Business Calculus Quiz: Continuous Cash Flow Models
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Continuous Cash Flow ModelsQuestion 1 of 20

A pharmaceutical company's research division generates patent royalties at a rate of R(t)=25000te−0.1tR(t) = 25000te^{-0.1t} dollars per year, where tt is years from now. The patents will expire after 15 years. Using a 5% annual discount rate compounded continuously, what percentage of the total present value comes from royalties earned in the first 5 years?

∫0525000te−0.15tdt∫01525000te−0.15tdt×100%≈72%\frac{\int_0^5 25000te^{-0.15t} dt}{\int_0^{15} 25000te^{-0.15t} dt} \times 100\% \approx 72\%
∫0525000te−0.1te−0.05tdt∫01525000te−0.1te−0.05tdt×100%≈68%\frac{\int_0^5 25000te^{-0.1t}e^{-0.05t} dt}{\int_0^{15} 25000te^{-0.1t}e^{-0.05t} dt} \times 100\% \approx 68\%
∫0525000te−0.05tdt∫01525000te−0.05tdt×100%≈45%\frac{\int_0^5 25000te^{-0.05t} dt}{\int_0^{15} 25000te^{-0.05t} dt} \times 100\% \approx 45\%
∫0525000te−0.1tdt∫01525000te−0.1tdt×100%≈52%\frac{\int_0^5 25000te^{-0.1t} dt}{\int_0^{15} 25000te^{-0.1t} dt} \times 100\% \approx 52\%
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Business Calculus Quiz

Business Calculus Quiz: Continuous Cash Flow Models

Practice Continuous Cash Flow Models in Business Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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Question 1

A pharmaceutical company's research division generates patent royalties at a rate of R(t)=25000te−0.1tR(t) = 25000te^{-0.1t} dollars per year, where tt is years from now. The patents will expire after 15 years. Using a 5% annual discount rate compounded continuously, what percentage of the total present value comes from royalties earned in the first 5 years?

  1. ∫0525000te−0.15tdt∫01525000te−0.15tdt×100%≈72%\frac{\int_0^5 25000te^{-0.15t} dt}{\int_0^{15} 25000te^{-0.15t} dt} \times 100\% \approx 72\%
  2. ∫0525000te−0.1te−0.05tdt∫01525000te−0.1te−0.05tdt×100%≈68%\frac{\int_0^5 25000te^{-0.1t}e^{-0.05t} dt}{\int_0^{15} 25000te^{-0.1t}e^{-0.05t} dt} \times 100\% \approx 68\% (correct answer)
  3. ∫0525000te−0.05tdt∫01525000te−0.05tdt×100%≈45%\frac{\int_0^5 25000te^{-0.05t} dt}{\int_0^{15} 25000te^{-0.05t} dt} \times 100\% \approx 45\%
  4. ∫0525000te−0.1tdt∫01525000te−0.1tdt×100%≈52%\frac{\int_0^5 25000te^{-0.1t} dt}{\int_0^{15} 25000te^{-0.1t} dt} \times 100\% \approx 52\%
Explanation: When you encounter present value problems with continuous cash flows, you need to discount future income streams back to today's dollars using the formula PV=∫0TR(t)e−rtdtPV = \int_0^T R(t)e^{-rt} dt, where R(t)R(t) is the income rate, rr is the discount rate, and TT is the time horizon. Here, the royalty rate is R(t)=25000te−0.1tR(t) = 25000te^{-0.1t} and the discount rate is 5% (0.05). To find the present value at any time tt, you multiply the income rate by the continuous discount factor e−0.05te^{-0.05t}. This gives you 25000te−0.1t⋅e−0.05t=25000te−0.15t25000te^{-0.1t} \cdot e^{-0.05t} = 25000te^{-0.15t} for the integrand. The percentage from the first 5 years is the ratio of the 5-year present value to the total 15-year present value, which is exactly what option B shows: ∫0525000te−0.1te−0.05tdt∫01525000te−0.1te−0.05tdt×100%\frac{\int_0^5 25000te^{-0.1t}e^{-0.05t} dt}{\int_0^{15} 25000te^{-0.1t}e^{-0.05t} dt} \times 100\%. Option A incorrectly combines the discount rate with the existing exponential term, writing e−0.15te^{-0.15t} directly instead of showing the separate discount factor. Option C ignores the original e−0.1te^{-0.1t} term entirely, using only the discount factor. Option D forgets to apply any discounting at all, calculating with undiscounted future values rather than present values. Study tip: In present value problems, always identify the cash flow function first, then multiply by e−rte^{-rt} as a separate discount factor. Don't combine exponential terms until after you've clearly shown both components.

Question 2

A company is considering two 10-year contracts. Contract A offers a continuous income stream at a constant rate of $50,000 per year. Contract B offers a stream at a rate of R(t) = 80,000e^(-0.1t) dollars per year. Assuming a continuous discount rate of 4%, which contract has a higher present value, and by approximately how much?

