Business Calculus Quiz: Antiderivatives And Indefinite Integrals

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Antiderivatives And Indefinite IntegralsQuestion 1 of 15

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?

150 \ln(10) + 5000

150 \ln(9) + 5000

150 \ln(5) + 5000

150 \ln(2030) + 5000

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Business Calculus Quiz

Business Calculus Quiz: Antiderivatives And Indefinite Integrals

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

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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.

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Question 1

$150 \ln(10) + 5000$ (correct answer)
$150 \ln(9) + 5000$
$150 \ln(5) + 5000$
$150 \ln(2030) + 5000$

Explanation: 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

:

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$.

Question 2

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 $P(x)$?

$P(x) = -0.2x^3 + 40x^2 - 100x + 3500$
$P(x) = -0.2x^3 + 40x^2 - 100x + 700$ (correct answer)
$P(x) = -0.6x^3 + 80x^2 - 100x - 2900$
$P(x) = -0.2x^3 + 40x^2 - 100x + 6300$

Explanation: 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

Question 3

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?

$10,000
$12,000
$14,000 (correct answer)
$6,240

Explanation: 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

Question 4

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?

44
64
144 (correct answer)
176

Explanation: 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

Question 5

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?

$7,400
$8,000 (correct answer)
$10,400
$13,000

Explanation: 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$.

Question 6

Let $F(x)$ be an antiderivative of a function $f(x)$ . Which of the following expressions represents a general antiderivative of the function $g(x) = f(x) - 7$ ?

$F(x) - 7$
$F(x-7)$
$-7F(x)$
$F(x) - 7x$ (correct answer)

Explanation: 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)

(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

Question 7

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.

$48.00
$68.59
$70.67 (correct answer)
$212.00

Explanation: 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

Question 8

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?

5,167
5,667 (correct answer)
6,167
53,000

Explanation: 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.

Question 9

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?

$D(t) = -40000e^{-0.05t} + 50000$ (correct answer)
$D(t) = -40000e^{-0.05t} + 10000$
$D(t) = 2000e^{-0.05t} + 8000$
$D(t) = -100e^{-0.05t} + 10100$

Explanation: 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

Question 10

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.

$228
$1,378 (correct answer)
$1,578
$1,778

Explanation: 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.

Question 11

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?

$4,800 (correct answer)
$4,850
$4,900
$19,650

Explanation: 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$.

Question 12

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 $s(t)$ ?

$s(t) = 2t^3 - 4t^2 + 24\sqrt{t} + 7$
$s(t) = 2t^3 - 4t^2 + 24\sqrt{t} - 7$ (correct answer)
$s(t) = 2t^3 - 4t^2 + 6\sqrt{t} + 11$
$s(t) = 6t^3 - 8t^2 + 24\sqrt{t} + 15$

Explanation: 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.

Question 13

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 $F(10) - G(10)$ ?

-7 (correct answer)
7
0
The value cannot be determined from the information given.

Explanation: 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

Question 14

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 $\int f'(x)dx$ ?

$\frac{x^2}{2} - \frac{3}{2}\ln(x^2 + 1) + C$ (correct answer)
$\frac{x^2}{2} - \frac{3}{2}\ln|x^2 + 1| + C$
$\frac{x^2}{2} - 3\ln(x^2 + 1) + C$
$\frac{x^2}{2} - x - \ln(x^2 + 1) + C$

Explanation: 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.

Question 15

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?

$P(t) = t^3 + 2e^{2t} - 8\ln|t+1| + 98$ (correct answer)
$P(t) = t^3 + 2e^{2t} - 8\ln|t+1| + 102$
$P(t) = t^3 + 4e^{2t} - 8\ln|t+1| + 96$
$P(t) = 3t^3 + 8e^{2t} - \frac{8}{t+1} + 100$

Explanation: 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.