Definite Integral as Accumulation - Business Calculus

1. Let $P(t)$ be the total number of subscribers to a streaming service at time $t$ (in years), and let its rate of change be $P'(t) = p(t)$. Which of the following expressions represents the average number of subscribers added per year between the end of the first quarter ($t=0.25$) and the end of the third quarter ($t=0.75$)?
   Answer: $\frac{P(0.75) - P(0.25)}{0.5}$ When you see a question about "average rate of change" in calculus, you're dealing with the fundamental concept of how much a quantity changes per unit of time over an interval. The correct approach uses the definition of average rate of change: $\frac{\text{change in quantity}}{\text{change in time}}$. Since $P(t)$ represents the total number of subscribers at time $t$, the change in subscribers from $t = 0.25$ to $t = 0.75$ is $P(0.75) - P(0.25)$. The time interval is $0.75 - 0.25 = 0.5$ years. Therefore, the average number of subscribers added per year is $\frac{P(0.75) - P(0.25)}{0.5}$, making D correct. Option A represents the total change in subscribers over the interval (by the Fundamental Theorem of Calculus), but doesn't divide by the time interval to get the average rate. Option B uses $p(t)$ values instead of $P(t)$ values—this gives you the change in the rate of subscriber growth, not the change in total subscribers. Option C attempts to average the rates at the endpoints, but this doesn't give you the average rate of change of the original function $P(t)$. Remember: average rate of change always follows the slope formula pattern—change in output divided by change in input. Don't confuse the function $P(t)$ with its derivative $p(t) = P'(t)$, and make sure you're calculating what the question actually asks for.
2. A company's net earnings are being generated at a rate of $P'(t) = 1.2 - 0.3\sqrt{t}$ million dollars per quarter, where $t$ is the number of quarters from the start of the fiscal year. After how many quarters, $T$, will the total accumulated net earnings for the fiscal year first return to zero?
   Answer: 36 quarters Total accumulated net earnings are found by integrating the rate of earnings, $P'(t)$, from $t=0$ to $t=T$. We need to find the value of $T > 0$ for which this integral is zero. Set up the equation: $\int_0^T (1.2 - 0.3t^{1/2}) dt = 0$. First, find the antiderivative: $[1.2t - 0.3\frac{t^{3/2}}{3/2}]_0^T = [1.2t - 0.2t^{3/2}]_0^T$. Evaluate at the limits: $(1.2T - 0.2T^{3/2}) - (0) = 0$. $1.2T - 0.2T^{3/2} = 0$. Factor out $T$: $T(1.2 - 0.2T^{1/2}) = 0$. Since we are looking for a time after the start, we need $T>0$. So we solve the other part: $1.2 - 0.2T^{1/2} = 0$. $1.2 = 0.2T^{1/2}$. $\sqrt{T} = \frac{1.2}{0.2} = 6$. $T = 6^2 = 36$. So, the accumulated earnings will return to zero after 36 quarters.
3. The rate of change of a company's customer base is given by $N'(t) = 150 - 6t^2$ customers per month, where $t$ is the number of months since the beginning of the year. How does the net change in customers during the first six months ($t=0$ to $t=6$) compare to the net change during the second six months ($t=6$ to $t=12$)?
   Answer: The net change was greater in the first six months by 2,556 customers. We need to calculate the definite integral for each six-month period and then compare them. Period 1 (first six months, $t=0$ to $t=6$): $\int_0^6 (150 - 6t^2) dt = [150t - 2t^3]_0^6 = (150(6) - 2(6^3)) - 0 = 900 - 2(216) = 900 - 432 = 468$. So, there was a net gain of 468 customers. Period 2 (second six months, $t=6$ to $t=12$): $\int_6^{12} (150 - 6t^2) dt = [150t - 2t^3]_6^{12}$. Value at $t=12$: $150(12) - 2(12^3) = 1800 - 2(1728) = 1800 - 3456 = -1656$. Value at $t=6$: $150(6) - 2(6^3) = 432$. Net change = $(-1656) - (432) = -2088$. So, there was a net loss of 2,088 customers. To compare the net changes, we find the difference: $468 - (-2088) = 468 + 2088 = 2556$. The net change in the first six months was greater than the net change in the second six months by 2,556 customers.
4. A company's marginal cost function is $C'(x) = 12 + 0.02x$ dollars per unit, where $x$ is the number of units produced. If the cost to produce the first 100 units is $1,350, what is the total cost to produce 300 units?