  1. Contract A, by approximately $18,388.
  2. Contract B, by approximately $18,388. (correct answer)
  3. Contract B, by approximately $5,699, which is the difference in their total undiscounted incomes.
  4. Contract A, by approximately $20,387, which is the difference if B's decay rate is ignored.
Explanation: We need to calculate and compare the present values (PV) of both contracts. PV = ∫_0^T R(t) e^(-rt) dt with T=10 and r=0.04. For Contract A: R_A(t) = 50,000. PV_A = ∫_0^10 50,000 e^(-0.04t) dt = 50,000/0.04 * (1 - e^(-0.04*10)) = 1,250,000 * (1 - e^(-0.4)) ≈ 1,250,000 * (0.32968) ≈ $412,100. For Contract B: R_B(t) = 80,000e^(-0.1t). The integrand is R_B(t)e^(-rt) = 80,000e^(-0.1t)e^(-0.04t) = 80,000e^(-0.14t). PV_B = ∫_0^10 80,000 e^(-0.14t) dt = 80,000/0.14 * (1 - e^(-0.14*10)) = 571,428.57 * (1 - e^(-1.4)) ≈ 571,428.57 * (0.7534) ≈ $430,513. Comparing the two: PV_B is higher than PV_A. The difference is 430,513 - 412,100 = $18,413. The closest answer is $18,388.

Question 3

A business generates a continuous income stream of $30,000 per year. This income is invested at 6% interest compounded continuously. Approximately how many years will it take for the future value of this investment to reach $500,000?

  1. 16.7 years, the time it would take if there were no interest.
  2. 11.6 years, derived from the correct future value formula. (correct answer)
  3. 46.9 years, derived from using a lump-sum investment formula.
  4. 10.0 years, assuming the interest rate directly determines the growth factor.
Explanation: The future value (FV) of a continuous income stream R(t) = K is given by the formula FV = ∫_0^T K e^(r(T-t)) dt, which simplifies to FV = (K/r) * (e^(rT) - 1). We are given FV = 500,000, K = 30,000, and r = 0.06. We need to solve for T. 500,000 = (30,000 / 0.06) * (e^(0.06T) - 1) 500,000 = 500,000 * (e^(0.06T) - 1) 1 = e^(0.06T) - 1 2 = e^(0.06T) ln(2) = 0.06T T = ln(2) / 0.06 Using ln(2) ≈ 0.6931, we get: T ≈ 0.6931 / 0.06 ≈ 11.55 years. The closest answer is 11.6 years.

Question 4

A startup's rate of profit is projected to be R(t) = P_0 e^(kt) for T years, where t is the time in years. The profit is to be discounted at a nominal rate of r compounded continuously. Which integral represents the present value of the company's total profit over the T-year period?

  1. ∫_0^T P_0 e^((k-r)t) dt (correct answer)
  2. ∫_0^T P_0 e^(kt) e^(r(T-t)) dt
  3. ∫_0^T P_0 e^(kt) dt
  4. ∫_0^T P_0 e^((k+r)t) dt
Explanation: The formula for the present value (PV) of a continuous income stream R(t) over a period T with a continuous discount rate r is PV = ∫_0^T R(t) e^(-rt) dt. In this scenario, the rate of profit is R(t) = P_0 e^(kt). Substituting this into the PV formula gives: PV = ∫_0^T (P_0 e^(kt)) e^(-rt) dt Using the property of exponents e^a * e^b = e^(a+b), we combine the exponential terms: PV = ∫_0^T P_0 e^(kt - rt) dt = ∫_0^T P_0 e^((k-r)t) dt. Choice B represents the future value. Choice C represents the total profit without any discounting. Choice D incorrectly adds the rates in the exponent.

Question 5

An asset is expected to generate a continuous income stream for 20 years with a present value of $100,000. Assuming a continuous discount rate of 5%, what constant rate of income K (in dollars per year) would the asset need to generate?

  1. $2,910
  2. $5,000
  3. $7,910 (correct answer)
  4. $15,820
Explanation: The present value (PV) of a continuous income stream at a constant rate K is given by PV = ∫_0^T K e^(-rt) dt. We are given PV = 100,000, T = 20, and r = 0.05. We need to solve for K. 100,000 = ∫_0^20 K e^(-0.05t) dt 100,000 = K ∫_0^20 e^(-0.05t) dt 100,000 = K [(-1/0.05) e^(-0.05t)]_0^20 100,000 = K * (-20) * [e^(-0.05*20) - e^0] 100,000 = K * (-20) * [e^(-1) - 1] 100,000 = K * 20 * (1 - e^(-1)) Using e^(-1) ≈ 0.36788, we get: 100,000 ≈ K * 20 * (1 - 0.36788) = K * 20 * (0.63212) 100,000 ≈ K * 12.6424 K ≈ 100,000 / 12.6424 ≈ $7,909.8. The closest answer is $7,910. Distractor B (5,000)isthesimpleaverage‘100,000/20‘.DistractorA(5,000) is the simple average `100,000 / 20`. Distractor A (2,910) results from incorrectly using the future value formula. Distractor D ($15,820) results from a calculation error where the 1/r term is omitted from the integration.

