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These are complete exercise solutions for Samacheer Kalvi 12th Business Maths — Integral Calculus II (Chapter 3). This chapter focuses on applications of integration — area under curves, consumer/producer surplus, and other business applications. Solve each question yourself first, then verify below.
Key Concepts (Quick Recap)
- Area under a curve: Area between y = f(x), the x-axis, and lines x = a, x = b is ∫ₐᵇ f(x) dx (when f(x) ≥ 0).
- Area between two curves: If f(x) ≥ g(x) on [a,b], Area = ∫ₐᵇ [f(x) − g(x)] dx.
- Area enclosed by a curve and y-axis: ∫ₐᵇ x dy where x is expressed as a function of y.
- Consumer Surplus (CS): CS = ∫₀^q₀ D(q) dq − p₀q₀, where (q₀, p₀) is the equilibrium point.
- Producer Surplus (PS): PS = p₀q₀ − ∫₀^q₀ S(q) dq.
- Equilibrium point: Found by solving D(q) = S(q). The intersection gives (q₀, p₀).
- Demand function D(q): Price consumers are willing to pay as a function of quantity. Downward sloping.
- Supply function S(q): Price at which producers are willing to supply as a function of quantity. Upward sloping.
- Area bounded by parametric curves: If x = f(t), y = g(t), Area = ∫ y (dx/dt) dt with appropriate limits.
- Properties used: All definite integral properties from Chapter 2 apply here.
I. Choose the Best Answer (MCQs)
Q1. The equilibrium price and quantity for demand p = 50 − 2q and supply p = 10 + 2q are: (a) p₀ = 30, q₀ = 10 ✓ (b) p₀ = 20, q₀ = 15 (c) p₀ = 40, q₀ = 5 (d) p₀ = 25, q₀ = 12
Answer(a) ✓ — Set D(q) = S(q): 50 − 2q = 10 + 2q → 40 = 4q → q₀ = 10. Then p₀ = 50 − 2(10) = 30. Equilibrium point: (10, 30).
Q2. The area enclosed by y = x², the x-axis, and the lines x = 0 and x = 2 is: (a) 4/3 (b) 8/3 ✓ (c) 2 (d) 16/3
Answer(b) ✓ — Area = ∫₀² x² dx = [x³/3]₀² = 8/3 − 0 = 8/3 square units.
Q3. Consumer surplus at equilibrium (5, 30) with demand p = 60 − 6q is: (a) 75 ✓ (b) 150 (c) 300 (d) 50
Answer(a) ✓ — CS = ∫₀⁵ (60 − 6q) dq − 30 × 5 = [60q − 3q²]₀⁵ − 150 = (300 − 75) − 150 = 225 − 150 = 75.
Q4. The area bounded by y = x and y = x² between x = 0 and x = 1 is: (a) 1/6 ✓ (b) 1/3 (c) 1/2 (d) 1/12
Answer(a) ✓ — x ≥ x² on [0,1]. Area = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6.
Q5. If the demand function is p = 100 − q² and the supply function is p = 20 + q, the equilibrium quantity is: (a) 6 (b) 8 ✓ (c) 10 (d) 12
Answer(b) ✓ — Set D = S: 100 − q² = 20 + q → q² + q − 80 = 0. Using the quadratic formula: q = (−1 ± √(1+320))/2 = (−1 ± √321)/2. Since √321 ≈ 17.9, q ≈ (−1 + 17.9)/2 ≈ 8.45. Actually, let me recalculate: q² + q − 80 = 0 → q = (−1 ± √321)/2 ≈ 8.4. The closest answer is 8. Alternatively, if the equation is q² + q − 80 = 0, checking q = 8: 64 + 8 − 80 = −8 ≠ 0. Let me try p = 100 − q² and p = 20 + q: 100 − q² = 20 + q → q² + q − 80 = 0. q = 8 gives 64 + 8 − 80 = −8, not zero. The exact answer is q = (−1 + √321)/2 ≈ 8.4. The closest integer is 8.
Q6. The area of the region bounded by y = 4 − x² and the x-axis is: (a) 8/3 (b) 16/3 ✓ (c) 32/3 (d) 4
Answer(b) ✓ — The curve meets the x-axis when 4 − x² = 0, i.e., x = ±2. Area = ∫₋₂² (4 − x²) dx = [4x − x³/3]₋₂² = (8 − 8/3) − (−8 + 8/3) = 16 − 16/3 = 16/3.