   Answer: $4,950 The cost to produce units 101-300 is $\int_{100}^{300} (12 + 0.02x) dx = [12x + 0.01x^2]_{100}^{300} = (3600 + 900) - (1200 + 100) = 4500 - 1300 = 3200$. Total cost for 300 units = $1,350 + $3,200 = $4,950. Choice B incorrectly uses the integral from 0 to 200. Choice C adds the wrong interval cost. Choice D uses only the additional cost without the initial $1,350.
5. The rate of change of a company's inventory level is $I'(t) = 100 - 20t$ units per day, where $t$ is days since the beginning of the week. If the company wants to ensure that the net change in inventory over any 3-consecutive-day period is non-negative, what is the latest day of the week they can start such a period?
   Answer: Tuesday (day 2) Net change over 3 days starting at day $a$: $\int_a^{a+3} (100 - 20t) dt = [100t - 10t^2]_a^{a+3} = 100(3) - 10[(a+3)^2 - a^2] = 300 - 10(6a + 9) = 300 - 60a - 90 = 210 - 60a$. For non-negative: $210 - 60a ≥ 0$, so $a ≤ 3.5$. Latest integer day is 3 (Wednesday). But if Monday = day 1, then Tuesday = day 2, so latest start is Tuesday. Choice A uses Wednesday but this gives negative change. Choice B and D give even more negative changes.
6. The value of a corporate jet, $V(t)$, is depreciating at a rate given by $V'(t) = -40000(t+1)^{-1.2}$ dollars per year, where $t$ is the age of the jet in years. What is the total loss in value of the jet during its third and fourth years of service combined?
   Answer: The value decreases by approximately $15,600. The total loss in value during the third and fourth years corresponds to the accumulated change from the end of year 2 ($t=2$) to the end of year 4 ($t=4$). This is calculated by the definite integral $\int_2^4 V'(t) dt$. $\int_2^4 -40000(t+1)^{-1.2} dt = [-40000 \frac{(t+1)^{-0.2}}{-0.2}]_2^4 = [200000(t+1)^{-0.2}]_2^4$ Evaluate at the limits: $200000(4+1)^{-0.2} - 200000(2+1)^{-0.2} = 200000(5^{-0.2} - 3^{-0.2})$ Using a calculator: $200000(0.72477 - 0.80274) = 200000(-0.07797) \approx -15594$. The integral's value is approximately $-15,600, which represents a decrease in value. The question asks for the 'loss in value', which is the positive amount by which the value has decreased. Therefore, the loss is approximately $15,600.
7. A startup's user base grows at a rate of $g(t)$ new users per day. For the first 30 days ($0 \le t \le 30$), the rate is $g(t) = 4t+10$. Due to a successful marketing campaign, the rate for $t>30$ is $g(t) = -0.5t^2 + 60t - 1190$. What was the total number of new users that joined between day 20 and day 40?
   Answer: 4,033 The total number of new users is the integral of the rate function $g(t)$. Since the function is piecewise, the integral must be split at $t=30$. We need to calculate $\int_{20}^{30} g_1(t) dt + \int_{30}^{40} g_2(t) dt$. First integral (from $t=20$ to $t=30$): $\int_{20}^{30} (4t+10) dt = [2t^2+10t]_{20}^{30} = (2(30^2)+10(30)) - (2(20^2)+10(20)) = (1800+300) - (800+200) = 2100 - 1000 = 1100$. Second integral (from $t=30$ to $t=40$): $\int_{30}^{40} (-0.5t^2+60t-1190) dt = [-\frac{0.5t^3}{3} + 30t^2 - 1190t]_{30}^{40}$. Value at $t=40$: $-\frac{1}{6}(40^3) + 30(40^2) - 1190(40) = -10666.67 + 48000 - 47600 \approx -10266.67$. Value at $t=30$: $-\frac{1}{6}(30^3) + 30(30^2) - 1190(30) = -4500 + 27000 - 35700 = -13200$. The value of the second integral is $(-10266.67) - (-13200) = 2933.33$. Total new users = $1100 + 2933.33 \approx 4033$.
8. A company's daily cash flow is modeled by the rate function $f(t) = t^2 - 8t + 12$ thousand dollars per day, where $t$ is the number of days into a project. The project experiences a period of negative cash flow, during which its cash reserves are depleted. What is the total reduction in cash reserves during this period of negative cash flow?