Question 6

You have two investment options. Option A is a lump-sum payment of $200,000 today. Option B is a continuous income stream for 5 years at a rate of R(t) = 40,000 + 1,000t dollars per year. If the prevailing interest rate is 6% compounded continuously, which option is better and by approximately how much?

  1. Option A is better by $16,756. (correct answer)
  2. Option B is better by $12,500.
  3. Option B is better by $183,244.
  4. The options are nearly equal in value.
Explanation: To compare the options, we must find the present value (PV) of Option B and compare it to the PV of Option A, which is $200,000. PV_B = ∫_0^5 (40,000 + 1,000t)e^(-0.06t) dt Using integration by parts with u = 40000 + 1000t and dv = e^(-0.06t)dt: du = 1000dt, v = (-1/0.06)e^(-0.06t) Antiderivative ∫udv = uv - ∫vdu is (40000+1000t)(-1/0.06)e^(-0.06t) - ∫(-1/0.06)e^(-0.06t)(1000dt) = (-1/0.06)(40000+1000t)e^(-0.06t) - (1000/($0.06^2$))e^(-0.06t) Evaluate from 0 to 5: At t=5: (-1/0.06)(45000)e^(-0.3) - (1000/0.0036)e^(-0.3) = (-750000 - 277778)e^(-0.3) ≈ -1027778 * 0.7408 ≈ -761394 At t=0: (-1/0.06)(40000)e^(0) - (1000/0.0036)e^(0) = -666667 - 277778 = -944445 PV_B = (-761394) - (-944445) = $183,051. The present value of Option B is approximately $183,051. Option A's present value is $200,000. Thus, Option A is better by 200,000 - 183,051 = $16,949. The closest answer is $16,756. Distractor B compares the total undiscounted income of B ($212,500) to A.

Question 7

A patent will generate a perpetual stream of royalties of $50,000 per year. However, the payments will not begin until 3 years from today. Using a continuous discount rate of 8%, what is the present value (at t=0) of this royalty stream?

  1. $625,000
  2. $504,032
  3. $491,644 (correct answer)
  4. $39,331
Explanation: This is a two-step problem.
  1. First, calculate the value of the perpetual stream at the time it begins, which is at t=3. The capital value of a perpetual stream R at a rate r is R/r. So, the value at t=3 is: V_3 = $50,000 / 0.08 = $625,000.
  2. Second, this value V_3 is a lump sum at t=3. We need to find its present value at t=0 by discounting it back 3 years using the formula PV = V * e^(-rt). PV = $625,000 * e^(-0.08 * 3) = $625,000 * e^(-0.24). Using e^(-0.24) ≈ 0.78663, we get: PV ≈ $625,000 * 0.78663 ≈ $491,644.
Alternatively, this can be computed with a definite integral from 3 to infinity: PV = ∫_3^∞ 50000e^(-0.08t) dt = [50000/(-0.08) * e^(-0.08t)]_3^∞ = 0 - (-625000 * e^(-0.24)) = 625000e^(-0.24) ≈ $491,644. Distractor A is the value at t=3, not t=0. Distractor B uses an incorrect simple interest discount.

Question 8

A rental property generates income at a rate of I(t)=2400+50tI(t) = 2400 + 50t dollars per month, where tt is months from now. The owner plans to sell the property after 36 months for $180,000. Using a discount rate of 8% per year compounded continuously, what is the total present value of this investment?

  1. ∫036(2400+50t)e−0.08t/12dt+180000e−0.08⋅3≈$245,200\int_0^{36} (2400 + 50t)e^{-0.08t/12} dt + 180000e^{-0.08 \cdot 3} \approx \$245,200 (correct answer)
  2. ∫036(2400+50t)e−0.08t/12dt+180000e−0.08⋅36≈$198,600\int_0^{36} (2400 + 50t)e^{-0.08t/12} dt + 180000e^{-0.08 \cdot 36} \approx \$198,600
  3. ∫03(2400+50t)e−0.08tdt+180000e−0.08⋅3≈$251,800\int_0^{3} (2400 + 50t)e^{-0.08t} dt + 180000e^{-0.08 \cdot 3} \approx \$251,800
  4. ∫036(2400+50t)e−0.08tdt+180000e−0.08⋅36≈$182,400\int_0^{36} (2400 + 50t)e^{-0.08t} dt + 180000e^{-0.08 \cdot 36} \approx \$182,400
Explanation: The total present value includes both the income stream and the sale proceeds. Income is monthly, so we convert the annual rate: r=0.08/12r = 0.08/12 per month, giving ∫036(2400+50t)e−0.08t/12dt\int_0^{36} (2400 + 50t)e^{-0.08t/12} dt. The sale occurs at t=36t = 36 months = 3 years, so its present value is 180000e−0.08⋅3180000e^{-0.08 \cdot 3}. Choice B incorrectly discounts the sale using months. Choice C converts to years but uses the wrong integral. Choice D uses the annual rate directly on monthly periods.

