 - 180t]_0^6 = -\frac{20}{3}(6^3) + 60(6^2) - 180(6) = -1440 + 2160 - 1080 = -360 units. The final inventory is the initial inventory plus the net change: I(6)=I(0)+Net Change=500+(360)=140I(6) = I(0) + \text{Net Change} = 500 + (-360) = 140 units.

Question 10

The value of a piece of industrial equipment, V(t)V(t), is decreasing at a rate given by V(t)=1000(t+2)2V'(t) = -1000(t+2)^{-2} dollars per year, where tt is the number of years since its purchase. What is the total loss in value of the equipment during its third year of operation?

  1. $300
  2. $50 (correct answer)
  3. $-50
  4. $250
Explanation: The 'third year' corresponds to the time interval from t=2t=2 to t=3t=3. The net change in value during this period is 23V(t)dt=231000(t+2)2dt\int_{2}^{3} V'(t) \,dt = \int_{2}^{3} -1000(t+2)^{-2} \,dt. The antiderivative of 1000(t+2)2-1000(t+2)^{-2} is 1000(t+2)11000(t+2)^{-1}. Evaluating the definite integral: [1000(t+2)1]23=10003+210002+2=1000510004=200250=50[1000(t+2)^{-1}]_{2}^{3} = \frac{1000}{3+2} - \frac{1000}{2+2} = \frac{1000}{5} - \frac{1000}{4} = 200 - 250 = -50. The value decreased by $50, so the loss in value is a positive quantity, $50.

Question 11

Let A(t)A(t) be the amount of oil, in barrels, in a storage tank at time tt in hours. Oil flows into the tank at a rate of I(t)I(t) barrels per hour and flows out at a rate of O(t)O(t) barrels per hour. Which expression represents the net change in the amount of oil in the tank from t=2t=2 to t=5t=5?

  1. 25(I(t)+O(t))dt\int_{2}^{5} (I(t) + O(t)) \,dt
  2. 1325(I(t)O(t))dt\frac{1}{3}\int_{2}^{5} (I(t) - O(t)) \,dt
  3. (I(5)O(5))(I(2)O(2))(I(5)-O(5)) - (I(2)-O(2))
  4. 25(I(t)O(t))dt\int_{2}^{5} (I(t) - O(t)) \,dt (correct answer)
Explanation: The net rate of change of the amount of oil in the tank is the rate in minus the rate out, which is A(t)=I(t)O(t)A'(t) = I(t) - O(t). The net change in the amount of oil between two times is the definite integral of this net rate of change over that time interval. Therefore, the net change from t=2t=2 to t=5t=5 is 25A(t)dt=25(I(t)O(t))dt\int_{2}^{5} A'(t) \,dt = \int_{2}^{5} (I(t) - O(t)) \,dt.

Question 12

The rate of production at a factory is given by P(t)P(t) units per hour. The selling price for each unit produced at time tt is given by S(t)S(t) dollars per unit. Which of the following expressions represents the total revenue generated by the factory from time t=at=a to t=bt=b?

  1. (abP(t)dt)(abS(t)dt)\left(\int_{a}^{b} P(t) \,dt\right) \cdot \left(\int_{a}^{b} S(t) \,dt\right)
  2. abP(t)S(t)dt\int_{a}^{b} P(t)S(t) \,dt (correct answer)
  3. P(b)S(b)P(a)S(a)P(b)S(b) - P(a)S(a)
  4. ab(P(t)+S(t))dt\int_{a}^{b} (P(t) + S(t)) \,dt
Explanation: To find the total revenue, we must first determine the rate at which revenue is generated. At any given time tt, the rate of revenue is (units per hour) × (dollars per unit), which is P(t)S(t)P(t)S(t) in dollars per hour. The total revenue accumulated over the interval [a,b][a, b] is the definite integral of this rate of revenue. Therefore, the total revenue is abP(t)S(t)dt\int_{a}^{b} P(t)S(t) \,dt.

Question 13

The rate of change of profit for Company A is PA(t)=10t+5P_A'(t) = 10t + 5 and for Company B is PB(t)=t2+8P_B'(t) = t^2 + 8, where profit is in thousands of dollars and tt is the number of years from now (t0t \ge 0). Over the time interval from t=0t=0 to t=6t=6, by how much does the net change in profit for Company A exceed the net change in profit for Company B?

  1. $90,000 in favor of Company A (correct answer)
  2. $330,000 in favor of Company B
  3. $90,000 in favor of Company B
  4. $120,000 in favor of Company A
Explanation: We need to find the difference between the net changes in profit: 06PA(t)dt06PB(t)dt\int_{0}^{6} P_A'(t) \,dt - \int_{0}^{6} P_B'(t) \,dt. This can be computed as a single integral of the difference of the rates: 06(PA(t)PB(t))dt=06((10t+5)(t2+8))dt=06(t2+10t3)dt\int_{0}^{6} (P_A'(t) - P_B'(t)) \,dt = \int_{0}^{6} ((10t + 5) - (t^2 + 8)) \,dt = \int_{0}^{6} (-t^2 + 10t - 3) \,dt. Evaluating the integral: [t33+5t23t]06=(633+5(62)3(6))0=72+18018=90[-\frac{t^3}{3} + 5t^2 - 3t]_{0}^{6} = (-\frac{6^3}{3} + 5(6^2) - 3(6)) - 0 = -72 + 180 - 18 = 90. Since the units are in thousands of dollars and the result is positive, Company A's net change exceeds Company B's by $90,000.

