Net Change & Total Change - Business Calculus

1. 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 $S'(t) = 20t^2 - 120t + 1180$ units per week, for $0 \le t \le 6$. If the initial inventory at $t=0$ was 500 units, what is the inventory level at the end of week 6?
   Answer: 140 units The rate of change of inventory is $I'(t) = (\text{rate in}) - (\text{rate out}) = 1000 - (20t^2 - 120t + 1180) = -20t^2 + 120t - 180$. The net change in inventory over 6 weeks is $\int_0^6 I'(t) dt = [-\frac{20}{3}t^3 + 60t^2 - 180t]_0^6 = -\frac{20}{3}(6^3) + 60(6^2) - 180(6) = -1440 + 2160 - 1080 = -360$ units. The final inventory is the initial inventory plus the net change: $I(6) = I(0) + \text{Net Change} = 500 + (-360) = 140$ units.
2. A company's cash reserves change based on its rate of income, $I'(t) = 25e^{-0.05t}$, and its rate of expenditure, $E'(t) = 10 + 0.2t$, both in thousands of dollars per month. What is the net change in the company's cash reserves over the first six months (from $t=0$ to $t=6$)?
   Answer: An increase of $66.0 thousand The rate of change of cash reserves, $S'(t)$, is the rate of income minus the rate of expenditure: $S'(t) = I'(t) - E'(t) = 25e^{-0.05t} - (10 + 0.2t)$. The net change is the integral of this rate from $t=0$ to $t=6$: $\int_0^6 (25e^{-0.05t} - 10 - 0.2t) dt$. The antiderivative is $[\frac{25}{-0.05}e^{-0.05t} - 10t - 0.1t^2]_0^6 = [-500e^{-0.05t} - 10t - 0.1t^2]_0^6$. Evaluating at the limits: $(-500e^{-0.3} - 60 - 3.6) - (-500e^0) = -500e^{-0.3} - 63.6 + 500 \approx -500(0.7408) - 63.6 + 500 = -370.4 - 63.6 + 500 = 66.0$. The net change is an increase of $66.0 thousand.
3. The rate at which water is draining from a tank is given by $R(t)$ in liters per minute, where $R(t) > 0$ for all $t \ge 0$. Which statement provides the most accurate interpretation of the value calculated from $\int_{5}^{10} R(t) \,dt$?
   Answer: The total volume of water, in liters, that drained from the tank between the 5-minute mark and the 10-minute mark. 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, $R(t)$ represents the rate of water drainage in liters per minute. When you integrate this rate from $t = 5$ to $t = 10$, 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 $R(10) - R(5)$, 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 $\frac{1}{5}\int_{5}^{10} R(t) \,dt$, 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.
4. The value of a piece of industrial equipment, $V(t)$, is decreasing at a rate given by $V'(t) = -1000(t+2)^{-2}$ dollars per year, where $t$ is the number of years since its purchase. What is the total loss in value of the equipment during its third year of operation?
   Answer: $50 The 'third year' corresponds to the time interval from $t=2$ to $t=3$. The net change in value during this period is $\int_{2}^{3} V'(t) \,dt = \int_{2}^{3} -1000(t+2)^{-2} \,dt$. The antiderivative of $-1000(t+2)^{-2}$ is $1000(t+2)^{-1}$. Evaluating the definite integral: $[1000(t+2)^{-1}]_{2}^{3} = \frac{1000}{3+2} - \frac{1000}{2+2} = \frac{1000}{5} - \frac{1000}{4} = 200 - 250 = -50$. The value decreased by $50, so the loss in value is a positive quantity, $50.
5. The net rate of change of the number of employees at a corporation is modeled by $E'(t) = 20 - 4t$ people per month, where $t$ is the number of months since the start of the year. When $E'(t) > 0$, there is net hiring, and when $E'(t) < 0$, 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 ($0 \le t \le 8$)?
   Answer: 68 This question asks for the total change, which means we need to calculate $\int_{0}^{8} |E'(t)| \,dt$. First, find where the rate changes sign: $20 - 4t = 0 \implies t=5$. For $t \in [0, 5]$, $E'(t) \ge 0$. For $t \in [5, 8]$, $E'(t) \le 0$. The integral is split: $\int_{0}^{5} (20-4t) \,dt + \int_{5}^{8} -(20-4t) \,dt$. First integral: $[20t-2t^2]_0^5 = 100-50 = 50$. This represents 50 net hires. Second integral: $\int_{5}^{8} (4t-20) \,dt = [2t^2-20t]_5^8 = (128-160) - (50-100) = -32 - (-50) = 18$. This represents 18 net departures. The total number of people moved (hired or departed) is the sum of the absolute values of these changes: $50 + 18 = 68$.