Q7. Producer surplus is given by the formula: (a) ∫₀^q₀ D(q) dq − p₀q₀ (b) p₀q₀ − ∫₀^q₀ S(q) dq ✓ (c) ∫₀^q₀ S(q) dq − p₀q₀ (d) p₀q₀ − ∫₀^q₀ D(q) dq
Answer(b) ✓ — Producer surplus is the area above the supply curve and below the equilibrium price line: PS = p₀q₀ − ∫₀^q₀ S(q) dq.
Q8. The area bounded by the curve y = √x, the x-axis, and x = 4 is: (a) 16/3 ✓ (b) 8/3 (c) 4 (d) 2√2
Answer(a) ✓ — Area = ∫₀⁴ √x dx = ∫₀⁴ x^(1/2) dx = [x^(3/2)/(3/2)]₀⁴ = (2/3)(4)^(3/2) = (2/3)(8) = 16/3.
II. Fill in the Blanks
1. The equilibrium point is the intersection of the ______ and ______ curves. demand, supply
2. Consumer surplus = ∫₀^q₀ D(q) dq − ______. p₀q₀
3. The area bounded by y = f(x) and y = g(x) from x = a to x = b (where f(x) ≥ g(x)) is given by ∫ₐᵇ ______ dx. [f(x) − g(x)]
4. Producer surplus = p₀q₀ − ∫₀^q₀ ______ dq. S(q)
5. The area enclosed between y = x and y = x³ in the first quadrant is ______. 1/4 (Area = ∫₀¹ (x − x³) dx = 1/2 − 1/4 = 1/4)
6. If demand p = 20 − q and supply p = 5 + q, then the equilibrium quantity q₀ = ______. 7.5 (20 − q = 5 + q → 15 = 2q → q₀ = 7.5)
7. The area enclosed by y² = 4x and x = 9 is ______. 36 (y = ±2√x; Area = 2∫₀⁹ 2√x dx = 4 · [x^(3/2)/(3/2)]₀⁹ = 4 · 18 = 36)
III. True or False
1. Consumer surplus is always greater than producer surplus. → False. There is no such guarantee. The relative sizes depend on the specific demand and supply functions.
2. The area between two curves is always ∫ₐᵇ [f(x) − g(x)] dx where f(x) ≥ g(x). → True. The area between curves is always computed using the absolute difference, but we arrange it so the upper curve minus the lower curve ensures a positive result.
3. At the equilibrium point, demand price equals supply price. → True. The equilibrium point is where D(q₀) = S(q₀) = p₀.
4. The area bounded by y = x² and y = x is the same as the area bounded by y = x and y = x³ from x = 0 to x = 1. → False. Area between x² and x on [0,1] = ∫₀¹ (x − x²) dx = 1/6. Area between x and x³ on [0,1] = ∫₀¹ (x − x³) dx = 1/2 − 1/4 = 1/4. These are different.
5. If the supply curve shifts upward, producer surplus increases. → False. An upward shift in supply (higher costs) actually reduces producer surplus at the same equilibrium. Producer surplus depends on the gap between equilibrium price and the supply curve.
| 6. Area enclosed by a curve below the x-axis is negative unless we take the absolute value. → True. ∫ₐᵇ f(x) dx gives a negative value when f(x) < 0. The geometric area is | ∫ₐᵇ f(x) dx | . |
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IV. Match the Following
| Left | Right |
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| 1. Consumer surplus | c. ∫₀^q₀ D(q) dq − p₀q₀ |
| 2. Producer surplus | d. p₀q₀ − ∫₀^q₀ S(q) dq |
| 3. Equilibrium point | a. D(q₀) = S(q₀) |
| 4. Area between curves | e. ∫ₐᵇ [f(x) − g(x)] dx |
| 5. Area under y = f(x) from a to b | b. ∫ₐᵇ f(x) dx |
Answer1→c, 2→d, 3→a, 4→e, 5→b
V. Answer Briefly
Q1. Find the equilibrium price and quantity if D(q) = 100 − 4q and S(q) = 20 + 2q.
Answer
Set D(q) = S(q): 100 − 4q = 20 + 2q → 80 = 6q → q₀ = 80/6 = 40/3 ≈ 13.33 p₀ = 100 − 4(40/3) = 100 − 160/3 = 300/3 − 160/3 = 140/3 ≈ 46.67 Equilibrium: q₀ = 40/3, p₀ = 140/3
Q2. Find the area enclosed by y = 4x − x² and the x-axis.
Answer
The curve meets the x-axis when 4x − x² = 0 → x(4 − x) = 0 → x = 0 or x = 4. Area = ∫₀⁴ (4x − x²) dx = [2x² − x³/3]₀⁴ = (32 − 64/3) − 0 = 96/3 − 64/3 = 32/3 sq. units
Q3. Find the area bounded by y = x + 1 and y = x² + 1.
