Antiderivatives & Indefinite Integrals - Business Calculus

1. A new streaming service finds that the net rate of change of its subscribers is modeled by $S'(t) = (t+10)(40-t)$ subscribers per month, where $t$ is the number of months after launch. At launch ($t=0$), they had 500 'early adopter' subscribers. How many subscribers, to the nearest whole number, do they have after 10 months?
   Answer: 5,667 First, expand the expression for the rate of change: $S'(t) = 40t - t^2 + 400 - 10t = -t^2 + 30t + 400$. Next, integrate $S'(t)$ to find the subscriber function $S(t)$: $S(t) = \int (-t^2 + 30t + 400) \,dt = -\frac{t^3}{3} + \frac{30t^2}{2} + 400t + C = -\frac{1}{3}t^3 + 15t^2 + 400t + C$. We are given that $S(0) = 500$, so we can find $C$: $S(0) = 0 + 0 + 0 + C = 500$, which means $C=500$. The subscriber function is $S(t) = -\frac{1}{3}t^3 + 15t^2 + 400t + 500$. Finally, evaluate at $t=10$: $S(10) = -\frac{1}{3}(10)^3 + 15(10)^2 + 400(10) + 500 = -\frac{1000}{3} + 1500 + 4000 + 500 = -333.33... + 6000 = 5666.66...$. To the nearest whole number, there are 5,667 subscribers.
2. The rate of change of the marginal cost for producing a certain electronic component is constant at $C''(x) = 12$. For this component, the marginal cost at a production level of 10 units is $150, and fixed costs are $5,000. What is the total cost of producing 20 units?
   Answer: $8,000 This problem requires two integrations. First, integrate $C''(x)$ to find the marginal cost function $C'(x)$: $C'(x) = \int 12 \,dx = 12x + K$. Use the condition $C'(10) = 150$ to find the constant $K$: $12(10) + K = 150 \implies 120 + K = 150 \implies K = 30$. So, the marginal cost function is $C'(x) = 12x + 30$. Next, integrate $C'(x)$ to find the total cost function $C(x)$: $C(x) = \int (12x + 30) \,dx = 6x^2 + 30x + C$. The fixed costs are $5,000, which means $C(0) = 5000$. This gives $6(0)^2 + 30(0) + C = 5000$, so $C = 5000$. The total cost function is $C(x) = 6x^2 + 30x + 5000$. Finally, evaluate at $x=20$: $C(20) = 6(20)^2 + 30(20) + 5000 = 6(400) + 600 + 5000 = 2400 + 600 + 5000 = 8000$.
3. The marginal profit for a product is given by $P'(x) = \frac{4x^5 - 6x^3}{2x^2}$ dollars per unit. If the company incurs a net loss of $50 when no units are sold, what is the profit from selling 10 units?
   Answer: $4,800 First, simplify the expression for the marginal profit $P'(x)$ before integrating: $P'(x) = \frac{4x^5}{2x^2} - \frac{6x^3}{2x^2} = 2x^3 - 3x$. Now, integrate to find the profit function $P(x)$: $P(x) = \int (2x^3 - 3x)\,dx = \frac{2x^4}{4} - \frac{3x^2}{2} + C = 0.5x^4 - 1.5x^2 + C$. A loss of $50 when no units are sold means $P(0) = -50$. Using this initial condition: $0.5(0)^4 - 1.5(0)^2 + C = -50$, which implies $C = -50$. So, the profit function is $P(x) = 0.5x^4 - 1.5x^2 - 50$. To find the profit from selling 10 units, calculate $P(10)$: $P(10) = 0.5(10)^4 - 1.5(10)^2 - 50 = 0.5(10000) - 1.5(100) - 50 = 5000 - 150 - 50 = 4800$.
4. A company's marginal revenue from selling $x$ units of a product is $R'(x) = 20e^{0.1x} + 5$ dollars per unit. Assuming the revenue from selling zero units is $0, what is the total revenue from selling 20 units? Round your answer to the nearest dollar.
