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Calculus I  ·  Chapter 5  ·  Section 5.5

Additional Applications of
Integration to Business & Economics

Income Streams · Consumer & Producer Surplus

Calculus doesn't just compute areas — it answers real financial questions: How much will a continuous income stream be worth in 5 years? Which investment is actually better? Who benefits most from a market price?

Learning Objectives

What You Will Learn

1

Use integration to compute the future value and present value of a continuous income stream.

2

Define consumer willingness to spend as a definite integral, then use it to compute consumer's surplus and producer's surplus.

§ 1 — Income Streams

Money Flowing Continuously
Into an Account

Think about a shop that earns sales revenue every hour of every day. The money doesn't come in one big lump — it trickles in continuously. If this revenue is deposited into a bank account that earns interest, it grows over time.

We call this a continuous income stream. The big question: at the end of N years, how much money has accumulated — including all the interest that kept compounding?

Key idea to remember: Money deposited early earns more interest than money deposited late. A rupee deposited today is worth more than a rupee deposited next year. The integral will automatically account for this.

Understanding with a Simple Story

Step 1 — Slice Time into Tiny Pieces

Divide the full term [0, T] into n equal slices, each of width Δt = T/n years. During the j-th slice (around time tj), your business earns income at rate f(tj), so that slice deposits f(tj)·Δt rupees into the account.

f(tⱼ)·Δt0ΔttⱼTΔt···n equal slices of width Δt = T/n
Step 2 — Each Deposit Earns Interest Until Year T

The deposit from slice j goes into the bank at time tj. It then earns continuously compounded interest for the remaining (T − tj) years — all the way until the end of the term.

deposit(T − tⱼ) years of interest0tⱼTgrownDeposit sits in bank, growing with e^(r·(T−tⱼ))
Step 3 — Earlier Deposits Grow More

By the continuous compounding formula, the slice deposited at tj becomes f(tj)·er(T−tj)·Δt by year T. Earlier deposits (small tj) have more time → they grow taller. Later deposits barely earn any interest.

t₁t₂t₃t₄t₅t₆→Tmostgrowthleastgrowth
Step 4 — Add All Slices → Take the Limit → Integral

Sum the future values of all n slices: Σ f(tj)·er(T−tj)·Δt. As n → ∞ (slices get infinitely thin), this Riemann sum converges to the definite integral:

$$\text{FV} = \int_0^T f(t)\,e^{r(T-t)}\,dt$$
Σ∫0Tn→∞ thin bars → smooth integralf(t)·e^r(T-t)
§ 2 — Future Value

Future Value of
an Income Stream

Future Value Formula

If money flows continuously into an account at rate $f(t)$ (PKR/year) and the account earns interest at annual rate $r$ compounded continuously, the future value at the end of year $T$ is:

$$\text{FV} = \int_0^T f(t)\,e^{r(T-t)}\,dt = e^{rT}\int_0^T f(t)\,e^{-rt}\,dt$$

The factor $e^{r(T-t)}$ represents how much each rupee deposited at time t grows by the end of year T.

Example 1 — Imran's Superstore Annuity

Imran owns a superstore in Lahore that generates revenue at a steady rate of PKR 120,000 per year. He deposits this continuously into a savings account earning 8% per year compounded continuously. How much will the account be worth at the end of 2 years?

How each small deposit grows to year T=2
012ttⱼ=0.4tⱼ=0.8tⱼ=1.2tⱼ=1.6grows e^(0.08(2−tⱼ))× (deposit)years

What we know: $f(t) = 120{,}000$ (constant), $r = 0.08$, $T = 2$.

Set up the integral:

$$\text{FV} = \int_0^2 120{,}000\cdot e^{0.08(2-t)}\,dt$$

Factor out the constant $e^{0.08\times 2} = e^{0.16}$:

$$= 120{,}000\cdot e^{0.16}\int_0^2 e^{-0.08t}\,dt$$

Evaluate the integral (antiderivative of $e^{-0.08t}$ is $\frac{e^{-0.08t}}{-0.08}$):

$$= 120{,}000\cdot e^{0.16}\cdot\left[\frac{e^{-0.08t}}{-0.08}\right]_0^2 = 120{,}000\cdot e^{0.16}\cdot\frac{e^{-0.16}-1}{-0.08}$$

$$= \frac{120{,}000}{0.08}\cdot e^{0.16}\cdot(1-e^{-0.16}) = 1{,}500{,}000\cdot(e^{0.16}-1)$$

$$= 1{,}500{,}000\cdot(1.17351-1) \approx \boxed{\text{PKR }260{,}266}$$

Interpretation: Imran deposited a total of PKR 120,000 × 2 = PKR 240,000. The account is worth PKR 260,266 — the extra PKR 20,266 is the interest earned on the continuously compounding deposits.