Question 9

A small business has cash outflows of C(t)=1200e0.03tC(t) = 1200e^{0.03t} dollars per month and cash inflows of I(t)=2000−40tI(t) = 2000 - 40t dollars per month, where tt is months from now. Using a 9% annual discount rate compounded continuously, what is the net present value over the next 24 months?

  1. ∫024[(2000−40t)+1200e0.03t]e−0.0075tdt\int_0^{24} [(2000 - 40t) + 1200e^{0.03t}]e^{-0.0075t} dt
  2. ∫024[(2000−40t)−1200e0.03t]e−0.09tdt\int_0^{24} [(2000 - 40t) - 1200e^{0.03t}]e^{-0.09t} dt
  3. ∫024[(2000−40t)e−0.0075t−1200e0.03te−0.09t]dt\int_0^{24} [(2000 - 40t)e^{-0.0075t} - 1200e^{0.03t}e^{-0.09t}] dt
  4. ∫024[(2000−40t)−1200e0.03t]e−0.0075tdt\int_0^{24} [(2000 - 40t) - 1200e^{0.03t}]e^{-0.0075t} dt (correct answer)
Explanation: Net present value (NPV) problems require you to discount future cash flows back to their present value using a continuous discount rate. The key is setting up the integral with the correct net cash flow and proper discount factor. To find NPV, you need the net cash flow at each time multiplied by the discount factor. Net cash flow equals inflows minus outflows: (2000−40t)−1200e0.03t(2000 - 40t) - 1200e^{0.03t}. For continuous compounding at 9% annual rate, the discount factor is e−0.09te^{-0.09t}. However, since cash flows occur monthly, you must convert the annual rate: 0.0912=0.0075\frac{0.09}{12} = 0.0075 per month, giving discount factor e−0.0075te^{-0.0075t}. The correct NPV integral is ∫024[(2000−40t)−1200e0.03t]e−0.0075tdt\int_0^{24} [(2000 - 40t) - 1200e^{0.03t}]e^{-0.0075t} dt, which is answer D. Answer A incorrectly adds outflows instead of subtracting them - this would give you total cash flow, not net cash flow. Answer B uses the wrong discount rate (e−0.09te^{-0.09t} instead of e−0.0075te^{-0.0075t}) by applying the annual rate directly to monthly periods. Answer C separates the discount factor application, which is mathematically equivalent to the correct form but unnecessarily complicated and not the standard NPV format. Study tip: Always remember the NPV formula structure: integrate (net cash flow × discount factor) over the time period. Convert annual rates to match your time units, and ensure outflows are subtracted from inflows, not added.

Question 10

An analyst is evaluating a project where the rate of income is a constant R dollars per year and the continuous discount rate is r. The analyst writes the integral ∫_5^10 R e^(-rt) dt. What does this integral represent?

  1. The total income generated from the end of year 5 to the end of year 10.
  2. The future value at year 10 of the income generated between years 5 and 10.
  3. The present value (at time t=0) of the income generated between years 5 and 10. (correct answer)
  4. The present value of the income generated over the first 5 years of the project.
Explanation: The formula for the present value of a continuous income stream R(t) from time t=T_1 to t=T_2 at a discount rate r is PV = ∫_(T_1)^($T_2$) R(t) e^(-rt) dt. The term e^(-rt) is the discount factor that converts a future cash flow at time t to its value at time t=0 (the present). In this case, R(t) = R, T_1 = 5, and T_2 = 10. The integral ∫_5^10 R e^(-rt) dt therefore calculates the sum of all discounted income flows occurring between the 5-year mark and the 10-year mark, giving their combined value at t=0. This is the present value of that portion of the income stream.

Question 11

A project requires an initial investment of $250,000 and is expected to generate a continuous income stream at the rate R(t) = 75,000e^(-0.05t) dollars per year. If the continuous discount rate is 10%, how many years will it take for the present value of the income stream to equal the initial investment?