6. Let $A(t)$ be the amount of oil, in barrels, in a storage tank at time $t$ in hours. Oil flows into the tank at a rate of $I(t)$ barrels per hour and flows out at a rate of $O(t)$ barrels per hour. Which expression represents the net change in the amount of oil in the tank from $t=2$ to $t=5$?
   Answer: $\int_{2}^{5} (I(t) - O(t)) \,dt$ The net rate of change of the amount of oil in the tank is the rate in minus the rate out, which is $A'(t) = I(t) - O(t)$. 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 $t=2$ to $t=5$ is $\int_{2}^{5} A'(t) \,dt = \int_{2}^{5} (I(t) - O(t)) \,dt$.
7. The rate of change of profit for Company A is $P_A'(t) = 10t + 5$ and for Company B is $P_B'(t) = t^2 + 8$, where profit is in thousands of dollars and $t$ is the number of years from now ($t \ge 0$). Over the time interval from $t=0$ to $t=6$, by how much does the net change in profit for Company A exceed the net change in profit for Company B?
   Answer: $90,000 in favor of Company A We need to find the difference between the net changes in profit: $\int_{0}^{6} P_A'(t) \,dt - \int_{0}^{6} P_B'(t) \,dt$. This can be computed as a single integral of the difference of the rates: $\int_{0}^{6} (P_A'(t) - P_B'(t)) \,dt = \int_{0}^{6} ((10t + 5) - (t^2 + 8)) \,dt = \int_{0}^{6} (-t^2 + 10t - 3) \,dt$. Evaluating the integral: $[-\frac{t^3}{3} + 5t^2 - 3t]_{0}^{6} = (-\frac{6^3}{3} + 5(6^2) - 3(6)) - 0 = -72 + 180 - 18 = 90$. 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.
8. The rate of change of inventory in a warehouse is modeled by the function $I'(t)$ in items per day, where $t$ 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 $\int_{0}^{30} |I'(t)| \,dt$?
   Answer: The total number of items that were either received into inventory or shipped out of inventory during the 30-day period. The integral of the absolute value of a rate function, $\int_{a}^{b} |f'(t)| \,dt$, represents the total change or total magnitude of change over the interval $[a, b]$. In this context, $I'(t)$ is the rate of inventory change. Integrating $|I'(t)|$ sums the total items that came in (when $I'(t)>0$) and the total items that went out (when $I'(t)<0$), giving the total 'traffic' or movement of items.
9. A new technology startup has a continuous revenue stream with a rate of flow modeled by $R'(t) = 150e^{-0.1t}$ thousands of dollars per month, where $t$ is in months. The project had initial start-up costs, resulting in a total revenue of -$50$ thousand (a loss) at time $t=0$. What is the total accumulated revenue, in thousands of dollars, after the first 10 months?
   Answer: $898.2 Total accumulated revenue at time $t=10$ is given by $R(10) = R(0) + \int_0^{10} R'(t) dt$. First, calculate the net change in revenue over the 10 months: $\int_0^{10} 150e^{-0.1t} dt = 150 [\frac{e^{-0.1t}}{-0.1}]_0^{10} = -1500[e^{-1} - e^0] = -1500(e^{-1} - 1) \approx 948.18$. Now, add the initial revenue: $R(10) = -50 + 948.18 = 898.18$. The total revenue is approximately $898.2 thousand.
10. The rate of production at a factory is given by $P(t)$ units per hour. The selling price for each unit produced at time $t$ is given by $S(t)$ dollars per unit. Which of the following expressions represents the total revenue generated by the factory from time $t=a$ to $t=b$?
    Answer: $\int_{a}^{b} P(t)S(t) \,dt$ To find the total revenue, we must first determine the rate at which revenue is generated. At any given time $t$, the rate of revenue is (units per hour) × (dollars per unit), which is $P(t)S(t)$ in dollars per hour. The total revenue accumulated over the interval $[a, b]$ is the definite integral of this rate of revenue. Therefore, the total revenue is $\int_{a}^{b} P(t)S(t) \,dt$.
11. A company's rate of profit is modeled by $P'(t) = 1.2 - 0.6t$ million dollars per year, where $t$ 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?
    Answer: $0 million The net change in profit is the definite integral of the rate of profit from $t=0$ to $t=4$. We calculate $\int_0^4 (1.2 - 0.6t) dt$. The antiderivative is $P(t) = 1.2t - 0.3t^2$. Evaluating the integral: $[1.2t - 0.3t^2]_0^4 = (1.2(4) - 0.3(4^2)) - (0) = 4.8 - 0.3(16) = 4.8 - 4.8 = 0$. 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.
12. A company's rate of revenue is $R'(t)$ and its rate of cost is $C'(t)$, both in thousands of dollars per quarter, where $t$ is the number of quarters since the beginning of the fiscal year. The company's profit at the beginning of the year ($t=0$) was $P_0$ thousand dollars. Which expression represents the company's total profit, in thousands of dollars, at the end of the second quarter ($t=2$)?