Example 2 — Growing Revenue Stream

A Karachi tech startup generates revenue at the rate $f(t) = 50{,}000e^{0.1t}$ PKR/year (revenue grows at 10%/year). Interest rate is 6% compounded continuously. Find the future value over 3 years.

Here $f(t)=50{,}000e^{0.1t}$, $r=0.06$, $T=3$.

$$\text{FV} = e^{0.18}\int_0^3 50{,}000e^{0.1t}\cdot e^{-0.06t}\,dt = 50{,}000e^{0.18}\int_0^3 e^{0.04t}\,dt$$

$$= 50{,}000e^{0.18}\left[\frac{e^{0.04t}}{0.04}\right]_0^3 = \frac{50{,}000e^{0.18}}{0.04}(e^{0.12}-1)$$

$$= 1{,}250{,}000\cdot e^{0.18}\cdot(e^{0.12}-1) \approx 1{,}250{,}000\times 1.1972\times 0.1275 \approx \boxed{\text{PKR }190{,}726}$$

💰 FV & PV Visualiser
Income Stream FV: Each slice deposited at t_j grows for only (T−t_j) remaining years. First slice = most growth → tallest. Last slice = barely grows → shortest. Bars DECREASE left→right (opposite of lump sum!).
f(t)
Rate C (PKR 000)
r = 8%
T = 4yr
n = 6 slices
Hover a bar → see deposit size & time to grow/discount
Exact FV
PKR 471.4k
e^(rT)·∫f·e^(−rt)dt
Exact PV
PKR 342.3k
∫f(t)·e^(−rt)dt
Deposited
PKR 400.0k
raw cash (no interest)
Interest
PKR 71.4k
FV − deposits
Riemann err
0.06k
n=6 vs exact
FV=e^(rT)·PV
471=e^0.3·342
always true
💡 Income stream FV bars DECREASE — early deposits earn MORE interest because they have longer to grow. This is the opposite of a lump sum. Switch to Lump Sum tab to compare!
§ 3 — Present Value

Present Value: What Is That
Future Income Worth Today?

Suppose someone offers you a business that will generate income for the next 5 years. What is a fair price to pay for it today? This is the present value question.

Intuition: PKR 100 today is worth more than PKR 100 a year from now — because you could invest today's PKR 100 and have more than PKR 100 next year. Present value works backwards: it asks "how much money do I need today so it grows to match the future income stream?"
Present Value Formula

The present value of an income stream with rate $f(t)$ over $[0,T]$ at interest rate $r$ compounded continuously is:

$$\text{PV} = \int_0^T f(t)\,e^{-rt}\,dt$$

The discount factor $e^{-rt}$ shrinks future money back to today's value. Note: $\text{FV} = e^{rT}\times\text{PV}$.

FV and PV — the relationship
PVToday (t=0)FVEnd of Year T× e^(rT) — invest and grow× e^(−rT) — discount back

Example 3 — Fair Price for a Business

A small factory in Faisalabad is expected to generate income at a constant rate of PKR 80,000/year for 5 years. If the prevailing interest rate is 6% compounded continuously, what is the fair present value of this income stream?

$f(t)=80{,}000$, $r=0.06$, $T=5$.

$$\text{PV} = \int_0^5 80{,}000\cdot e^{-0.06t}\,dt = 80{,}000\left[\frac{e^{-0.06t}}{-0.06}\right]_0^5$$

$$= \frac{80{,}000}{0.06}(1-e^{-0.30}) = 1{,}333{,}333\cdot(1-0.7408) \approx \boxed{\text{PKR }345{,}600}$$

Interpretation: If you invest PKR 345,600 today at 6%, it would grow to exactly match the income stream. So PKR 345,600 is a fair price to pay for this business today.
§ 4 — Comparing Two Investments

Which Investment Is
Actually Better?

