Business Calculus Quiz: Fundamental Theorem Of Calculus
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Fundamental Theorem Of CalculusQuestion 1 of 14

The rate of change of a population P(t)P(t) (in thousands) is given by P′(t)=3t2−6t+2P'(t) = 3t^2 - 6t + 2 people per year, where tt is years since 2020. If ∫04P′(t)dt=8\int_0^4 P'(t) dt = 8, what does this value represent in the context of the problem?

The population increased by 8,000 people from 2020 to 2024
The average rate of population change from 2020 to 2024 was 8,000 people per year
The population in 2024 was 8,000 people greater than twice the population in 2020
The total accumulated population over the 4-year period was 8,000 people
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Business Calculus Quiz

Business Calculus Quiz: Fundamental Theorem Of Calculus

Practice Fundamental Theorem Of Calculus 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 Fundamental Theorem Of Calculus, 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

The rate of change of a population P(t)P(t) (in thousands) is given by P′(t)=3t2−6t+2P'(t) = 3t^2 - 6t + 2 people per year, where tt is years since 2020. If ∫04P′(t)dt=8\int_0^4 P'(t) dt = 8, what does this value represent in the context of the problem?

  1. The population increased by 8,000 people from 2020 to 2024 (correct answer)
  2. The average rate of population change from 2020 to 2024 was 8,000 people per year
  3. The population in 2024 was 8,000 people greater than twice the population in 2020
  4. The total accumulated population over the 4-year period was 8,000 people
Explanation: By the Fundamental Theorem of Calculus, ∫04P′(t)dt=P(4)−P(0)\int_0^4 P'(t) dt = P(4) - P(0), which represents the net change in population from t=0t = 0 (2020) to t=4t = 4 (2024). Since the result is 8 and P(t)P(t) is in thousands, this means the population increased by 8,000 people. Choice B confuses net change with average rate. Choice C misinterprets what the integral represents. Choice D confuses net change with total accumulation.

Question 2

The function h(x)=∫1xt2−4t+1dth(x) = \int_1^x \frac{t^2 - 4}{t + 1} dt is defined for x>1x > 1. What is h′(3)h'(3)?

  1. 54\frac{5}{4} (correct answer)
  2. 134\frac{13}{4}
  3. 94\frac{9}{4}
  4. 14\frac{1}{4}
Explanation: By the Fundamental Theorem of Calculus, h′(x)=x2−4x+1h'(x) = \frac{x^2 - 4}{x + 1}. At x=3x = 3: h′(3)=32−43+1=9−44=54h'(3) = \frac{3^2 - 4}{3 + 1} = \frac{9 - 4}{4} = \frac{5}{4}. Choice B incorrectly computes the numerator as 9+4=139 + 4 = 13. Choice C forgets to subtract 4 from x2x^2. Choice D incorrectly computes the numerator as 4−3=14 - 3 = 1.

Question 3

A warehouse is being filled with stock at a rate of A(t)=100−8tA(t) = 100 - 8t units per hour. At the same time, stock is shipped out at a constant rate of 40 units per hour. If the warehouse contained 500 units at time t=0t=0, how many units are in the warehouse after 5 hours?

  1. 600 units
  2. 900 units
  3. 700 units (correct answer)
  4. 200 units
Explanation: The net rate of change of stock in the warehouse is the rate in minus the rate out: N′(t)=A(t)−R(t)=(100−8t)−40=60−8tN'(t) = A(t) - R(t) = (100 - 8t) - 40 = 60 - 8t units per hour. To find the total change in the number of units over the first 5 hours, we integrate this net rate from t=0t=0 to t=5t=5: ΔN=∫05(60−8t) dt=[60t−4t2]05\Delta N = \int_{0}^{5} (60 - 8t) \, dt = [60t - 4t^2]_{0}^{5} Evaluating this gives (60(5)−4(52))−(0)=300−4(25)=300−100=200(60(5) - 4(5^2)) - (0) = 300 - 4(25) = 300 - 100 = 200 units. This is the net increase in stock. The total number of units after 5 hours is the initial amount plus this change: 500+200=700500 + 200 = 700 units.