  1. 3.33 years
  2. 3.65 years
  3. 4.05 years
  4. 4.62 years (correct answer)
Explanation: We need to find the time T such that the present value (PV) of the income stream equals the investment cost. PV = ∫_0^T R(t) e^(-rt) dt. We set PV = 250,000. R(t) = 75,000e^(-0.05t) and r = 0.10. 250,000 = ∫_0^T (75,000e^(-0.05t)) * e^(-0.10t) dt 250,000 = ∫_0^T 75,000e^(-0.15t) dt 250,000 = 75,000 [(-1/0.15)e^(-0.15t)]_0^T Divide by 75,000: 10/3 = (-1/0.15) * [e^(-0.15T) - 1] 10/3 = (1/0.15) * (1 - e^(-0.15T)) (10/3) * 0.15 = 1 - e^(-0.15T) 0.5 = 1 - e^(-0.15T) e^(-0.15T) = 0.5 -0.15T = ln(0.5) = -ln(2) T = ln(2) / 0.15 ≈ 0.6931 / 0.15 ≈ 4.62 years. Distractor B forgets to apply the external discount rate r. Distractor C ignores the time-varying part of R(t).

Question 12

An investment generates a continuous income stream at a constant rate of $15,000 per year for 8 years. If this income is continuously invested at an interest rate of 5% compounded continuously, what initial lump-sum investment P would yield the same total amount after 8 years?

  1. An investment of $98,910, which is the present value of the income stream. (correct answer)
  2. An investment of $147,547, which is the future value of the income stream.
  3. An investment of $120,000, which is the total income generated by the stream.
  4. An investment of $100,548, which is the value of the final year's income discounted to the present.
Explanation: The problem asks for a present-day lump-sum investment P that will grow to the same future value as the income stream. The future value of the lump sum is P * e^(rT). The future value of the income stream is FV = ∫_0^T R(t) e^(r(T-t)) dt. Setting them equal gives P * e^(rT) = ∫_0^T R(t) e^(r(T-t)) dt. Multiplying by e^(-rT) gives P = ∫_0^T R(t) e^(-rt) dt, which is the formula for the present value (PV) of the income stream. Here, R(t) = 15000, r = 0.05, and T = 8. PV = ∫_0^8 15000 e^(-0.05t) dt PV = 15000 [(-1/0.05) * e^(-0.05t)]_0^8 PV = 15000 * (-20) * [e^(-0.4) - e^0] PV = -300,000 * (e^(-0.4) - 1) = 300,000 * (1 - e^(-0.4)) Using e^(-0.4) ≈ 0.67032, we get: PV ≈ 300,000 * (1 - 0.67032) = 300,000 * 0.32968 ≈ $98,904. The closest answer is $98,910.

Question 13

The rate of revenue for a new product is modeled by R(t) = 1000 + 50t dollars per year. If the revenue is invested at a rate of 4% compounded continuously for 10 years, what is the total interest earned over the 10-year period?

  1. $15,164, which is the future value of the income stream.
  2. $12,500, which is the total principal deposited.
  3. $10,167, which is the present value of the income stream.
  4. $2,664, which is the future value minus the total principal. (correct answer)
Explanation: The total interest earned is the difference between the future value (FV) and the total income deposited (Principal).
  1. Calculate Total Income (Principal): TI = ∫_0^10 (1000 + 50t) dt = [1000t + 25t^2]_0^10 = (10,000 + 2,500) - 0 = $12,500.
  2. Calculate Future Value (FV): FV = ∫_0^10 (1000 + 50t)e^(0.04(10-t)) dt. Let u = 10-t, so t = 10-u and dt = -du. FV = ∫_10^0 (1000 + 50(10-u))e^(0.04u) (-du) = ∫_0^10 (1500 - 50u)e^(0.04u) du. Using integration by parts (∫ f dg = fg - ∫ g df): Let f = 1500 - 50u and dg = e^(0.04u) du. Then df = -50 du and g = 25e^(0.04u). FV = [(1500 - 50u)(25e^(0.04u))]_0^10 - ∫_0^10 (25e^(0.04u))(-50 du) FV = [(1500 - 50u)(25e^(0.04u)) + 1250(25e^(0.04u))]_0^10 FV = [25(1500 - 50u + 1250)e^(0.04u)]_0^10 = [25(2750 - 50u)e^(0.04u)]_0^10 FV = 25[(2250e^(0.4)) - ($2750e^0$)] = 25(2250 * 1.4918 - 2750) ≈ 25(3356.55 - 2750) = 25(606.55) ≈ $15,164.
  3. Calculate Interest: Interest = FV - TI = 15,164 - 12,500 = $2,664.

Question 14

A machine generates income at a continuous rate of $10,000 per year for its first 5 years. After an upgrade, it generates income at a rate of $15,000 per year for the next 5 years. What is the total present value of the income stream over the 10-year period, assuming a discount rate of 7% compounded continuously?