When comparing two investment options, the fair way is to compute the net value = PV of income − initial cost. The higher the net value, the better the investment.

Example 4 — Sana Compares Two Investment Schemes

Sana is deciding between two investment options:

OptionCostIncome Rate
Option A — Tech Startup StakePKR 900,000$f_1(t)=300{,}000e^{0.03t}$ / year
Option B — Fixed AnnuityPKR 1,200,000$f_2(t)=400{,}000$ / year (constant)

The prevailing annual interest rate is 5% compounded continuously. Which option is better over a 5-year term?

Strategy: Compute PV − cost for each. Larger net value = better investment.

Option A: $r=0.05$, $T=5$, $f_1(t)=300{,}000e^{0.03t}$.

$$\text{PV}_A = \int_0^5 300{,}000e^{0.03t}\cdot e^{-0.05t}\,dt = 300{,}000\int_0^5 e^{-0.02t}\,dt$$

$$= 300{,}000\left[\frac{e^{-0.02t}}{-0.02}\right]_0^5 = \frac{300{,}000}{0.02}(1-e^{-0.10}) = 15{,}000{,}000\cdot(1-0.9048) \approx 1{,}427{,}500$$

$$\text{Net A} = 1{,}427{,}500 - 900{,}000 = \text{PKR }527{,}500$$

Option B: $f_2(t)=400{,}000$ (constant).

$$\text{PV}_B = 400{,}000\int_0^5 e^{-0.05t}\,dt = \frac{400{,}000}{0.05}(1-e^{-0.25}) = 8{,}000{,}000\cdot(1-0.7788) \approx 1{,}769{,}600$$

$$\text{Net B} = 1{,}769{,}600 - 1{,}200{,}000 = \text{PKR }569{,}600$$

Conclusion: Option B (net PKR 569,600) is better than Option A (net PKR 527,500), even though it costs more — the higher constant income more than compensates for the larger upfront cost.

Example 5 — Three-Way Comparison

Three investment options are available at 7% interest compounded continuously over a 4-year term:

OptionCostRate f(t)
AlphaPKR 500,000$200{,}000$ / yr
BetaPKR 600,000$150{,}000e^{0.05t}$ / yr
GammaPKR 400,000$180{,}000e^{-0.02t}$ / yr (declining)

Let $r=0.07$, $T=4$. General PV formula: $\text{PV}=\int_0^4 f(t)e^{-0.07t}\,dt$.

Alpha: $\text{PV}_{\alpha}=200{,}000\cdot\frac{1-e^{-0.28}}{0.07}=200{,}000\cdot\frac{0.2442}{0.07}\approx 697{,}714$. Net $= 697{,}714-500{,}000 = \text{PKR }197{,}714$.

Beta: $\text{PV}_{\beta}=150{,}000\int_0^4 e^{-0.02t}\,dt=150{,}000\cdot\frac{1-e^{-0.08}}{0.02}\approx 150{,}000\cdot 3.847=577{,}050$. Net $=577{,}050-600{,}000 = -\text{PKR }22{,}950$. ❌

Gamma: $\text{PV}_{\gamma}=180{,}000\int_0^4 e^{-0.09t}\,dt=180{,}000\cdot\frac{1-e^{-0.36}}{0.09}\approx 180{,}000\cdot 3.027=544{,}860$. Net $=544{,}860-400{,}000 = \text{PKR }144{,}860$.

Ranking: Alpha (PKR 197,714) > Gamma (PKR 144,860) > Beta (−PKR 22,950, avoid!)
§ 5 — Consumer Willingness to Spend

How Much Are Consumers
Actually Willing to Pay?

The TV Set Story

Imagine a family is willing to pay PKR 50,000 for their first TV. For a second TV (maybe for a different room), they'd only pay PKR 30,000 — it's less urgent. For a third TV, maybe just PKR 5,000. Their demand function captures this declining willingness.

Total willingness to spend for 3 TVs = PKR 50,000 + 30,000 + 5,000 = PKR 85,000. But for a continuous commodity (like grain, fuel, or electricity), we can't just add up a few values — we need to integrate.