   Answer: $1,378 To find the revenue function $R(x)$, integrate the marginal revenue function: $R(x) = \int (20e^{0.1x} + 5) \,dx = \frac{20}{0.1}e^{0.1x} + 5x + C = 200e^{0.1x} + 5x + C$. Use the condition $R(0) = 0$ to find $C$: $200e^{0.1(0)} + 5(0) + C = 0 \implies 200e^0 + C = 0 \implies 200(1) + C = 0$, so $C = -200$. The revenue function is $R(x) = 200e^{0.1x} + 5x - 200$. To find the revenue from 20 units, calculate $R(20)$: $R(20) = 200e^{0.1(20)} + 5(20) - 200 = 200e^2 + 100 - 200 = 200e^2 - 100$. Using $e^2 \approx 7.389056$, we get $R(20) \approx 200(7.389056) - 100 = 1477.81 - 100 = 1377.81$. Rounded to the nearest dollar, the revenue is $1,378.
5. The rate of change of a company's debt is modeled by the function $D'(t) = 2000e^{-0.05t}$, where $t$ is time in years and $D(t)$ is the debt in dollars. If the company's initial debt is $10,000, which function correctly models the company's debt over time?
   Answer: $D(t) = -40000e^{-0.05t} + 50000$ To find the debt function $D(t)$, we integrate the rate of change $D'(t)$: $D(t) = \int 2000e^{-0.05t} \,dt$. Using the rule for integrating exponential functions, we get $D(t) = \frac{2000}{-0.05}e^{-0.05t} + C = -40000e^{-0.05t} + C$. The initial debt is $D(0) = 10000$. We use this to solve for $C$: $-40000e^{-0.05(0)} + C = 10000 \implies -40000e^0 + C = 10000 \implies -40000(1) + C = 10000 \implies C = 10000 + 40000 = 50000$. Therefore, the debt function is $D(t) = -40000e^{-0.05t} + 50000$.
6. The marginal cost for a manufacturer is given by $C'(x) = 12\sqrt{x} + 20$ dollars per unit, where $x$ is the number of units produced. If the company's fixed costs are $4,000, what is the total cost of producing 100 units?
   Answer: $14,000 First, find the total cost function $C(x)$ by integrating the marginal cost function $C'(x)$. Rewrite $C'(x) = 12x^{1/2} + 20$. The antiderivative is $C(x) = \int (12x^{1/2} + 20) \,dx = 12\frac{x^{3/2}}{3/2} + 20x + K = 8x^{3/2} + 20x + K$. The fixed costs represent the cost when $x=0$, so $C(0) = 4000$. Plugging this in: $8(0)^{3/2} + 20(0) + K = 4000$, which gives $K = 4000$. Thus, the total cost function is $C(x) = 8x^{3/2} + 20x + 4000$. Finally, evaluate this function at $x=100$: $C(100) = 8(100)^{3/2} + 20(100) + 4000 = 8(1000) + 2000 + 4000 = 8000 + 2000 + 4000 = 14,000$.
7. The rate of change of a worker's productivity is given by $P'(t) = 12 - 3t$ units per hour, per hour, where $t$ is the number of hours into an 8-hour shift. At the beginning of the shift ($t=0$), the worker's production rate is 20 units per hour. Assuming the worker has produced zero units at $t=0$, what is the total number of units produced after 4 hours?
   Answer: 144 This is a two-step integration problem. First, find the productivity function $P(t)$ by integrating its rate of change $P'(t)$: $P(t) = \int (12 - 3t)\,dt = 12t - \frac{3}{2}t^2 + C_1$. We are given $P(0)=20$, so $12(0) - 1.5(0)^2 + C_1 = 20$, which means $C_1 = 20$. Thus, the productivity rate is $P(t) = 12t - 1.5t^2 + 20$. The total number of units produced, $N(t)$, is the integral of the productivity rate $P(t)$. So, $N(t) = \int (12t - 1.5t^2 + 20)\,dt = \frac{12t^2}{2} - \frac{1.5t^3}{3} + 20t + C_2 = 6t^2 - 0.5t^3 + 20t + C_2$. Since $N(0)=0$, we find $C_2 = 0$. The total production function is $N(t) = 6t^2 - 0.5t^3 + 20t$. Finally, we evaluate $N(4)$: $N(4) = 6(4)^2 - 0.5(4)^3 + 20(4) = 6(16) - 0.5(64) + 80 = 96 - 32 + 80 = 144$.