  1. $86,779 (correct answer)
  2. $89,895
  3. $105,465
  4. $125,000
Explanation: The total present value (PV) is the sum of the present values of the two separate income streams. PV = ∫_0^5 10000e^(-0.07t) dt + ∫_5^10 15000e^(-0.07t) dt Part 1: ∫_0^5 10000e^(-0.07t) dt = 10000 * [(-1/0.07)e^(-0.07t)]_0^5 = (10000/0.07) * (1 - e^(-0.35)) ≈ 142857 * (1 - 0.70469) ≈ 142857 * (0.29531) ≈ $42,187 Part 2: ∫_5^10 15000e^(-0.07t) dt = 15000 * [(-1/0.07)e^(-0.07t)]_5^10 = (15000/0.07) * (e^(-0.35) - e^(-0.70)) ≈ 214286 * (0.70469 - 0.49659) ≈ 214286 * (0.2081) ≈ $44,592 Total PV = $42,187 + $44,592 = $86,779. Distractor D is the total undiscounted income. Distractor B uses an average income rate over the whole period. Distractor C incorrectly calculates the PV of the second stream as if it started at t=0.

Question 15

The projected income from an oil well is given by the continuous stream R(t) = (9000 + 1000t)e^(-0.02t) dollars per year. Find the capital value of the well, assuming the income continues indefinitely and the prevailing interest rate is 8% compounded continuously.

  1. $90,000, from only considering the constant part of the income stream.
  2. $190,000, from integrating the discounted combined income stream. (correct answer)
  3. $268,750, from discounting only by the prevailing interest rate.
  4. $427,778, from an incorrect combination of the decay and discount rates.
Explanation: The capital value is the present value of a perpetual income stream. The formula is PV = ∫_0^∞ R(t) e^(-rt) dt. Here, r = 0.08 and R(t) = (9000 + 1000t)e^(-0.02t). First, combine the exponential terms in the integrand: R(t)e^(-rt) = (9000 + 1000t)e^(-0.02t)e^(-0.08t) = (9000 + 1000t)e^(-0.1t). So, we need to evaluate PV = ∫_0^∞ (9000 + 1000t)e^(-0.1t) dt. This requires integration by parts. Let u = 9000 + 1000t and dv = e^(-0.1t) dt. Then du = 1000 dt and v = -10e^(-0.1t). The antiderivative is uv - ∫v du = (9000 + 1000t)(-10e^(-0.1t)) - ∫(-10e^(-0.1t))(1000 dt) = -10(9000 + 1000t)e^(-0.1t) + 10000 ∫e^(-0.1t) dt = -10(9000 + 1000t)e^(-0.1t) - 100,000e^(-0.1t). Now, evaluate this from 0 to ∞: As t → ∞, the expression approaches 0. At t = 0, the expression is -10(9000)e^0 - 100,000e^0 = -90,000 - 100,000 = -190,000. The value of the definite integral is (limit at ∞) - (value at 0) = 0 - (-190,000) = $190,000.

Question 16

A manufacturing plant generates revenue at a continuous rate of R(t)=50000e0.02tR(t) = 50000e^{0.02t} dollars per year, where tt is years from now. The plant will operate for 10 years before closure. If money has a time value of 4% per year compounded continuously, which expression gives the net present value of the revenue stream minus an initial investment of $400,000?

  1. ∫01050000e0.02te−0.04tdt−400000\int_0^{10} 50000e^{0.02t}e^{-0.04t} dt - 400000 (correct answer)
  2. ∫01050000e0.02te−0.04tdt+400000\int_0^{10} 50000e^{0.02t}e^{-0.04t} dt + 400000
  3. ∫01050000e−0.02te−0.04tdt−400000\int_0^{10} 50000e^{-0.02t}e^{-0.04t} dt - 400000
  4. ∫01050000e0.06te−0.04tdt−400000\int_0^{10} 50000e^{0.06t}e^{-0.04t} dt - 400000
Explanation: Net present value equals the present value of future cash flows minus initial investment. The present value of continuous revenue R(t)R(t) with discount rate rr is ∫0TR(t)e−rtdt\int_0^T R(t)e^{-rt} dt. Here we subtract the $400,000 initial cost. Choice B incorrectly adds the investment. Choice C has the wrong sign in the revenue growth rate. Choice D incorrectly combines the growth and discount rates before applying the discount factor.

Question 17

A company's profit rate follows P(t)=5000sin⁡(πt/6)+8000P(t) = 5000\sin(\pi t/6) + 8000 dollars per month, where tt is months from now. This represents seasonal variation with a 12-month cycle. If the company uses a 6% annual discount rate compounded continuously, what is the present value of profits over one complete cycle?