Question 4

Let H(x)=∫2x31+t2 dtH(x) = \int_{2}^{x^3} \sqrt{1+t^2} \, dt. Find the value of H′(2)H'(2).

  1. 12512\sqrt{5}
  2. 00
  3. 65\sqrt{65}
  4. 126512\sqrt{65} (correct answer)
Explanation: According to the Fundamental Theorem of Calculus Part 1, combined with the Chain Rule, if H(x)=∫ag(x)f(t) dtH(x) = \int_{a}^{g(x)} f(t) \, dt, then H′(x)=f(g(x))⋅g′(x)H'(x) = f(g(x)) \cdot g'(x). In this problem, f(t)=1+t2f(t) = \sqrt{1+t^2} and the upper limit is g(x)=x3g(x) = x^3. The derivative of the upper limit is g′(x)=3x2g'(x) = 3x^2. Therefore, H′(x)=1+(x3)2⋅(3x2)=3x21+x6H'(x) = \sqrt{1+(x^3)^2} \cdot (3x^2) = 3x^2\sqrt{1+x^6}. To find H′(2)H'(2), we substitute x=2x=2: H′(2)=3(22)1+26=3(4)1+64=1265H'(2) = 3(2^2)\sqrt{1+2^6} = 3(4)\sqrt{1+64} = 12\sqrt{65}.

Question 5

The marginal profit function for a certain product is given by P′(x)=150−0.2xP'(x) = 150 - 0.2x dollars per unit, where xx is the number of units produced and sold. What is the best interpretation of the value of the integral ∫200250(150−0.2x) dx\int_{200}^{250} (150 - 0.2x) \, dx?

  1. The total profit generated from selling the first 250 units.
  2. The marginal profit when production is increased to 250 units.
  3. The average profit per unit for production levels between 200 and 250 units.
  4. The additional profit gained from increasing production from 200 units to 250 units. (correct answer)
Explanation: The definite integral of a rate of change function (like marginal profit, P′(x)P'(x)) over an interval [a,b][a, b] gives the total net change in the original function (profit, P(x)P(x)) from aa to bb. Therefore, ∫200250P′(x) dx=P(250)−P(200)\int_{200}^{250} P'(x) \, dx = P(250) - P(200), which represents the change in profit, or the additional profit, when production increases from 200 to 250 units. Choice A represents P(250)P(250). Choice B represents P′(250)P'(250). Choice C would be related to the average value of the profit function, not the marginal profit function.

Question 6

Let G(x)=∫xx21ln⁡(t) dtG(x) = \int_{x}^{x^2} \frac{1}{\ln(t)} \, dt for x>1x > 1. Find G′(x)G'(x).

  1. 1ln⁡(x2)−1ln⁡x\frac{1}{\ln(x^2)} - \frac{1}{\ln x}
  2. x−1ln⁡x\frac{x-1}{\ln x} (correct answer)
  3. x+1ln⁡x\frac{x+1}{\ln x}
  4. 2xln⁡(x2)\frac{2x}{\ln(x^2)}
Explanation: To differentiate an integral with variable limits in both the upper and lower bounds, we first split the integral: G(x)=∫xa1ln⁡tdt+∫ax21ln⁡tdt=−∫ax1ln⁡tdt+∫ax21ln⁡tdtG(x) = \int_{x}^{a} \frac{1}{\ln t} dt + \int_{a}^{x^2} \frac{1}{\ln t} dt = -\int_{a}^{x} \frac{1}{\ln t} dt + \int_{a}^{x^2} \frac{1}{\ln t} dt. Now, we apply the Fundamental Theorem of Calculus Part 1 with the Chain Rule. The derivative of the first term is −1ln⁡x-\frac{1}{\ln x}. The derivative of the second term is 1ln⁡(x2)⋅ddx(x2)=12ln⁡x⋅2x=xln⁡x\frac{1}{\ln(x^2)} \cdot \frac{d}{dx}(x^2) = \frac{1}{2\ln x} \cdot 2x = \frac{x}{\ln x}. Combining them, G′(x)=−1ln⁡x+xln⁡x=x−1ln⁡xG'(x) = -\frac{1}{\ln x} + \frac{x}{\ln x} = \frac{x-1}{\ln x}.