8. The velocity of a particle is given by $v(t) = 6t^2 - 8t + \frac{12}{\sqrt{t}}$ for $t > 0$. If the particle's position at $t = 1$ is 15 units, what is the position function $s(t)$?
   Answer: $s(t) = 2t^3 - 4t^2 + 24\sqrt{t} - 7$ Integrate v(t): ∫(6t² - 8t + $12t^{-1/2}$)dt = 2t³ - 4t² + 12($2t^{1/2}$) + C = 2t³ - 4t² + 24√t + C. Using s(1) = 15: 2(1) - 4(1) + 24(1) + C = 15, so 2 - 4 + 24 + C = 15, giving 22 + C = 15, so C = -7. Choice A has wrong sign for constant, Choice C uses wrong coefficient for √t term, Choice D fails to integrate properly and uses wrong constant.
9. The rate of change of a population $P(t)$ is given by $\frac{dP}{dt} = 3t^2 + 4e^{2t} - \frac{8}{t+1}$. If $P(0) = 100$, which expression represents the population function?
   Answer: $P(t) = t^3 + 2e^{2t} - 8\ln|t+1| + 98$ Integrate dP/dt: ∫(3t² + $4e^{2t}$ - 8/(t+1))dt = t³ + $2e^{2t}$ - 8ln|t+1| + C. Using P(0) = 100: 0 + 2e⁰ - 8ln(1) + C = 100, so 2 + 0 + C = 100, giving C = 98. Choice B miscalculates the constant. Choice C has wrong coefficient for $e^{2t}$. Choice D fails to integrate correctly.
10. The value of an asset is changing at a rate of $V'(t) = \frac{150}{t+1}$ dollars per year, where $t$ is the number of years after its purchase in 2020. If the asset was purchased for $5,000, what is its value in the year 2029?
    Answer: $150 \ln(10) + 5000$ To find the value function $V(t)$, we integrate the rate of change $V'(t)$. The integral is $V(t) = \int \frac{150}{t+1} \,dt = 150 \ln|t+1| + C$. Since $t \ge 0$, we can write $V(t) = 150 \ln(t+1) + C$. The initial value at purchase ($t=0$, year 2020) was $5,000. So, $V(0) = 5000$. We use this to find $C$: $150 \ln(0+1) + C = 5000 \implies 150 \ln(1) + C = 5000 \implies 0 + C = 5000$, so $C=5000$. The value function is $V(t) = 150 \ln(t+1) + 5000$. The year 2029 corresponds to $t = 2029 - 2020 = 9$. We need to find $V(9)$: $V(9) = 150 \ln(9+1) + 5000 = 150 \ln(10) + 5000$.
11. Suppose $F(x)$ and $G(x)$ are two different antiderivatives of the same function $f(x)$. Given that $F(1) = 5$ and $G(1) = 12$, what is the value of the expression $F(10) - G(10)$?
    Answer: -7 A fundamental property of antiderivatives is that any two antiderivatives of the same function $f(x)$ on an interval differ by a constant. This means that $G(x) = F(x) + C$ for some constant $C$. We can find the value of $C$ using the given information at $x=1$: $G(1) = F(1) + C \implies 12 = 5 + C \implies C = 7$. Thus, for any value of $x$, we have $G(x) = F(x) + 7$. The question asks for the value of $F(10) - G(10)$. We can rearrange the relationship to $F(x) - G(x) = -C$. Since $C=7$, the expression $F(x) - G(x)$ is always equal to $-7$, regardless of the value of $x$. Therefore, $F(10) - G(10) = -7$.
12. A function $f(x)$ has the property that $f'(x) = \frac{x^3 - 2x}{x^2 + 1}$. Using polynomial long division, which of the following represents $\int f'(x)dx$?
