9 min readConcept sheetShort integral; 10-mark total cost or consumer surplus problem

Integration and its business uses

Integration reverses differentiation. Integrate a marginal function to rebuild the total; a definite integral gives the area under a curve.

Power rule∫xⁿ dx = xⁿ⁺¹/(n + 1) + C, n ≠ −1

Reciprocal∫(1/x) dx = ln|x| + C

Definite integral∫ₐᵇ f(x) dx = F(b) − F(a)

Total costC(x) = ∫MC dx; constant = fixed cost

Rules of integration

Rule	Formula	Example
Constant	∫k dx = kx + C	∫5 dx = 5x + C
Power (n ≠ −1)	∫xⁿ dx = xⁿ⁺¹/(n + 1) + C	∫x³ dx = x⁴/4 + C
Constant multiple	∫kf dx = k∫f dx	∫6x² dx = 2x³ + C
Sum / difference	∫(f ± g) dx = ∫f dx ± ∫g dx	∫(x + 1) dx = x²/2 + x + C
Reciprocal	∫(1/x) dx = ln\|x\| + C	∫(4/x) dx = 4 ln\|x\| + C
Exponential	∫eˣ dx = eˣ + C	∫2eˣ dx = 2eˣ + C
Definite	∫ₐᵇ f dx = F(b) − F(a)	∫₀² 3x² dx = 8 − 0 = 8

Business uses

Total cost from marginal costC(x) = ∫MC dx + fixed cost

Total revenue from marginal revenueR(x) = ∫MR dx (R(0) = 0)

Change in cost from a to b units∫ₐᵇ MC dx

Consumer surplus (demand D, price p₀, quantity q₀)∫₀^q₀ D(q) dq − p₀q₀

Producer surplus (supply S)p₀q₀ − ∫₀^q₀ S(q) dq

Indefinite integrals

Problem

Find (a) ∫(6x² − 4x + 5) dx, (b) ∫(3/x² + √x) dx.

Working

(a) Integrate term by term: 6x³/3 − 4x²/2 + 5x + C = 2x³ − 2x² + 5x + C.
(b) Write with powers: 3x⁻² + x^(1/2).
∫3x⁻² dx = 3x⁻¹/(−1) = −3/x.
∫x^(1/2) dx = x^(3/2) ÷ (3/2) = (2/3)x^(3/2).
Check (a) by differentiating: 6x² − 4x + 5. ✓

Answer

(a) 2x³ − 2x² + 5x + C (b) −3/x + (2/3)x^(3/2) + C

Definite integrals

Problem

Evaluate (a) ∫₁³ (2x + 1) dx, (b) ∫₀² 3x² dx.

Working

(a) Antiderivative x² + x. [x² + x] from 1 to 3 = (9 + 3) − (1 + 1) = 12 − 2 = 10.
(b) Antiderivative x³. [x³] from 0 to 2 = 8 − 0 = 8.
No + C is needed: it cancels in F(b) − F(a).

Answer

(a) 10 (b) 8

Total cost from marginal cost

Problem

A Kurunegala factory's marginal cost is MC = 30 + 0.6x (Rs. per unit) and its fixed cost is Rs. 8,000. Find (a) the total cost function, (b) the cost of 100 units, (c) the extra cost of raising output from 100 to 200 units.

Working

(a) C(x) = ∫(30 + 0.6x) dx = 30x + 0.3x² + K. At x = 0, C = fixed cost, so K = 8,000.
C(x) = 8,000 + 30x + 0.3x².
(b) C(100) = 8,000 + 3,000 + 3,000 = 14,000.
(c) ∫ from 100 to 200 of (30 + 0.6x) dx = [30x + 0.3x²] = (6,000 + 12,000) − (3,000 + 3,000) = 12,000.

Answer

(a) C(x) = 8,000 + 30x + 0.3x² (b) Rs. 14,000 (c) Rs. 12,000

Consumer and producer surplus

Problem

Demand is p = 120 − 2q and supply is p = 20 + 0.5q. Find the equilibrium, the consumer surplus and the producer surplus.

Working

Equilibrium: 120 − 2q = 20 + 0.5q, so 100 = 2.5q, q₀ = 40 and p₀ = 40.
CS = ∫₀⁴⁰ (120 − 2q) dq − 40 × 40 = [120q − q²]₀⁴⁰ − 1,600 = (4,800 − 1,600) − 1,600 = 1,600.
PS = 40 × 40 − ∫₀⁴⁰ (20 + 0.5q) dq = 1,600 − [20q + 0.25q²]₀⁴⁰ = 1,600 − (800 + 400) = 400.
Check with triangles: CS = ½ × 40 × 80 = 1,600 and PS = ½ × 40 × 20 = 400. ✓

Answer

q₀ = 40, p₀ = 40; consumer surplus 1,600; producer surplus 400.

In Sri Lankan business

If a Sri Lankan factory knows its marginal cost from its records, integrating it rebuilds the total cost function; the constant of integration is the fixed cost (rent, salaries, loan interest).

Exam angle

Short answer: “Find ∫(4x³ + 6x) dx.” x⁴ + 3x² + C.

Structured question: “Demand p = 120 − 2q, supply p = 20 + 0.5q. Find the consumer surplus at equilibrium.” q₀ = 40, p₀ = 40; CS = ∫₀⁴⁰(120 − 2q)dq − 1,600 = 1,600.

Essay: “MC = 30 + 0.6x, fixed cost Rs. 8,000. Find C(x), the cost of 100 units and the extra cost from 100 to 200 units.” C = 8,000 + 30x + 0.3x²; Rs. 14,000; Rs. 12,000.

Remember it

Integration adds one to the power, then divides by the new power. Never forget + C: in business, that C is the fixed cost.