Question 7

The marginal cost to operate a factory is C′(t)=10t−t2C'(t) = 10t - t^2 dollars per hour, where tt is the number of hours after opening at 8 AM. Which of the following is the best interpretation of the expression 14∫04(10t−t2) dt\frac{1}{4} \int_{0}^{4} (10t - t^2) \, dt?

  1. The total increase in cost during the first four hours of operation (8 AM to 12 PM).
  2. The average marginal cost, in dollars per hour, during the first four hours of operation. (correct answer)
  3. The average rate of change of the cost, in dollars per hour per hour, from 8 AM to 12 PM.
  4. The average hourly cost during the first four hours of operation.
Explanation: When you encounter an integral with a fraction coefficient like 1b−a\frac{1}{b-a}, you're looking at the average value formula. The expression 14∫04(10t−t2) dt\frac{1}{4} \int_{0}^{4} (10t - t^2) \, dt represents the average value of the function C′(t)=10t−t2C'(t) = 10t - t^2 over the interval from t=0t = 0 to t=4t = 4. Since C′(t)C'(t) is the marginal cost in dollars per hour, the average value of this marginal cost function gives you the average marginal cost during those four hours. This makes answer choice B correct. Let's examine why the other options miss the mark. Choice A describes ∫04(10t−t2) dt\int_{0}^{4} (10t - t^2) \, dt without the 14\frac{1}{4} coefficient—that integral would give the total increase in cost. Choice C confuses marginal cost with the rate of change of marginal cost. The marginal cost C′(t)C'(t) is already a rate (dollars per hour), so its average isn't a "rate of change of rate." Choice D refers to average hourly cost, but we're dealing with marginal cost (the cost of producing one additional unit), not the total cost divided by time. The key insight is recognizing the average value formula: 1b−a∫abf(x) dx\frac{1}{b-a} \int_{a}^{b} f(x) \, dx. When you see this pattern, you're finding the average value of whatever function is being integrated. Always identify what the original function represents—here it's marginal cost—and that tells you what the average represents.

Question 8

The marginal revenue for a product is given by R′(q)=100(q+1)2R'(q) = \frac{100}{(q+1)^2}, where qq is the quantity sold. If the total revenue from selling 9 items is R(9) = \100, what is the total revenue from selling 19 items?

  1. $5
  2. $85
  3. $95
  4. $105 (correct answer)
Explanation: By the Fundamental Theorem of Calculus, the change in revenue from q=9q=9 to q=19q=19 is ∫919R′(q) dq\int_{9}^{19} R'(q) \, dq. We can find R(19)R(19) using the relationship R(19)=R(9)+∫919100(q+1)2 dqR(19) = R(9) + \int_{9}^{19} \frac{100}{(q+1)^2} \, dq. First, evaluate the integral. Let u=q+1u=q+1, du=dqdu=dq. The antiderivative of 100u−2100u^{-2} is −100u−1=−100q+1-100u^{-1} = -\frac{100}{q+1}. Now evaluate the definite integral: [−100q+1]919=(−10019+1)−(−1009+1)=−10020−(−10010)=−5−(−10)=5[-\frac{100}{q+1}]_{9}^{19} = (-\frac{100}{19+1}) - (-\frac{100}{9+1}) = -\frac{100}{20} - (-\frac{100}{10}) = -5 - (-10) = 5. This is the increase in revenue. The total revenue at q=19q=19 is R(19)=R(9)+5=100+5=105R(19) = R(9) + 5 = 100 + 5 = 105.