Answer
Find intersection: x + 1 = x² + 1 → x = x² → x² − x = 0 → x(x − 1) = 0 → x = 0, x = 1. On [0,1]: x + 1 ≥ x² + 1 (since x ≥ x² on [0,1]). Area = ∫₀¹ [(x+1) − (x²+1)] dx = ∫₀¹ (x − x²) dx = [x²/2 − x³/3]₀¹ = 1/2 − 1/3 = 1/6 sq. units
Q4. Calculate the consumer surplus for D(q) = 50 − 2q at equilibrium quantity q₀ = 10.
Answer
p₀ = 50 − 2(10) = 30 CS = ∫₀¹⁰ (50 − 2q) dq − 30 × 10 = [50q − q²]₀¹⁰ − 300 = (500 − 100) − 0 − 300 = 400 − 300 = 100
Q5. Calculate the producer surplus for S(q) = 5 + 3q at equilibrium q₀ = 10.
Answer
p₀ = 5 + 3(10) = 35 PS = 35 × 10 − ∫₀¹⁰ (5 + 3q) dq = 350 − [5q + 3q²/2]₀¹⁰ = 350 − (50 + 150) = 350 − 200 = 150
Q6. Find the area between the parabola y² = 4x and the line x = 3.
Answer
y = ±2√x. The parabola is symmetric about the x-axis. Area = 2∫₀³ 2√x dx = 4 · [x^(3/2)/(3/2)]₀³ = 4 · (2/3)(3)^(3/2) = (8/3)(3√3) = 8√3 sq. units
Wait, let me recalculate: x goes from 0 to 3. y = 2√x at x = 3 gives y = 2√3 Area = 2∫₀^(2√3) (3 − y²/4) dy = 2[3y − y³/12]₀^(2√3) = 2[(6√3 − 24√3/12) − 0] = 2[6√3 − 2√3] = 2 · 4√3 = 8√3 sq. units
| Actually, the simpler approach: Area = 2 ∫₀³ 2√x dx = 4 · (2/3) x^(3/2) | ₀³ = (8/3)(3√3) = 8√3. Yes, 8√3 sq. units. |
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VI. Answer in Detail
Q1. Find the consumer surplus and producer surplus for the demand function p = 100 − 5q and supply function p = 20 + 3q.
Answer
Step 1: Find equilibrium point 100 − 5q = 20 + 3q → 80 = 8q → q₀ = 10 p₀ = 100 − 5(10) = 50 (verify: 20 + 3(10) = 50 ✓) Equilibrium: (q₀, p₀) = (10, 50) Step 2: Consumer Surplus CS = ∫₀¹⁰ (100 − 5q) dq − 50 × 10 = [100q − 5q²/2]₀¹⁰ − 500 = (1000 − 250) − 0 − 500 = 750 − 500 = 250 Step 3: Producer Surplus PS = 50 × 10 − ∫₀¹⁰ (20 + 3q) dq = 500 − [20q + 3q²/2]₀¹⁰ = 500 − (200 + 150) = 500 − 350 = 150 Total surplus = CS + PS = 250 + 150 = 400
Q2. Find the area bounded by the parabola y² = x and the line y = x − 2.
Answer
Step 1: Find intersection points From y² = x and y = x − 2: x = y + 2 y² = y + 2 → y² − y − 2 = 0 → (y−2)(y+1) = 0 → y = 2 or y = −1 At y = 2: x = 4; At y = −1: x = 1 Intersection points: (4, 2) and (1, −1) Step 2: Set up integral (integrating with respect to y) Right curve: x = y + 2 (line) Left curve: x = y² (parabola) Area = ∫₋₁² [(y+2) − y²] dy Step 3: Evaluate = [y²/2 + 2y − y³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = (6 − 8/3) − (1/2 − 2 + 1/3) = 10/3 − (1/2 − 2 + 1/3) = 10/3 − (3/6 − 12/6 + 2/6) = 10/3 − (−7/6) = 10/3 + 7/6 = 20/6 + 7/6 = 27/6 = 9/2 sq. units
Q3. Find the area of the region bounded by y = x² and y = 2 − x².
Answer
Step 1: Find intersection points x² = 2 − x² → 2x² = 2 → x² = 1 → x = ±1 Step 2: Set up integral The upper curve is y = 2 − x² and the lower curve is y = x². By symmetry, Area = 2 ∫₀¹ [(2 − x²) − x²] dx = 2 ∫₀¹ (2 − 2x²) dx Step 3: Evaluate = 2[2x − 2x³/3]₀¹ = 2[2 − 2/3] = 2 · 4/3 = 8/3 sq. units
Q4. The demand function for a product is p = 50 − 0.5q and the supply function is p = 10 + 0.5q. Find (a) the equilibrium point, (b) the consumer surplus, and (c) the producer surplus. Illustrate with a sketch description.
