Fundamental Theorem of Calculus - Business Calculus

The marginal profit function for a certain product is given by $P'(x) = 150 - 0.2x$ dollars per unit, where $x$ is the number of units produced and sold. What is the best interpretation of the value of the integral $\int_{200}^{250} (150 - 0.2x) \, dx$ ?
Answer: The additional profit gained from increasing production from 200 units to 250 units. The definite integral of a rate of change function (like marginal profit, $P'(x)$ ) over an interval $[a, b]$ gives the total net change in the original function (profit, $P(x)$ ) from $a$ to $b$ . Therefore, $\int_{200}^{250} P'(x) \, dx = P(250) - P(200)$ , which represents the change in profit, or the additional profit, when production increases from 200 to 250 units. Choice A represents $P(250)$ . Choice B represents $P'(250)$ . Choice C would be related to the average value of the profit function, not the marginal profit function.
The rate of change of a population $P(t)$ (in thousands) is given by $P'(t) = 3t^2 - 6t + 2$ people per year, where $t$ is years since 2020. If $\int_0^4 P'(t) dt = 8$ , what does this value represent in the context of the problem?
Answer: The population increased by 8,000 people from 2020 to 2024 By the Fundamental Theorem of Calculus, $\int_0^4 P'(t) dt = P(4) - P(0)$ , which represents the net change in population from $t = 0$ (2020) to $t = 4$ (2024). Since the result is 8 and $P(t)$ 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.
Let $H(x) = \int_{2}^{x^3} \sqrt{1+t^2} \, dt$ . Find the value of $H'(2)$ .
Answer: $12\sqrt{65}$ According to the Fundamental Theorem of Calculus Part 1, combined with the Chain Rule, if $H(x) = \int_{a}^{g(x)} f(t) \, dt$ , then $H'(x) = f(g(x)) \cdot g'(x)$ . In this problem, $f(t) = \sqrt{1+t^2}$ and the upper limit is $g(x) = x^3$ . The derivative of the upper limit is $g'(x) = 3x^2$ . Therefore, $H'(x) = \sqrt{1+(x^3)^2} \cdot (3x^2) = 3x^2\sqrt{1+x^6}$ . To find $H'(2)$ , we substitute $x=2$ : $H'(2) = 3(2^2)\sqrt{1+2^6} = 3(4)\sqrt{1+64} = 12\sqrt{65}$ .
A company's marginal revenue function is $R'(x) = 120 - 0.4x$ dollars per unit, where $x$ 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?
Answer: $9,000 Using the Fundamental Theorem of Calculus, $R(x) = \int (120 - 0.4x) dx = 120x - 0.2x^2 + C$ . Given $R(50) = 2000$ : $2000 = 120(50) - 0.2(50)^2 + C = 6000 - 500 + C$ , so $C = -3500$ . Therefore $R(x) = 120x - 0.2x^2 - 3500$ . At $x = 100$ : $R(100) = 120(100) - 0.2(100)^2 - 3500 = 12000 - 2000 - 3500 = 9000$ . Choice A incorrectly uses $C = -4000$ . Choice C makes an arithmetic error in calculating $0.2(100)^2$ . Choice D incorrectly adds rather than subtracts the constant.
A warehouse is being filled with stock at a rate of $A(t) = 100 - 8t$ 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 $t=0$ , how many units are in the warehouse after 5 hours?
Answer: 700 units The net rate of change of stock in the warehouse is the rate in minus the rate out: $N'(t) = A(t) - R(t) = (100 - 8t) - 40 = 60 - 8t$ units per hour. To find the total change in the number of units over the first 5 hours, we integrate this net rate from $t=0$ to $t=5$ : $\Delta N = \int_{0}^{5} (60 - 8t) \, dt = [60t - 4t^2]_{0}^{5}$ Evaluating this gives $(60(5) - 4(5^2)) - (0) = 300 - 4(25) = 300 - 100 = 200$ units. This is the net increase in stock. The total number of units after 5 hours is the initial amount plus this change: $500 + 200 = 700$ units.
