Case Study 2 — Pricing a Subscription: Surplus, Discounting, and the Lifetime Value of a Customer

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Case Study 2 — Pricing a Subscription: Surplus, Discounting, and the Lifetime Value of a Customer

Field: Microeconomics / managerial economics Techniques used: $u$-substitution (Sections 15.1–15.2), integration by parts (Section 15.5), and the rotating idea of choosing the right $u$ (Section 15.4).

The Setting

Marcus Webb is the head of monetization at a streaming startup. Three questions sit on his desk, and all three are integrals that Chapter 15 can crack:

How much value do our subscribers get above what they pay — the consumer surplus?
What is a customer worth over their lifetime, once we discount future revenue back to today — the customer lifetime value (LTV)?
If churn isn't constant but climbs over time, how does that change the answer?

Marcus does not need a forecasting team for the math; he needs the product rule and the chain rule, read backwards. We follow his afternoon.

Part 1 — Consumer Surplus from an Exponential Demand Curve (a $u$-Substitution)

The startup's demand curve — how many subscribers would pay a given monthly price $p$ — fits an exponential well:

$$D(p) = a\, e^{-bp}, \qquad a = 50{,}000, \quad b = 0.08\ \text{per dollar}.$$

The market clears at the chosen price $p^* = \$10$. Consumer surplus is the total value subscribers receive above the price they actually pay, which (Section 15.10) is the area to the left of the demand curve above the clearing price:

$$\text{CS} = \int_{p^*}^{\infty} D(p)\, dp = \int_{10}^{\infty} a\, e^{-bp}\, dp.$$

This is a clean $u$-substitution. Let $u = -bp$, $du = -b\,dp$, so $dp = -\tfrac{1}{b}\,du$:

$$\int a\,e^{-bp}\,dp = -\frac{a}{b}\,e^{-bp} + \text{const}.$$

Evaluating the (improper, Chapter 17) integral, the antiderivative decays to zero at the top:

$$\text{CS} = \left[-\frac{a}{b}e^{-bp}\right]_{10}^{\infty} = 0 - \left(-\frac{a}{b}e^{-10b}\right) = \frac{a}{b}\,e^{-bp^*}.$$

Numerically, $\tfrac{a}{b} = 50{,}000/0.08 = 625{,}000$ and $e^{-0.8}\approx 0.4493$, so $\text{CS} \approx 280{,}800$ dollars of surplus value per period. The surplus is the demand intercept-rate $a/b$ shrunk by the exponential factor $e^{-bp^*}$: every dollar Marcus adds to the price slides $p^*$ rightward and multiplies surplus by $e^{-b}\approx 0.923$ — a 7.7% surplus loss per dollar. That elasticity intuition came straight out of the substitution.

Part 2 — Lifetime Value with Constant Churn and Discounting (a Pure Substitution)

A subscriber paying $m = \$10$/month does not stay forever, and a dollar next year is worth less than a dollar today. Model the *probability the customer is still subscribed* at time $t$ (years) as exponential survival $S(t) = e^{-ct}$ with annual churn $c = 0.5$, and discount future money at continuous rate $r = 0.1$. The expected, discounted lifetime revenue is

$$\text{LTV} = \int_0^\infty (12m)\, e^{-ct}\, e^{-rt}\, dt = 12m \int_0^\infty e^{-(c+r)t}\, dt.$$

The two exponentials combine, and what remains is the Model-1 substitution again, $u = -(c+r)t$:

$$\text{LTV} = 12m \cdot \frac{1}{c + r} = \frac{12\cdot 10}{0.6} = \$200.$$

LTV is annual revenue divided by the combined churn-plus-discount rate. Doubling churn from $0.5$ to $1.0$ drops the denominator-driven LTV from \$200 to $120/1.1 \approx \$109$. Marcus now has a hard number for what he can spend to acquire a customer: anything under \$200 in acquisition cost is profitable at these rates — a decision that rests entirely on one exponential integral.

Part 3 — When Churn Rises Over Time (Integration by Parts)

Constant churn is a convenient fiction. In reality, the hazard of cancelling often grows as a promotional discount expires — customers who were "on the fence" leak out faster as time passes. Suppose the instantaneous value the firm collects is modeled as a rate that rises then is eroded by churn, giving a revenue-density of the form $t\,e^{-\alpha t}$. The expected discounted contribution becomes

$$V = \int_0^\infty K\, t\, e^{-\alpha t}\, dt,$$

with $\alpha = c + r = 0.6$ and $K$ a revenue constant. Now there is a stray $t$ multiplying the exponential — a product of different types, the unmistakable signal for integration by parts (Section 15.4). LIATE puts the algebraic factor $t$ as $u$. Take $u = t$, $dv = e^{-\alpha t}\,dt$, so $du = dt$, $v = -\tfrac{1}{\alpha}e^{-\alpha t}$:

$$\int_0^\infty t\, e^{-\alpha t}\, dt = \left[-\frac{t}{\alpha}e^{-\alpha t}\right]_0^\infty + \frac{1}{\alpha}\int_0^\infty e^{-\alpha t}\, dt.$$