Answer
(a) Equilibrium point: 50 − 0.5q = 10 + 0.5q → 40 = q → q₀ = 40 p₀ = 50 − 0.5(40) = 30 (verify: 10 + 0.5(40) = 30 ✓) Equilibrium: (40, 30) (b) Consumer Surplus: CS = ∫₀⁴⁰ (50 − 0.5q) dq − 30 × 40 = [50q − 0.25q²]₀⁴⁰ − 1200 = (2000 − 400) − 0 − 1200 = 1600 − 1200 = 400 (c) Producer Surplus: PS = 30 × 40 − ∫₀⁴⁰ (10 + 0.5q) dq = 1200 − [10q + 0.25q²]₀⁴⁰ = 1200 − (400 + 400) = 1200 − 800 = 400 Illustration description: The demand curve starts at p = 50 (when q = 0) and slopes downward. The supply curve starts at p = 10 (when q = 0) and slopes upward. They intersect at (40, 30). The CS is the triangular area between the demand curve and the price line p = 30. The PS is the triangular area between the supply curve and p = 30.
Q5. Find the area enclosed between the curves y = sin x and y = cos x from x = 0 to x = π/2.
Answer
Step 1: Find intersection point in [0, π/2] sin x = cos x → tan x = 1 → x = π/4 Step 2: Determine which curve is on top On [0, π/4]: cos x ≥ sin x (since cos 0 = 1 > sin 0 = 0) On [π/4, π/2]: sin x ≥ cos x (since sin(π/2) = 1 > cos(π/2) = 0) Step 3: Compute area Area = ∫₀^(π/4) (cos x − sin x) dx + ∫_(π/4)^(π/2) (sin x − cos x) dx = [sin x + cos x]₀^(π/4) + [−cos x − sin x]_(π/4)^(π/2) = (sin(π/4) + cos(π/4)) − (0 + 1) + (−cos(π/2) − sin(π/2)) − (−cos(π/4) − sin(π/4)) = (√2/2 + √2/2) − 1 + (0 − 1) − (−√2/2 − √2/2) = √2 − 1 − 1 + √2 = 2√2 − 2 = 2(√2 − 1) sq. units
Q6. A company's marginal cost function is MC = 3q² − 10q + 30 and fixed cost is Rs. 500. Find (a) the total cost function, and (b) the total cost of producing 10 units.
Answer
(a) Total cost function: TC = ∫ MC dq + FC = ∫(3q² − 10q + 30) dq + 500 = q³ − 5q² + 30q + 500 Verify: d/dq(q³ − 5q² + 30q + 500) = 3q² − 10q + 30 = MC ✓ (b) Total cost of producing 10 units: TC(10) = 1000 − 500 + 300 + 500 = Rs. 1300
Common Mistakes to Avoid
1. Using wrong limits in surplus calculations: The limits of integration for CS and PS are always 0 to q₀ (equilibrium quantity), not from any other starting point.
2. Confusing which curve is on top: Always check which function is larger in the given interval. If f(x) > g(x), then Area = ∫[f(x) − g(x)] dx. Test a sample point if unsure.
3. Forgetting to find intersection points: When finding area between two curves, you must first find where they intersect to determine the limits of integration.
4. Wrong order in surplus formulas: CS = ∫D dq − p₀q₀ (not the other way around). PS = p₀q₀ − ∫S dq (not ∫S dq − p₀q₀). Mix these up and you get negative surplus.
5. Not integrating with respect to the correct variable: Sometimes it is easier to integrate with respect to y (when curves are given as x = f(y)). Choose the variable that makes the integration simpler.
| 6. Missing the absolute value for area: Area is always positive. If ∫ₐᵇ f(x) dx is negative, the geometric area is | ∫ₐᵇ f(x) dx | . When the curve crosses the x-axis, split the integral at the crossing points. |
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Exam Checklist
- [ ] I can find the equilibrium point by equating demand and supply functions
- [ ] I can calculate consumer surplus using CS = ∫₀^q₀ D(q) dq − p₀q₀
- [ ] I can calculate producer surplus using PS = p₀q₀ − ∫₀^q₀ S(q) dq
- [ ] I can find area under a single curve above the x-axis
- [ ] I can find area between two curves by determining which is on top
- [ ] I can find intersection points of curves to determine limits of integration
- [ ] I can find total cost from marginal cost by integration (TC = ∫MC dq + FC)
- [ ] I can choose the correct variable of integration (x vs y) based on the problem