Question 9

The rate of change of revenue from a new product is given by R′(t)=300e−0.1tR'(t) = 300e^{-0.1t} dollars per week, where tt is the number of weeks since the product's launch. The total cost to produce the items sold over the first 10 weeks is $5000. Assuming the revenue is zero at $t=0$, what is the total profit from this product over the first 10 weeks?

  1. 3000(1−e−1)−50003000(1 - e^{-1}) - 5000 (correct answer)
  2. 3000(1−e−1)3000(1 - e^{-1})
  3. 300e−1−5000300e^{-1} - 5000
  4. 3000(e−1−1)−50003000(e^{-1} - 1) - 5000
Explanation: Total profit is Total Revenue minus Total Cost. The total revenue over the first 10 weeks is the definite integral of the marginal revenue function from 0 to 10: R(10)=∫010300e−0.1t dtR(10) = \int_{0}^{10} 300e^{-0.1t} \, dt. The antiderivative of 300e−0.1t300e^{-0.1t} is 300−0.1e−0.1t=−3000e−0.1t\frac{300}{-0.1}e^{-0.1t} = -3000e^{-0.1t}. Evaluating the definite integral: [−3000e−0.1t]010=(−3000e−1)−(−3000e0)=−3000e−1+3000=3000(1−e−1)[-3000e^{-0.1t}]_{0}^{10} = (-3000e^{-1}) - (-3000e^{0}) = -3000e^{-1} + 3000 = 3000(1 - e^{-1}). This is the total revenue. The total profit is this revenue minus the total cost of $5000, so Profit = 3000(1 - e−1e^{-1}) - 5000.

Question 10

A function f(x)f(x) is continuous everywhere. It is known that ∫13f(x) dx=5\int_{1}^{3} f(x) \, dx = 5 and ∫37f(x) dx=−2\int_{3}^{7} f(x) \, dx = -2. If F(x)F(x) is an antiderivative of f(x)f(x) and F(1)=10F(1) = 10, what is the value of F(7)F(7)?

  1. 3
  2. 7
  3. 13 (correct answer)
  4. 17
Explanation: First, use the property of definite integrals that ∫acf(x) dx=∫abf(x) dx+∫bcf(x) dx\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx. So, ∫17f(x) dx=∫13f(x) dx+∫37f(x) dx=5+(−2)=3\int_{1}^{7} f(x) \, dx = \int_{1}^{3} f(x) \, dx + \int_{3}^{7} f(x) \, dx = 5 + (-2) = 3. Next, by the Fundamental Theorem of Calculus, ∫17f(x) dx=F(7)−F(1)\int_{1}^{7} f(x) \, dx = F(7) - F(1). We are given that the integral equals 3 and F(1)=10F(1) = 10. So, 3=F(7)−103 = F(7) - 10. Solving for F(7)F(7) gives F(7)=3+10=13F(7) = 3 + 10 = 13.

Question 11

A company's marginal revenue function is R′(x)=120−0.4xR'(x) = 120 - 0.4x dollars per unit, where xx is the number of units sold. If the company's revenue is $2,000 when 50 units are sold, what is the total revenue when 100 units are sold?

  1. $8,000
  2. $9,000 (correct answer)
  3. $8,500
  4. $7,500
Explanation: Using the Fundamental Theorem of Calculus, R(x)=∫(120−0.4x)dx=120x−0.2x2+CR(x) = \int (120 - 0.4x) dx = 120x - 0.2x^2 + C. Given R(50)=2000R(50) = 2000: 2000=120(50)−0.2(50)2+C=6000−500+C2000 = 120(50) - 0.2(50)^2 + C = 6000 - 500 + C, so C=−3500C = -3500. Therefore R(x)=120x−0.2x2−3500R(x) = 120x - 0.2x^2 - 3500. At x=100x = 100: R(100)=120(100)−0.2(100)2−3500=12000−2000−3500=9000R(100) = 120(100) - 0.2(100)^2 - 3500 = 12000 - 2000 - 3500 = 9000. Choice A incorrectly uses C=−4000C = -4000. Choice C makes an arithmetic error in calculating 0.2(100)20.2(100)^2. Choice D incorrectly adds rather than subtracts the constant.