Consumer Willingness to Spend (WS)

If $p = D(q)$ is the demand function (price consumers are willing to pay for the $q$-th unit), then the total willingness to spend for up to $q_0$ units is:

$$\text{WS} = \int_0^{q_0} D(q)\,dq$$

Geometrically, this is the entire area under the demand curve from 0 to q₀.

Willingness to Spend = Area Under Demand Curve

Example 6 — Rashid's Grain Market

Rashid, a farm manager in Punjab, finds that buyers are willing to pay $p = D(q) = 10(25-q^2)$ rupees per kg when $q$ kg of grain is available. Find the total amount buyers are willing to spend for up to 3 kg.

Check the price at $q=3$: $D(3)=10(25-9)=10(16)=160$ rupees/kg. This is the market price when 3 kg are sold.

$$\text{WS} = \int_0^3 10(25-q^2)\,dq = 10\int_0^3(25-q^2)\,dq = 10\left[25q-\frac{q^3}{3}\right]_0^3$$

$$= 10\left[(75-9)-(0)\right] = 10\times 66 = \boxed{\text{PKR }660\text{ per kg}}$$

Buyers are collectively willing to spend PKR 660 for 3 kg of grain at this demand schedule.

Example 7 — Electricity Demand

The demand for electricity (in units) in a neighbourhood follows $D(q) = 200 - 0.5q^2$ PKR per unit. Find total willingness to spend for up to 15 units.

$$\text{WS}=\int_0^{15}(200-0.5q^2)\,dq=\left[200q-\frac{0.5q^3}{3}\right]_0^{15}=(3000-562.5)-0=\boxed{\text{PKR }2{,}437.5}$$

§ 6 — Consumer's Surplus

Consumers' Surplus:
The "Happy Bargain" Measure

When you go to the market and buy something for less than you were willing to pay, you feel like you got a bargain. That savings — summed across all buyers — is the consumers' surplus.

Example: You were willing to pay PKR 5,000 for a textbook. It only costs PKR 3,500 in the market. Your consumers' surplus is PKR 1,500 — the amount you "saved" compared to your maximum willingness to pay.
Consumers' Surplus (CS)

If $p_0 = D(q_0)$ is the market price at which $q_0$ units are sold, the consumers' surplus is:

$$\text{CS} = \int_0^{q_0} D(q)\,dq - p_0 q_0$$

This equals: what consumers were willing to pay minus what they actually paid.

Geometrically: CS = area under the demand curve above the price line (the teal triangular region in the diagram).

Consumers' Surplus = Willingness to Spend − Actual Expenditure
∫D(q)dqp₀q₀
Total WS
p₀·q₀p₀q₀
− Actual Cost
CSp₀q₀
= CS

Example 8 — Grain Market CS

Using Rashid's demand function $D(q)=10(25-q^2)$, find the consumers' surplus when 3 kg of grain are sold at the market price.

Step 1 — Market price: $p_0=D(3)=10(25-9)=160$ PKR/kg.

Step 2 — CS formula:

$$\text{CS} = \int_0^3 10(25-q^2)\,dq - 160\times 3 = 660 - 480 = \boxed{\text{PKR }180}$$

Interpretation: Buyers collectively paid PKR 480 for 3 kg, but were willing to pay PKR 660. Their total savings (surplus) is PKR 180.

Example 9 — Electronics Bazaar

The demand for smartphones at a Saddar market follows $D(q)=500-q^2$ (PKR hundreds/unit). The market price is set at PKR 400 hundred. Find (a) the equilibrium quantity $q_0$, and (b) the consumers' surplus.

(a) Find q₀: $D(q_0)=400 \Rightarrow 500-q_0^2=400 \Rightarrow q_0^2=100 \Rightarrow q_0=10$.

(b) Consumers' surplus:

$$\text{CS}=\int_0^{10}(500-q^2)\,dq - 400\times 10 = \left[500q-\frac{q^3}{3}\right]_0^{10} - 4{,}000$$

$$= (5{,}000-333.33)-4{,}000 = \boxed{\text{PKR }666.67\text{ hundred} \approx 66{,}667}$$

Example 10 — Hyperbolic Demand

A commodity has demand function $D(q) = \dfrac{100}{q+1}$ PKR/unit. Find consumers' surplus when the market price is PKR 20.