    Answer: $\frac{x^2}{2} - \frac{3}{2}\ln(x^2 + 1) + C$ Using polynomial long division: (x³ - 2x) ÷ (x² + 1) = x - (3x)/(x² + 1). So f'(x) = x - 3x/(x² + 1). Integrating: ∫[x - 3x/(x² + 1)]dx = x²/2 - (3/2)∫(2x)/(x² + 1)dx = x²/2 - (3/2)ln(x² + 1) + C. Since x² + 1 > 0 for all real x, we don't need absolute value bars. Choice B unnecessarily includes absolute value, Choice C has wrong coefficient, Choice D incorrectly performs the division.
13. The marginal profit for a product is given by $P'(x) = -0.6x^2 + 80x - 100$ dollars per unit. The company found that the total profit from selling the first 10 units is $3,500. Which of the following is the profit function $P(x)$?
    Answer: $P(x) = -0.2x^3 + 40x^2 - 100x + 700$ First, find the general antiderivative of the marginal profit function: $P(x) = \int (-0.6x^2 + 80x - 100) \,dx = -\frac{0.6x^3}{3} + \frac{80x^2}{2} - 100x + C = -0.2x^3 + 40x^2 - 100x + C$. Next, use the given condition $P(10) = 3500$ to solve for the constant of integration $C$. $P(10) = -0.2(10)^3 + 40(10)^2 - 100(10) + C = 3500$. This simplifies to $-0.2(1000) + 40(100) - 1000 + C = 3500$, which is $-200 + 4000 - 1000 + C = 3500$. Combining terms gives $2800 + C = 3500$, so $C = 700$. Therefore, the profit function is $P(x) = -0.2x^3 + 40x^2 - 100x + 700$.
14. Let $F(x)$ be an antiderivative of a function $f(x)$. Which of the following expressions represents a general antiderivative of the function $g(x) = f(x) - 7$?
    Answer: $F(x) - 7x$ To find an antiderivative of $g(x)$, we must compute the indefinite integral $\int g(x) \,dx$. Substituting the expression for $g(x)$, we have $\int (f(x) - 7) \,dx$. By the difference rule for integration, this is equivalent to $\int f(x) \,dx - \int 7 \,dx$. We are given that $F(x)$ is an antiderivative of $f(x)$, so $\int f(x) \,dx = F(x) + C_1$. The integral of the constant is $\int 7 \,dx = 7x + C_2$. Combining these, the general antiderivative of $g(x)$ is $(F(x) + C_1) - (7x + C_2) = F(x) - 7x + (C_1 - C_2)$. Since $C_1 - C_2$ is just another arbitrary constant, we can write the family of antiderivatives as $F(x) - 7x + C$. One such antiderivative is $F(x) - 7x$.
15. A company's marginal cost function is $C'(x) = 50 + \frac{18}{x^2}$ dollars per unit, where $x$ is the number of units produced. The total cost to produce 1 unit is $100. What is the average cost per unit when 3 units are produced? Round to the nearest cent.
    Answer: $70.67 First, find the cost function $C(x)$ by integrating $C'(x) = 50 + 18x^{-2}$. $C(x) = \int (50 + 18x^{-2}) \,dx = 50x + \frac{18x^{-1}}{-1} + K = 50x - \frac{18}{x} + K$. Use the condition $C(1)=100$ to find the constant $K$: $50(1) - \frac{18}{1} + K = 100 \implies 32 + K = 100 \implies K = 68$. The cost function is $C(x) = 50x - \frac{18}{x} + 68$. Now, find the total cost of producing 3 units: $C(3) = 50(3) - \frac{18}{3} + 68 = 150 - 6 + 68 = 212$. The average cost, $\bar{C}(x)$, is the total cost divided by the number of units, $\bar{C}(x) = \frac{C(x)}{x}$. So, the average cost for 3 units is $\bar{C}(3) = \frac{C(3)}{3} = \frac{212}{3} \approx 70.67$.

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