Question 14

The net rate of change of the number of employees at a corporation is modeled by E(t)=204tE'(t) = 20 - 4t people per month, where tt is the number of months since the start of the year. When E(t)>0E'(t) > 0, there is net hiring, and when E(t)<0E'(t) < 0, net departures exceed hires. What was the total number of employees who were either hired or who departed during the first 8 months of the year (0t80 \le t \le 8)?

  1. 18
  2. 32
  3. 50
  4. 68 (correct answer)
Explanation: This question asks for the total change, which means we need to calculate 08E(t)dt\int_{0}^{8} |E'(t)| \,dt. First, find where the rate changes sign: 204t=0    t=520 - 4t = 0 \implies t=5. For t[0,5]t \in [0, 5], E(t)0E'(t) \ge 0. For t[5,8]t \in [5, 8], E(t)0E'(t) \le 0. The integral is split: 05(204t)dt+58(204t)dt\int_{0}^{5} (20-4t) \,dt + \int_{5}^{8} -(20-4t) \,dt. First integral: [20t2t2]05=10050=50[20t-2t^2]_0^5 = 100-50 = 50. This represents 50 net hires. Second integral: 58(4t20)dt=[2t220t]58=(128160)(50100)=32(50)=18\int_{5}^{8} (4t-20) \,dt = [2t^2-20t]_5^8 = (128-160) - (50-100) = -32 - (-50) = 18. This represents 18 net departures. The total number of people moved (hired or departed) is the sum of the absolute values of these changes: 50+18=6850 + 18 = 68.

Question 15

The rate of change of inventory in a warehouse is modeled by the function I(t)I'(t) in items per day, where tt is the number of days since the beginning of the month. The function is positive when new stock arrives and negative when orders are shipped out. What is the correct business interpretation of the quantity 030I(t)dt\int_{0}^{30} |I'(t)| \,dt?

  1. The net change in the number of items in the warehouse during the 30-day period.
  2. The total number of items in the warehouse at the end of the 30-day period.
  3. The total number of items that were either received into inventory or shipped out of inventory during the 30-day period. (correct answer)
  4. The average rate at which items were moved (received or shipped) during the 30-day period.
Explanation: The integral of the absolute value of a rate function, abf(t)dt\int_{a}^{b} |f'(t)| \,dt, represents the total change or total magnitude of change over the interval [a,b][a, b]. In this context, I(t)I'(t) is the rate of inventory change. Integrating I(t)|I'(t)| sums the total items that came in (when I(t)>0I'(t)>0) and the total items that went out (when I(t)<0I'(t)<0), giving the total 'traffic' or movement of items.

Question 16

The rate at which water is draining from a tank is given by R(t)R(t) in liters per minute, where R(t)>0R(t) > 0 for all t0t \ge 0. Which statement provides the most accurate interpretation of the value calculated from 510R(t)dt\int_{5}^{10} R(t) \,dt?

  1. The total volume of water, in liters, that drained from the tank between the 5-minute mark and the 10-minute mark. (correct answer)
  2. The rate at which water is draining at t=10t=10 minutes minus the rate at which it was draining at t=5t=5 minutes.
  3. The average rate of drainage, in liters per minute, over the 5-minute interval from t=5t=5 to t=10t=10.
  4. The total amount of water remaining in the tank at the 10-minute mark.
Explanation: When you encounter a definite integral involving a rate function, you're dealing with the Fundamental Theorem of Calculus, which connects rates of change to accumulated quantities. The key insight is that integrating a rate gives you the total amount of change over that interval. Here, R(t)R(t) represents the rate of water drainage in liters per minute. When you integrate this rate from t=5t = 5 to t=10t = 10, you're finding the total amount of water that drained during those 5 minutes. Think of it this way: if water drains at varying rates over time, the integral "adds up" all those infinitesimal amounts of water that left the tank each instant. This makes choice A correct. Let's examine why the other options are wrong. Choice B describes R(10)R(5)R(10) - R(5), which would simply be the difference between two rate values—not an integral at all. Choice C confuses the integral with the average value formula; the average rate would be 15510R(t)dt\frac{1}{5}\int_{5}^{10} R(t) \,dt, requiring division by the interval length. Choice D misinterprets what the integral tells us—it gives the amount that drained out, not what remains in the tank. Remember this pattern: when you see a definite integral of a rate function, you're always finding the net change or total accumulation of the quantity being measured. Rate × time = amount, and integration extends this concept to varying rates over continuous time intervals.

Question 17

A company's rate of revenue is R(t)R'(t) and its rate of cost is C(t)C'(t), both in thousands of dollars per quarter, where tt is the number of quarters since the beginning of the fiscal year. The company's profit at the beginning of the year (t=0t=0) was P0P_0 thousand dollars. Which expression represents the company's total profit, in thousands of dollars, at the end of the second quarter (t=2t=2)?

  1. 02(R(t)C(t))dt\int_{0}^{2} (R'(t) - C'(t)) \,dt
  2. P0+02R(t)dtP_0 + \int_{0}^{2} R'(t) \,dt
  3. P0+02(R(t)C(t))dtP_0 + \int_{0}^{2} (R'(t) - C'(t)) \,dt (correct answer)
  4. P0+R(2)C(2)P_0 + R'(2) - C'(2)
Explanation: The total profit at time t=2t=2 is the initial profit P0P_0 plus the net change in profit from t=0t=0 to t=2t=2. The rate of change of profit is P(t)=R(t)C(t)P'(t) = R'(t) - C'(t). Therefore, the net change in profit is the definite integral of the rate of change of profit, which is 02(R(t)C(t))dt\int_{0}^{2} (R'(t) - C'(t)) \,dt. Adding this change to the initial profit gives the final profit: P0+02(R(t)C(t))dtP_0 + \int_{0}^{2} (R'(t) - C'(t)) \,dt.