Find q₀: $\frac{100}{q_0+1}=20 \Rightarrow q_0+1=5 \Rightarrow q_0=4$.

$$\text{CS}=\int_0^4\frac{100}{q+1}\,dq - 20\times 4 = 100\Big[\ln(q+1)\Big]_0^4 - 80 = 100\ln 5 - 80$$

$$= 100\times 1.6094 - 80 = 160.94 - 80 = \boxed{\text{PKR }80.94}$$

📊 Consumer & Producer Surplus
CS = area under D(q) above the market price p₀ — the "savings" buyers get vs what they were willing to pay.
Demand curve
q₀ = 3.0 units
Consumer's Surplus (CS)
36.00
∫D(q)dq − p₀q₀
Producer's Surplus (PS)
18.00
p₀q₀ − ∫S(q)dq
Market Price p₀
26.00
at q₀ = 3.0 units
CS — consumers' surplus
p₀·q₀ — actual spend
D(q) demand
§ 7 — Producer's Surplus

Producer's Surplus:
The Seller's Windfall

The story works in reverse for sellers. A producer might be willing to sell the first unit for as low as PKR 100, the second for PKR 150, and so on — but if the market price is PKR 300, they sell all units at PKR 300. The extra they receive compared to their minimum asking price is the producer's surplus.

Producer's Surplus (PS)

If $p_0 = S(q_0)$ is the market price and $p = S(q)$ is the supply function (minimum price producers will accept for the $q$-th unit), the producers' surplus is:

$$\text{PS} = p_0 q_0 - \int_0^{q_0} S(q)\,dq$$

Geometrically: PS = area above the supply curve, below the price line (the gold region).

Producer's Surplus — Gold region above supply curve
p₀q₀PSS(q)qp

Example 11 — Wheat Farmers' Surplus

Wheat farmers in Sindh have the supply function $S(q)=q^2+10$ PKR/unit. The market price is set at PKR 35. Find the producers' surplus.

Find q₀: $S(q_0)=35 \Rightarrow q_0^2+10=35 \Rightarrow q_0=5$.

$$\text{PS}=35\times 5-\int_0^5(q^2+10)\,dq = 175 - \left[\frac{q^3}{3}+10q\right]_0^5 = 175-(41.67+50) = \boxed{\text{PKR }83.33}$$

Farmers receive PKR 175 total but would have accepted as little as PKR 91.67. Their surplus is PKR 83.33.

Example 12 — Both Surpluses Together

A commodity has demand $D(q)=40-2q$ and supply $S(q)=4+q$ (both in PKR/unit). Find the equilibrium price and quantity, then compute both CS and PS.

Equilibrium: $D(q)=S(q) \Rightarrow 40-2q=4+q \Rightarrow 3q=36 \Rightarrow q_0=12$, $p_0=4+12=16$.

Consumers' Surplus:

$$\text{CS}=\int_0^{12}(40-2q)\,dq - 16\times 12 = \left[40q-q^2\right]_0^{12} - 192 = (480-144)-192 = \boxed{144}$$

Producers' Surplus:

$$\text{PS}=16\times 12-\int_0^{12}(4+q)\,dq = 192-\left[4q+\frac{q^2}{2}\right]_0^{12} = 192-(48+72) = \boxed{72}$$

Total Social Welfare = CS + PS = 144 + 72 = PKR 216. This is the total economic benefit generated by this market at equilibrium — split roughly 2:1 between consumers and producers.

Example 13 — Square Root Supply

Supply function: $S(q)=2\sqrt{q}+8$ PKR/unit. Market price is PKR 16. Find producers' surplus.

Find q₀: $2\sqrt{q_0}+8=16 \Rightarrow \sqrt{q_0}=4 \Rightarrow q_0=16$.

$$\text{PS}=16\times 16-\int_0^{16}(2\sqrt{q}+8)\,dq = 256-\left[\frac{4q^{3/2}}{3}+8q\right]_0^{16}$$

$$= 256-\left(\frac{4\times 64}{3}+128\right) = 256-(85.33+128) = 256-213.33 = \boxed{\text{PKR }42.67}$$

§ 8 — Practice Problems

Test Yourself

Problem 1 — Future Value

A LUMS alumni's business generates PKR 200,000/year continuously. Interest rate is 5% compounded continuously. Find the future value after 4 years.