Let $f(x)$ be a continuous function and define an accumulation function $H(x) = \int_{c}^{x} f(t) \, dt$ , where $c$ is a constant. Which of the following statements must be true?
Answer: $H(c) = 0$ . The statement $H(c) = 0$ must be true because the definite integral from a point to itself is always zero: $H(c) = \int_{c}^{c} f(t) \, dt = 0$ . A is incorrect because $H(x)$ is an antiderivative, not the antiderivative (which refers to the family of functions $F(x)+C$ ). D is an incorrect statement of the FTC; the theorem states $H'(x) = f(x)$ . B is incorrect because $f(x) > 0$ implies $H'(x) > 0$ , meaning $H(x)$ is increasing. For $H(x)$ to be concave up, we need $H''(x) > 0$ . Since $H''(x) = f'(x)$ , this would require $f(x)$ to be an increasing function, which is not guaranteed by $f(x) > 0$ .
Let $G(x) = \int_{x}^{x^2} \frac{1}{\ln(t)} \, dt$ for $x > 1$ . Find $G'(x)$ .
Answer: $\frac{x-1}{\ln x}$ To differentiate an integral with variable limits in both the upper and lower bounds, we first split the integral: $G(x) = \int_{x}^{a} \frac{1}{\ln t} dt + \int_{a}^{x^2} \frac{1}{\ln t} dt = -\int_{a}^{x} \frac{1}{\ln t} dt + \int_{a}^{x^2} \frac{1}{\ln t} dt$ . Now, we apply the Fundamental Theorem of Calculus Part 1 with the Chain Rule. The derivative of the first term is $-\frac{1}{\ln x}$ . The derivative of the second term is $\frac{1}{\ln(x^2)} \cdot \frac{d}{dx}(x^2) = \frac{1}{2\ln x} \cdot 2x = \frac{x}{\ln x}$ . Combining them, $G'(x) = -\frac{1}{\ln x} + \frac{x}{\ln x} = \frac{x-1}{\ln x}$ .
The function $h(x) = \int_1^x \frac{t^2 - 4}{t + 1} dt$ is defined for $x > 1$ . What is $h'(3)$ ?
Answer: $\frac{5}{4}$ By the Fundamental Theorem of Calculus, $h'(x) = \frac{x^2 - 4}{x + 1}$ . At $x = 3$ : $h'(3) = \frac{3^2 - 4}{3 + 1} = \frac{9 - 4}{4} = \frac{5}{4}$ . Choice B incorrectly computes the numerator as $9 + 4 = 13$ . Choice C forgets to subtract 4 from $x^2$ . Choice D incorrectly computes the numerator as $4 - 3 = 1$ .
If $f(x)$ is continuous on $[a,b]$ and $g(x) = \int_a^x f(t) dt$ , which statement about $g'(x)$ is correct when $x$ is in the interior of $[a,b]$ ?
Answer: $g'(x) = f(x)$ regardless of whether $f(x)$ is positive, negative, or zero The Fundamental Theorem of Calculus states that if $f$ is continuous on $[a,b]$ , then $\frac{d}{dx}\int_a^x f(t) dt = f(x)$ for all $x$ in $(a,b)$ . This holds regardless of the sign of $f(x)$ . Choice A incorrectly adds a positivity condition. Choice C confuses continuity with differentiability - $f$ only needs to be continuous. Choice D is irrelevant since $f$ is continuous on a closed interval, making the integral automatically convergent.
The rate of change of revenue from a new product is given by $R'(t) = 300e^{-0.1t}$ dollars per week, where $t$ 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?