  1. ∫012[5000cos⁡(πt/6)+8000]e−0.005tdt≈$93,600\int_0^{12} [5000\cos(\pi t/6) + 8000]e^{-0.005t} dt \approx \$93,600
  2. ∫012[5000sin⁡(πt/6)+8000]e−0.06tdt≈$85,400\int_0^{12} [5000\sin(\pi t/6) + 8000]e^{-0.06t} dt \approx \$85,400
  3. ∫012[5000sin⁡(πt/6)+8000]e−0.5tdt≈$78,800\int_0^{12} [5000\sin(\pi t/6) + 8000]e^{-0.5t} dt \approx \$78,800
  4. ∫012[5000sin⁡(πt/6)+8000]e−0.005tdt≈$91,200\int_0^{12} [5000\sin(\pi t/6) + 8000]e^{-0.005t} dt \approx \$91,200 (correct answer)
Explanation: When you encounter present value problems with continuous cash flows, you need to set up an integral that discounts future profits back to today's dollars using the exponential decay factor e−rte^{-rt}. The present value formula requires three key components: the profit function P(t)=5000sin⁡(πt/6)+8000P(t) = 5000\sin(\pi t/6) + 8000, the correct discount factor, and proper integration limits. Since we want one complete 12-month cycle, we integrate from 0 to 12. The discount factor converts the annual rate to the proper form: a 6% annual rate compounded continuously means r=0.06r = 0.06, so our discount factor is e−0.06te^{-0.06t}. This gives us the integral: ∫012[5000sin⁡(πt/6)+8000]e−0.06tdt\int_0^{12} [5000\sin(\pi t/6) + 8000]e^{-0.06t} dt Answer A uses the wrong profit function—it has cosine instead of sine, which would represent the derivative of profit, not profit itself. Answer B has the correct setup but uses an incorrect discount rate of 6% instead of converting it properly for the exponential function. Answer C makes a decimal error, using 0.5 instead of 0.06, which would represent a 50% annual discount rate—unrealistically high. Answer D incorrectly converts the annual rate to 0.005 (which would be 0.5% annually), significantly underestimating the time value of money. Remember: in continuous discounting problems, always convert percentage rates to decimals (6% = 0.06) and double-check that your profit function matches exactly what's given in the problem statement.

Question 18

An oil well produces revenue at a declining rate given by R(t)=80000e−0.15tR(t) = 80000e^{-0.15t} dollars per year, where tt is years from now. If the well has an infinite operating life and the discount rate is 10% per year compounded continuously, what is the present value of all future revenues?

  1. ∫0∞80000e−0.15te−0.10tdt=800000.25=$320,000\int_0^{\infty} 80000e^{-0.15t}e^{-0.10t} dt = \frac{80000}{0.25} = \$320,000 (correct answer)
  2. ∫0∞80000e−0.15te−0.10tdt=800000.05=$1,600,000\int_0^{\infty} 80000e^{-0.15t}e^{-0.10t} dt = \frac{80000}{0.05} = \$1,600,000
  3. ∫0∞80000e−0.15te−0.10tdt=800000.15=$533,333\int_0^{\infty} 80000e^{-0.15t}e^{-0.10t} dt = \frac{80000}{0.15} = \$533,333
  4. ∫0∞80000e−0.15te−0.10tdt=800000.10=$800,000\int_0^{\infty} 80000e^{-0.15t}e^{-0.10t} dt = \frac{80000}{0.10} = \$800,000
Explanation: For the improper integral ∫0∞80000e−0.25tdt\int_0^{\infty} 80000e^{-0.25t} dt, we get 80000⋅10.25=32000080000 \cdot \frac{1}{0.25} = 320000. The combined exponent is −0.15−0.10=−0.25-0.15 - 0.10 = -0.25. Choice B uses ∣0.15−0.10∣=0.05|0.15 - 0.10| = 0.05. Choice C uses only the decline rate 0.15. Choice D uses only the discount rate 0.10.

Question 19

An investment fund receives contributions at a rate of C(t)=10000+2000tC(t) = 10000 + 2000t dollars per year and pays distributions at a rate of D(t)=8000e0.05tD(t) = 8000e^{0.05t} dollars per year, where tt is years from now. If the fund's required rate of return is 7% per year compounded continuously, what is the net present value of cash flows over the next 5 years?