$f(t)=200{,}000$, $r=0.05$, $T=4$.

$$\text{FV}=e^{0.20}\int_0^4 200{,}000\,e^{-0.05t}\,dt=200{,}000\cdot e^{0.20}\cdot\frac{1-e^{-0.20}}{0.05}\approx \boxed{\text{PKR }884{,}424}$$

Problem 2 — Present Value

A Quetta factory generates PKR 150,000/year for 6 years. At 4% interest compounded continuously, what is the present value?

$$\text{PV}=150{,}000\int_0^6 e^{-0.04t}\,dt=\frac{150{,}000}{0.04}(1-e^{-0.24})\approx 3{,}750{,}000\times 0.2127\approx \boxed{\text{PKR }797{,}625}$$

Problem 3 — Compare Two Options

Option X costs PKR 500,000 and earns $f(t)=120{,}000e^{0.02t}$ / year. Option Y costs PKR 400,000 and earns $f(t)=100{,}000$ / year. At 6% interest over 5 years, which is better?

$\text{PV}_X=120{,}000\int_0^5 e^{-0.04t}\,dt=120{,}000\cdot\frac{1-e^{-0.2}}{0.04}=120{,}000\cdot 4.524=542{,}880$. Net X $=42{,}880$.

$\text{PV}_Y=100{,}000\cdot\frac{1-e^{-0.3}}{0.06}=100{,}000\cdot 4.346=434{,}600$. Net Y $=34{,}600$.

Option X is better (higher net PV).

Problem 4 — Consumer's Surplus — Quadratic Demand

Demand function: $D(q)=100-2q-q^2$. Market sells 5 units. Find consumers' surplus.

$p_0=D(5)=100-10-25=65$.

$$\text{CS}=\int_0^5(100-2q-q^2)\,dq-65\times 5=\left[100q-q^2-\frac{q^3}{3}\right]_0^5-325=(500-25-41.67)-325=\boxed{108.33}$$

Problem 5 — Producer's Surplus — Exponential Supply

Supply function: $S(q)=e^{0.5q}$ PKR/unit. Market price is $e^2$ PKR. Find producers' surplus.

$S(q_0)=e^2 \Rightarrow e^{0.5q_0}=e^2 \Rightarrow q_0=4$.

$$\text{PS}=e^2\times 4-\int_0^4 e^{0.5q}\,dq=4e^2-\left[\frac{e^{0.5q}}{0.5}\right]_0^4=4e^2-2(e^2-1)=2e^2+2\approx\boxed{16.78}$$

Problem 6 — Growing Income Stream FV

Income rate $f(t)=10{,}000(1+0.1t)$ PKR/year, $r=5\%$, $T=3$ years. Find FV.

$$\text{FV}=e^{0.15}\int_0^3 10{,}000(1+0.1t)e^{-0.05t}\,dt$$

Use integration by parts: $\int(1+0.1t)e^{-0.05t}\,dt$. Let $u=1+0.1t$, $dv=e^{-0.05t}dt$. After IBP: $\approx 3.083$.

$$\text{FV}\approx 10{,}000\cdot e^{0.15}\cdot 3.083\times 3\approx 10{,}000\times 1.1618\times 9.249\approx\boxed{\text{PKR }107{,}484}$$

Problem 7 — Market Equilibrium + Both Surpluses

Demand $D(q)=60-3q$, Supply $S(q)=2q+10$. Find equilibrium, CS, and PS.

Equilibrium: $60-3q=2q+10 \Rightarrow q_0=10$, $p_0=30$.

$$\text{CS}=\int_0^{10}(60-3q)\,dq-30\times 10=\left[60q-\frac{3q^2}{2}\right]_0^{10}-300=(600-150)-300=\boxed{150}$$

$$\text{PS}=30\times 10-\int_0^{10}(2q+10)\,dq=300-[q^2+10q]_0^{10}=300-200=\boxed{100}$$

Chapter 5 Complete! 🎉 You have now mastered indefinite integration, substitution, the definite integral, the FTC, area between curves, Lorenz curves, average value, income streams, and consumer/producer surplus. Chapter 6 awaits: Integration by Parts, improper integrals, and continuous probability.
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