Question 12

Let f(x)f(x) be a continuous function and define an accumulation function H(x)=∫cxf(t) dtH(x) = \int_{c}^{x} f(t) \, dt, where cc is a constant. Which of the following statements must be true?

  1. H(x)H(x) is the antiderivative of f(x)f(x).
  2. If f(x)>0f(x) > 0 for all xx, then H(x)H(x) must be concave up.
  3. H(c)=0H(c) = 0. (correct answer)
  4. H′(x)=f(x)−f(c)H'(x) = f(x) - f(c).
Explanation: The statement H(c)=0H(c) = 0 must be true because the definite integral from a point to itself is always zero: H(c)=∫ccf(t) dt=0H(c) = \int_{c}^{c} f(t) \, dt = 0. A is incorrect because H(x)H(x) is an antiderivative, not the antiderivative (which refers to the family of functions F(x)+CF(x)+C). D is an incorrect statement of the FTC; the theorem states H′(x)=f(x)H'(x) = f(x). B is incorrect because f(x)>0f(x) > 0 implies H′(x)>0H'(x) > 0, meaning H(x)H(x) is increasing. For H(x)H(x) to be concave up, we need H′′(x)>0H''(x) > 0. Since H′′(x)=f′(x)H''(x) = f'(x), this would require f(x)f(x) to be an increasing function, which is not guaranteed by f(x)>0f(x) > 0.

Question 13

If f(x)f(x) is continuous on [a,b][a,b] and g(x)=∫axf(t)dtg(x) = \int_a^x f(t) dt, which statement about g′(x)g'(x) is correct when xx is in the interior of [a,b][a,b]?

  1. g′(x)=f(x)g'(x) = f(x) only if f(x)>0f(x) > 0 on the entire interval [a,b][a,b]
  2. g′(x)=f(x)g'(x) = f(x) regardless of whether f(x)f(x) is positive, negative, or zero (correct answer)
  3. g′(x)=f(x)g'(x) = f(x) only if f(x)f(x) is differentiable at xx
  4. g′(x)=f(x)g'(x) = f(x) only if the integral ∫abf(t)dt\int_a^b f(t) dt converges
Explanation: The Fundamental Theorem of Calculus states that if ff is continuous on [a,b][a,b], then ddx∫axf(t)dt=f(x)\frac{d}{dx}\int_a^x f(t) dt = f(x) for all xx in (a,b)(a,b). This holds regardless of the sign of f(x)f(x). Choice A incorrectly adds a positivity condition. Choice C confuses continuity with differentiability - ff only needs to be continuous. Choice D is irrelevant since ff is continuous on a closed interval, making the integral automatically convergent.

Question 14

Evaluate the definite integral: ∫ee2(ln⁡x)3x dx\int_{e}^{e^2} \frac{(\ln x)^3}{x} \, dx

  1. 154\frac{15}{4} (correct answer)
  2. 1515
  3. 74\frac{7}{4}
  4. 32\frac{3}{2}
Explanation: This integral can be solved using u-substitution. Let u=ln⁡xu = \ln x. Then du=1x dxdu = \frac{1}{x} \, dx. We must also change the limits of integration. When x=ex=e, u=ln⁡(e)=1u = \ln(e) = 1. When x=e2x=e^2, u=ln⁡(e2)=2u = \ln(e^2) = 2. The integral becomes ∫12u3 du\int_{1}^{2} u^3 \, du Using the power rule for integration, we get [u44]12=244−144=164−14=154\left[ \frac{u^4}{4} \right]_{1}^{2} = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}