(a) Find the consumer's surplus for the demand function $p = 25 - x - x^2$ , where the prevailing market price $p_0 = 19$ .

OR

(b) Solve the following initial value differential equation : $(x - 1) \frac{dy}{dx} = 2xy$ , when $y(2) = 1$ .

Integration and its Applications / Differential Equations

Consumer's Surplus / Initial Value Differential Equation

Official Answer

This is an internal-choice long-answer question; either part is worth the full 5 marks.

Part (a) — Consumer's Surplus

The demand function is $p = 25 - x - x^2$ , and the market price is $p_0 = 19$ .

Find the quantity x₀ at price 19: $19 = 25 - x - x^2 \implies x^2 + x - 6 = 0 \implies (x + 3)(x - 2) = 0 \implies x_0 = 2$ (reject $x = -3$ , quantity cannot be negative).
Apply the formula: $CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0$ .
$\int_0^2 (25 - x - x^2) dx = [25x - x^2/2 - x^3/3]_0^2 = 50 - 2 - 8/3 = 136/3$ .
$CS = 136/3 - (19 \times 2) = 136/3 - 38 = 22/3$ .

Consumer's Surplus = $22/3 \approx 7.33$ (monetary units).

Part (b) — Initial Value Problem

$(x - 1) dy/dx = 2xy$ is variable-separable.

Separate: $dy/y = [2x/(x - 1)] dx$ .
Split the RHS: $2x/(x - 1) = 2 + 2/(x - 1)$ .
Integrate: $\ln y = 2x + 2 \ln|x - 1| + C$ .
Apply $y(2) = 1$ : $\ln 1 = 4 + 2 \ln 1 + C \implies 0 = 4 + C \implies C = -4$ .

Particular solution: $\ln y = 2x + 2 \ln|x - 1| - 4$ , i.e. $y = (x - 1)^2 \cdot e^{2x - 4}$ .

consumer's surplusdemand functiondefinite integralmarket pricevariable separableinitial value problemparticular solutionintegrating factor of rational

Marking Scheme

1Part (a): 1 mark for setting up and solving $19 = 25 - x - x^2$ to get $x_0 = 2$ (rejecting $x = -3$ ).
2Part (a): 1 mark for writing the correct surplus formula $CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0$ .
3Part (a): 2 marks for correct integration $[25x - x^2/2 - x^3/3]_0^2 = 136/3$ .
4Part (a): 1 mark for the final answer $CS = 22/3$ ≈ Rs 7.33.
5Part (b) alternative: 1 mark separation of variables; 2 marks for splitting $2x/(x-1)$ and integrating to $\ln y = 2x + 2 \ln|x-1| + C$ ; 1 mark applying $y(2)=1$ to get $C = -4$ ; 1 mark final solution $y = (x-1)^2 e^{2x-4}$ .

Hint

For (a): first put $p_0 = 19$ into the demand law to get the quantity x_0, then $CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0$ . For (b): it is variable-separable — write $dy/y = [2x/(x-1)] dx$ and split $2x/(x-1) = 2 + 2/(x-1)$ .

Quick Oral Answer

Consumer's surplus is the extra benefit buyers get because the market price is below what they were willing to pay — computed as the definite integral of the demand curve minus price times quantity; here it works out to $22/3$ , about Rs 7.33.

Analysis & Explanation

This question tests two distinct calculus applications through internal choice, and a student should attempt whichever they are more confident in.

Concept — Consumer's Surplus (a)

Consumer's surplus is the monetary gain to buyers who were willing to pay more than the actual market price. Geometrically it is the area between the demand curve and the horizontal price line $p = p_0$ , from $x = 0$ to the equilibrium quantity x₀.
The single formula $CS = \int_0^{x_0} f(x) dx - p_0 \cdot x_0$ captures this area; the definite integral is the total willingness-to-pay and p₀·x₀ is the amount actually paid.

Concept — Separable ODE (b)

An equation of the form $g(y) dy = h(x) dx$ is solved by integrating both sides. The algebraic trick of writing $2x/(x-1) = 2 + 2/(x-1)$ (splitting an improper rational function) is the key step and the most common place students get stuck.

Exam trap

In (a), students forget to reject the negative root $x = -3$ , or subtract p₀·x₀ before integrating.
In (b), forgetting the +C or applying the initial condition incorrectly loses the accuracy mark.

Real-world

Consumer's surplus is a core idea in welfare economics used to measure how much a market price benefits consumers; governments use it when analysing the impact of taxes and subsidies.

Common Mistakes

1In (a), keeping the negative root $x = -3$ as a quantity, or computing $p_0 \cdot x_0$ with the wrong x₀.
2In (b), failing to split $2x/(x-1)$ as $2 + 2/(x-1)$ and instead integrating incorrectly.
3Omitting the constant of integration C or mis-substituting the initial condition, so the particular solution is wrong.

Interesting Facts

The concept of consumer's surplus was introduced by French engineer Jules Dupuit in 1844 and popularised by Alfred Marshall in his 1890 'Principles of Economics'.

Separable differential equations are the oldest solved class of ODEs — Leibniz described the separation-of-variables method in 1691, just years after inventing calculus notation.

Consumer's surplus is still used today by regulators and antitrust bodies to estimate the welfare loss from monopolies and price increases.

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Frequently Asked Questions

Why do we reject $x = -3$ when finding the quantity in the consumer's surplus problem?

The variable x represents the quantity demanded, which cannot be negative in an economic context. Solving $x^2 + x - 6 = 0$ gives $x = 2$ and $x = -3$ ; only $x = 2$ is physically meaningful, so $x_0 = 2$ is used as the upper limit of integration.

How do you integrate $2x/(x - 1)$ in the differential equation part?

Since the numerator's degree is not lower than the denominator's, first do the division: $2x/(x - 1) = 2 + 2/(x - 1)$ . Then integrate term by term to get $2x + 2 \ln|x - 1|$ , which is the standard technique for improper rational functions.