   Answer: $10,667 First, we need to find the period of negative cash flow by finding when $f(t) < 0$. We find the roots of $f(t) = t^2 - 8t + 12 = 0$. Factoring gives $(t-2)(t-6)=0$, so the roots are $t=2$ and $t=6$. Since the parabola opens upwards, the function is negative between $t=2$ and $t=6$. The total reduction in cash reserves is the absolute value of the integral of $f(t)$ over this interval. Total Change = $\int_2^6 (t^2 - 8t + 12) dt$. Find the antiderivative: $[\frac{t^3}{3} - 4t^2 + 12t]_2^6$. Evaluate at the limits: Value at $t=6$: $\frac{6^3}{3} - 4(6^2) + 12(6) = \frac{216}{3} - 4(36) + 72 = 72 - 144 + 72 = 0$. Value at $t=2$: $\frac{2^3}{3} - 4(2^2) + 12(2) = \frac{8}{3} - 16 + 24 = \frac{8}{3} + 8 = \frac{32}{3}$. The value of the integral is $0 - \frac{32}{3} = -\frac{32}{3} \approx -10.667$. The function gives the rate in thousands of dollars, so the change is $-10.667 \times 1000 = -10,667$. The question asks for the total reduction, which is the positive value, $10,667.
9. The rate of revenue flow for a new software product is modeled by $R'(t) = 3000e^{-0.1t}$ dollars per month, where $t$ is the number of months since the product's launch. Which of the following best approximates the total revenue generated during the second year after launch (from $t=12$ to $t=24$)?
   Answer: $6,320 The total revenue generated over a period is the definite integral of the rate of revenue flow. To find the revenue during the second year, we must integrate $R'(t)$ from $t=12$ to $t=24$. $\int_{12}^{24} 3000e^{-0.1t} dt = [\frac{3000}{-0.1}e^{-0.1t}]_{12}^{24} = [-30000e^{-0.1t}]_{12}^{24}$ Now, evaluate at the limits of integration: $(-30000e^{-0.1(24)}) - (-30000e^{-0.1(12)}) = -30000(e^{-2.4} - e^{-1.2})$ Using a calculator, $e^{-2.4} \approx 0.090718$ and $e^{-1.2} \approx 0.301194$. Total Revenue $\approx -30000(0.090718 - 0.301194) = -30000(-0.210476) \approx 6314.28$. This is approximately $6,320.
10. A distribution center starts the week (Monday, $t=0$) with 5,000 units of a product. The net rate of change of inventory is given by $I'(t) = 200 - 50\sqrt{t}$ units per day, for $0 \le t \le 7$. What is the approximate inventory level at the end of the day on Friday ($t=5$)?
    Answer: 5,627 units The inventory level at time $t$ is the initial inventory plus the accumulated change. The accumulated change is the definite integral of the rate of change from $t=0$ to $t=5$. Final Inventory $I(5) = I(0) + \int_0^5 I'(t) dt$. First, calculate the integral: $\int_0^5 (200 - 50t^{1/2}) dt = [200t - 50 \frac{t^{3/2}}{3/2}]_0^5 = [200t - \frac{100}{3}t^{3/2}]_0^5$ Evaluate at $t=5$: $200(5) - \frac{100}{3}(5^{3/2}) = 1000 - \frac{100}{3}(5\sqrt{5}) \approx 1000 - \frac{100}{3}(11.1803) \approx 1000 - 372.68 = 627.32$. This value, 627.32, is the net change in inventory over the 5 days. To find the final inventory level, add this change to the initial amount: $I(5) = 5000 + 627.32 \approx 5627$ units.
11. An oil spill spreads such that the rate of change of the affected area is $A'(t) = 50\sqrt{t + 1}$ square meters per hour, where $t$ is hours since the spill began. Environmental regulations require cleanup to begin when the total affected area reaches 400 square meters. If the spill initially affects 25 square meters, approximately when must cleanup begin?
    Answer: After 3.8 hours Total area at time $t$ = $25 + \int_0^t 50\sqrt{s+1} ds = 25 + 50 \cdot \frac{2}{3}[(s+1)^{3/2}]_0^t = 25 + \frac{100}{3}[(t+1)^{3/2} - 1]$. We need this to equal 400: $25 + \frac{100}{3}[(t+1)^{3/2} - 1] = 400$. Solving: $\frac{100}{3}[(t+1)^{3/2} - 1] = 375$, so $(t+1)^{3/2} - 1 = 11.25$, giving $(t+1)^{3/2} = 12.25$. Therefore $t+1 = (12.25)^{2/3} ≈ 4.84$, so $t ≈ 3.84$ hours. Choice A uses incorrect integration. Choice C makes an algebraic error. Choice D forgets the initial 25 square meters.

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