Answer: $3000(1 - e^{-1}) - 5000$ 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: $R(10) = \int_{0}^{10} 300e^{-0.1t} \, dt$ . The antiderivative of $300e^{-0.1t}$ is $\frac{300}{-0.1}e^{-0.1t} = -3000e^{-0.1t}$ . Evaluating the definite integral: $[-3000e^{-0.1t}]_{0}^{10} = (-3000e^{-1}) - (-3000e^{0}) = -3000e^{-1} + 3000 = 3000(1 - e^{-1})$ . This is the total revenue. The total profit is this revenue minus the total cost of $5000, so Profit = 3000(1 - $e^{-1}$ ) - 5000.
The marginal revenue for a product is given by $R'(q) = \frac{100}{(q+1)^2}$ , where $q$ 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?
Answer: $105 By the Fundamental Theorem of Calculus, the change in revenue from $q=9$ to $q=19$ is $\int_{9}^{19} R'(q) \, dq$ . We can find $R(19)$ using the relationship $R(19) = R(9) + \int_{9}^{19} \frac{100}{(q+1)^2} \, dq$ . First, evaluate the integral. Let $u=q+1$ , $du=dq$ . The antiderivative of $100u^{-2}$ is $-100u^{-1} = -\frac{100}{q+1}$ . Now evaluate the definite integral: $[-\frac{100}{q+1}]_{9}^{19} = (-\frac{100}{19+1}) - (-\frac{100}{9+1}) = -\frac{100}{20} - (-\frac{100}{10}) = -5 - (-10) = 5$ . This is the increase in revenue. The total revenue at $q=19$ is $R(19) = R(9) + 5 = 100 + 5 = 105$ .
Evaluate the definite integral: $\int_{e}^{e^2} \frac{(\ln x)^3}{x} \, dx$
Answer: $\frac{15}{4}$ This integral can be solved using u-substitution. Let $u = \ln x$ . Then $du = \frac{1}{x} \, dx$ . We must also change the limits of integration. When $x=e$ , $u = \ln(e) = 1$ . When $x=e^2$ , $u = \ln(e^2) = 2$ . The integral becomes $\int_{1}^{2} u^3 \, du$ Using the power rule for integration, we get $\left[ \frac{u^4}{4} \right]_{1}^{2} = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$
A function $f(x)$ is continuous everywhere. It is known that $\int_{1}^{3} f(x) \, dx = 5$ and $\int_{3}^{7} f(x) \, dx = -2$ . If $F(x)$ is an antiderivative of $f(x)$ and $F(1) = 10$ , what is the value of $F(7)$ ?
Answer: 13 First, use the property of definite integrals that $\int_{a}^{c} f(x) \, dx = \int_{a}^{b} f(x) \, dx + \int_{b}^{c} f(x) \, dx$ . So, $\int_{1}^{7} f(x) \, dx = \int_{1}^{3} f(x) \, dx + \int_{3}^{7} f(x) \, dx = 5 + (-2) = 3$ . Next, by the Fundamental Theorem of Calculus, $\int_{1}^{7} f(x) \, dx = F(7) - F(1)$ . We are given that the integral equals 3 and $F(1) = 10$ . So, $3 = F(7) - 10$ . Solving for $F(7)$ gives $F(7) = 3 + 10 = 13$ .
The marginal cost to operate a factory is $C'(t) = 10t - t^2$ dollars per hour, where $t$ is the number of hours after opening at 8 AM. Which of the following is the best interpretation of the expression $\frac{1}{4} \int_{0}^{4} (10t - t^2) \, dt$ ?
Answer: The average marginal cost, in dollars per hour, during the first four hours of operation. When you encounter an integral with a fraction coefficient like $\frac{1}{b-a}$ , you're looking at the average value formula. The expression $\frac{1}{4} \int_{0}^{4} (10t - t^2) \, dt$ represents the average value of the function $C'(t) = 10t - t^2$ over the interval from $t = 0$ to $t = 4$ . Since $C'(t)$ 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 $\int_{0}^{4} (10t - t^2) \, dt$ without the $\frac{1}{4}$ 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 $C'(t)$ 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: $\frac{1}{b-a} \int_{a}^{b} f(x) \, dx$ . 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.

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