    Answer: $P_0 + \int_{0}^{2} (R'(t) - C'(t)) \,dt$ The total profit at time $t=2$ is the initial profit $P_0$ plus the net change in profit from $t=0$ to $t=2$. The rate of change of profit is $P'(t) = R'(t) - C'(t)$. Therefore, the net change in profit is the definite integral of the rate of change of profit, which is $\int_{0}^{2} (R'(t) - C'(t)) \,dt$. Adding this change to the initial profit gives the final profit: $P_0 + \int_{0}^{2} (R'(t) - C'(t)) \,dt$.
13. The marginal cost for a company to produce $q$ units of a product is given by $C'(q) = 0.03q^2 - 1.2q + 15$ dollars per unit. What is the total increase in cost if the company raises its production level from $q=20$ to $q=40$ units?
    Answer: $140 The total increase in cost is the net change in the cost function from $q=20$ to $q=40$. This is found by integrating the marginal cost function over this interval: $\int_{20}^{40} C'(q) dq = \int_{20}^{40} (0.03q^2 - 1.2q + 15) dq$. The antiderivative is $C(q) = 0.01q^3 - 0.6q^2 + 15q$. Evaluating the definite integral: $C(40) - C(20) = (0.01(40)^3 - 0.6(40)^2 + 15(40)) - (0.01(20)^3 - 0.6(20)^2 + 15(20)) = (640 - 960 + 600) - (80 - 240 + 300) = 280 - 140 = 140$. Thus, the cost increases by $140.
14. $V(t)$ represents the value of a commercial property in thousands of dollars, $t$ years after its purchase. Financial analysts determine that $\int_5^{10} V'(t) dt = -150$. Which of the following statements is the correct interpretation of this result?
    Answer: The property's value decreased by a total of $150,000 between the end of year 5 and the end of year 10. By the Net Change Theorem, the integral of a rate of change function $V'(t)$ from $t=a$ to $t=b$ gives the net change in the original function, $V(b) - V(a)$. Therefore, $\int_5^{10} V'(t) dt = V(10) - V(5) = -150$. This means the change in value from the end of year 5 to the end of year 10 is -$150$ 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.
15. The rate of change of items in a warehouse is given by $I'(t) = 3t^2 - 18t + 15$ items per hour for an 8-hour workday, $0 \le t \le 8$. 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?
    Answer: 34 items Total items moved is the integral of the absolute value of the rate, $\int_0^4 |I'(t)| dt$. First, find where $I'(t)$ is positive or negative by finding its roots: $3t^2 - 18t + 15 = 3(t-1)(t-5) = 0$, so roots are at $t=1$ and $t=5$. On the interval $[0, 4]$, $I'(t) > 0$ for $[0, 1)$ and $I'(t) < 0$ for $(1, 4]$. The integral becomes $\int_0^1 (3t^2 - 18t + 15) dt + \int_1^4 -(3t^2 - 18t + 15) dt$. Let $F(t) = t^3 - 9t^2 + 15t$. The first integral is $F(1) - F(0) = 7-0=7$. The second integral is $-(F(4)-F(1)) = -(-20 - 7) = -(-27) = 27$. The total number of items moved is the sum of the magnitudes of these changes: $7 + 27 = 34$.
16. A company's marginal profit function is $P'(q)$ dollars per unit, where $q$ is the number of units produced. Which of the following is the best interpretation of the statement $\int_{500}^{600} P'(q) dq = 4000$?
    Answer: The total profit increases by $4000 when production is increased from 500 to 600 units. 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, $\int_{500}^{600} P'(q) dq$ represents the total change in profit, $P(600) - P(500)$, when the production level goes from 500 to 600 units. The value of $4000 indicates that this change is an increase of $4000.
17. The population of subscribers to a streaming service is growing at a rate of $N'(t) = \frac{1000}{\sqrt{t+1}}$ subscribers per month, where $t$ is the number of months since the service launched. Three months after launch ($t=3$), the service had 2500 subscribers. How many subscribers, to the nearest whole number, did the service have after one year ($t=12$)?
    Answer: 5711 We need to find $N(12)$, given $N(3)=2500$. The value is $N(12) = N(3) + \int_3^{12} N'(t) dt$. First, calculate the integral representing the net change in subscribers from month 3 to month 12: $\int_3^{12} 1000(t+1)^{-1/2} dt = 1000[2(t+1)^{1/2}]_3^{12} = 2000[\sqrt{12+1} - \sqrt{3+1}] = 2000(\sqrt{13} - \sqrt{4}) \approx 2000(3.60555 - 2) = 2000(1.60555) \approx 3211.1$. This is the increase in subscribers. The total number is the initial amount at $t=3$ plus this increase: $N(12) = 2500 + 3211.1 = 5711.1$. To the nearest whole number, this is 5711.

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