The boundary term is zero at both limits: $t=0$ kills it below, and $t\,e^{-\alpha t}\to 0$ above because exponential decay outruns the linear $t$. The leftover is the Part-2 substitution:

$$\int_0^\infty t\, e^{-\alpha t}\, dt = 0 + \frac{1}{\alpha}\cdot\frac{1}{\alpha} = \frac{1}{\alpha^2}.$$

So $V = K/\alpha^2$. Compare this with the constant-rate version $\int_0^\infty e^{-\alpha t}\,dt = 1/\alpha$: the rising-then-falling profile contributes $1/\alpha^2$ instead of $1/\alpha$, an extra factor of $1/\alpha$ that, with $\alpha = 0.6 < 1$, increases the weight on the early high-engagement period. Integration by parts is exactly what turns a more realistic churn story into a closed-form LTV.

Choosing $u$ correctly (Section 15.4). Had Marcus set $u = e^{-\alpha t}$ and $dv = t\,dt$, the new integral would carry $\tfrac{t^2}{2}e^{-\alpha t}$ — the power of $t$ goes up, making things worse. LIATE's "let the algebraic factor be $u$" is the rule that keeps the polynomial marching toward zero.

Part 4 — A Surplus Integral That Needs Both Techniques

Finally Marcus considers a richer demand model where willingness-to-pay is linear-times-exponential, $D(p) = (A - Bp)\,e^{-bp}$, capturing both a saturation ceiling and exponential decay. The willingness-to-pay area over $[0, p_{\max}]$ is

$$\int_0^{p_{\max}} (A - Bp)\, e^{-bp}\, dp.$$

The exponential alone is a substitution; the linear factor $A - Bp$ in front forces one integration by parts on top of it. Split the integral: the $A\,e^{-bp}$ piece is the Part-1 substitution, and the $-Bp\,e^{-bp}$ piece is the Part-3 parts calculation. The two techniques of this chapter, used in sequence, dispatch the whole thing — a concrete instance of the combined-techniques skill of Section 15.8. Marcus does not even need the final number to make his point in the strategy meeting: he can say with confidence that the model has a clean closed form, which is itself the finding worth reporting (Section 15.11's portfolio note).

What Marcus Concludes

Every question on Marcus's desk collapsed to an exponential integral solved by substitution, sometimes with one layer of integration by parts on top:

Consumer surplus: $\text{CS} = \tfrac{a}{b}e^{-bp^*}$ — substitution.
Constant-churn LTV: $\text{LTV} = \tfrac{12m}{c+r}$ — substitution.
Rising-churn contribution: $V = K/\alpha^2$ — integration by parts.

The strategic upshot — how much to spend acquiring a customer, how a price change moves surplus — rests on antiderivatives no harder than the chapter's worked examples. That is the recurring theme of this book in microeconomic dress: calculus appears in every quantitative field, and the same two reverse-rules that physics uses for work, statistics uses for expected values, and economics uses for surplus and lifetime value.

Discussion Questions

In Part 1, show that raising the price $p^*$ by one dollar multiplies consumer surplus by exactly $e^{-b}$. What does this say about the elasticity of surplus with respect to price?
Part 2 found $\text{LTV} = 12m/(c+r)$. Holding revenue fixed, is LTV more sensitive to a change in churn $c$ or in the discount rate $r$? Justify with the formula.
Re-do the Part 3 integration by parts with the wrong choice $u = e^{-\alpha t}$, $dv = t\,dt$, and show explicitly how the resulting integral is harder than the original. Connect this to LIATE.
The rising-churn model gave $V = K/\alpha^2$ versus $K/\alpha$ for constant churn. For what values of $\alpha$ does the rising-churn model assign more total value, and what is the economic interpretation?
Part 4's integral splits into a substitution piece and a parts piece. Carry out the full calculation for $A = 100$, $B = 5$, $b = 0.1$, $p_{\max} = 20$, and compare with exercise F5.

A Short Annotated Reading

Varian, Intermediate Microeconomics (9th ed.), ch. 14–15. Consumer surplus as an integral under the demand curve; read it alongside Part 1 to see the area interpretation made rigorous.
Stewart, Calculus: Early Transcendentals (9th ed.), §7.1 and §8.4. §7.1 is integration by parts (Part 3); §8.4 applies integrals to economics, including consumer/producer surplus.
OpenStax, Calculus Volume 2, §2.8 and §3.1. §2.8 covers economic applications of integration (surplus); §3.1 is integration by parts. Freely available and a clean match to this case study's two techniques.

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