  1. ∫05[(10000+2000t)e−0.07t−8000e0.05te−0.05t]dt≈$12,650\int_0^5 [(10000 + 2000t)e^{-0.07t} - 8000e^{0.05t}e^{-0.05t}] dt \approx \$12,650
  2. ∫05[(10000+2000t)+8000e0.05t]e−0.07tdt≈$46,230\int_0^5 [(10000 + 2000t) + 8000e^{0.05t}]e^{-0.07t} dt \approx \$46,230
  3. ∫05[(10000+2000t)−8000e0.05t]e−0.07tdt≈$8,420\int_0^5 [(10000 + 2000t) - 8000e^{0.05t}]e^{-0.07t} dt \approx \$8,420 (correct answer)
  4. ∫05[(10000+2000t)−8000e0.05t]e−0.02tdt≈$11,880\int_0^5 [(10000 + 2000t) - 8000e^{0.05t}]e^{-0.02t} dt \approx \$11,880
Explanation: When you encounter a net present value problem with continuous cash flows, you need to discount each future cash flow back to present value using the continuously compounded discount rate. The key insight is that contributions are positive cash flows (money coming in) while distributions are negative cash flows (money going out). The net cash flow at time tt is: C(t)−D(t)=(10000+2000t)−8000e0.05tC(t) - D(t) = (10000 + 2000t) - 8000e^{0.05t}. To find the present value, you multiply this net flow by the discount factor e−rte^{-rt} where r=0.07r = 0.07, then integrate over the 5-year period. This gives you: ∫05[(10000+2000t)−8000e0.05t]e−0.07tdt\int_0^5 [(10000 + 2000t) - 8000e^{0.05t}]e^{-0.07t} dt. Answer A incorrectly applies the discount factor. It multiplies only the contributions by e−0.07te^{-0.07t} but incorrectly handles the distributions by using e−0.05te^{-0.05t} instead of the proper discount rate. Answer B treats distributions as positive cash flows by adding 8000e0.05t8000e^{0.05t} instead of subtracting it. This fundamental error ignores that distributions represent money leaving the fund. Answer D uses the wrong discount rate of 2% (e−0.02te^{-0.02t}) instead of the required 7% return. This appears to stem from incorrectly calculating 0.07−0.05=0.020.07 - 0.05 = 0.02, which has no basis in present value calculations. Study tip: In NPV problems, always identify cash inflows versus outflows first, then apply the single discount rate consistently to the entire net cash flow. Don't let different growth rates in the problem confuse you about which discount rate to use.

Question 20

A mining operation extracts ore at a rate that follows E(t)=1000(1+0.1t)e−0.05tE(t) = 1000(1 + 0.1t)e^{-0.05t} tons per year, where tt is years from now. The ore sells for $200 per ton. If the company's cost of capital is 8% per year compounded continuously, what is the present value of revenue from this operation over its 20-year life?

  1. ∫020200⋅1000(1+0.1t)e−0.05te−0.03tdt≈$1,680,000\int_0^{20} 200 \cdot 1000(1 + 0.1t)e^{-0.05t}e^{-0.03t} dt \approx \$1,680,000
  2. ∫020200⋅1000(1+0.1t)e−0.05te−0.08tdt≈$1,240,000\int_0^{20} 200 \cdot 1000(1 + 0.1t)e^{-0.05t}e^{-0.08t} dt \approx \$1,240,000 (correct answer)
  3. ∫020200⋅1000(1+0.1t)e0.05te−0.08tdt≈$1,520,000\int_0^{20} 200 \cdot 1000(1 + 0.1t)e^{0.05t}e^{-0.08t} dt \approx \$1,520,000
  4. ∫020200⋅1000(1+0.1t)e−0.13tdt≈$1,180,000\int_0^{20} 200 \cdot 1000(1 + 0.1t)e^{-0.13t} dt \approx \$1,180,000
Explanation: Present value problems require you to discount future cash flows back to today's dollars using the cost of capital as your discount rate. When you see continuous compounding and a continuous income stream, you'll need to integrate the revenue function multiplied by the continuous discount factor. The revenue at time tt equals the extraction rate times the price per ton: 200×1000(1+0.1t)e−0.05t200 \times 1000(1 + 0.1t)e^{-0.05t}. To find present value, you must discount this by multiplying by e−rte^{-rt} where rr is the cost of capital (8% = 0.08). This gives you the integral: ∫020200⋅1000(1+0.1t)e−0.05te−0.08tdt\int_0^{20} 200 \cdot 1000(1 + 0.1t)e^{-0.05t}e^{-0.08t} dt Notice how the two exponential terms combine: e−0.05t⋅e−0.08t=e−0.13te^{-0.05t} \cdot e^{-0.08t} = e^{-0.13t}. Answer A incorrectly uses 3% instead of 8% as the discount rate, perhaps confusing a risk-free rate with the company's cost of capital. Answer C makes a sign error in the extraction rate function, changing e−0.05te^{-0.05t} to e+0.05te^{+0.05t}, which would represent growing rather than declining extraction efficiency. Answer D correctly combines the exponents to get e−0.13te^{-0.13t} but omits the crucial discount factor e−0.08te^{-0.08t} entirely in the setup, making it appear as if no time value of money adjustment is needed. Answer B correctly applies the 8% discount rate and maintains the original extraction function, yielding approximately $1,240,000. Study tip: In present value problems, always identify the discount rate first, then multiply the cash flow function by $e^{-rt}$. The cost of capital is your discount rate unless stated otherwise.