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The Vault
RowQ
The Vault
CBSE Class 12 Maths · 11 questions · 26 marks
Once the definite integral is understood as a signed accumulation, it becomes the natural tool for measuring the area enclosed by curves. This chapter sets up the integrals for the area between a curve and an axis, between two intersecting curves, and for standard shapes like circles and ellipses, always taking care with sign — a curve dipping below the x-axis contributes a negative integral that must be corrected to a positive area.
The area enclosed by the circle x² + y² = 16 is:
Answer
The area is 16π sq units. The circle x²+y²=16 has radius a=4, and the enclosed area is πa² = π(4)² = 16π.
The area bounded by the curve y = x², the x-axis, and the lines x=0 and x=2 is:
Answer
The area is 8/3 sq units. Area = ∫[0,2] x² dx = [x³/3] from 0 to 2 = 8/3 − 0 = 8/3.
The area of the region bounded by y = √x, the x-axis, x=0 and x=4 is:
Answer
The area is 16/3 sq units. Area = ∫[0,4] √x dx = [(2/3)x^(3/2)] from 0 to 4 = (2/3)(8) − 0 = 16/3.
Using integration, the area of the region bounded by the line y = x, the x-axis, and x=4 is:
Answer
The area is 8 sq units. Area = ∫[0,4] x dx = [x²/2] from 0 to 4 = 16/2 − 0 = 8.
Assertion (A): The area bounded by y = sin x and the x-axis from x=0 to x=2π is 4 square units. Reason (R): Since sin x is negative on (π,2π), the total area is found by adding the absolute values of the integrals over (0,π) and (π,2π) separately.
Answer
Both A and R are true and R is the correct explanation of A. ∫[0,π] sinx dx = [-cosx] from 0 to π = -(-1)-(-1) = 2, a positive contribution. ∫[π,2π] sinx dx = [-cosx] from π to 2π = -1-1 = -2, which is negative because sinx<0 there; its absolute value is 2. Adding the absolute values as R describes: 2 + 2 = 4, confirming Assertion A and matching the reasoning in R.
Find the area of the region bounded by the curve y = x², the x-axis, and the lines x=1 and x=3.
Answer
Area = ∫[1,3] x² dx = [x³/3] from 1 to 3. = 27/3 − 1/3 = 26/3. So the area is 26/3 square units.
Find the area of the region enclosed by the circle x² + y² = 4 that lies in the first quadrant.
Answer
The first-quadrant region is bounded by y = √(4-x²) from x=0 to x=2. Area = ∫[0,2] √(4-x²) dx = [(x/2)√(4-x²) + 2 sin⁻¹(x/2)] from 0 to 2. At x=2: (2/2)(0) + 2 sin⁻¹(1) = 0 + 2(π/2) = π. At x=0: 0. So the area is π square units, matching the expected quarter-circle area πa²/4 = π(4)/4 = π.
Find the area of the region in the first quadrant bounded by the parabola y² = 4x and the line x = 4.
Answer
In the first quadrant, y = 2√x (the upper branch of the parabola). Area = ∫[0,4] 2√x dx = 2×(2/3)x^(3/2) from 0 to 4 = (4/3)(4)^(3/2). 4^(3/2) = (√4)³ = 2³ = 8. Area = (4/3)(8) = 32/3 square units.
Find the area of the region bounded by the parabola y = x² and the line y = x + 2.
Answer
Find the intersection points by solving x² = x + 2, i.e. x² − x − 2 = 0, which factors as (x-2)(x+1) = 0, giving x = -1 and x = 2. Check which curve is on top: at x=0, the line gives y=2 and the parabola gives y=0, so the line lies above the parabola on [-1,2]. Area = ∫[-1,2] [(x+2) - x²] dx = [x²/2 + 2x - x³/3] from -1 to 2. At x=2: 4/2 + 4 - 8/3 = 2+4-8/3 = 6-8/3 = 10/3. At x=-1: 1/2 - 2 - (-1/3) = 1/2 - 2 + 1/3 = 3/6 - 12/6 + 2/6 = -7/6. Area = 10/3 - (-7/6) = 20/6 + 7/6 = 27/6 = 9/2. So the area is 9/2 square units.
Find the area of the region bounded by the ellipse x²/16 + y²/9 = 1 using integration.
Answer
Here a=4 and b=3. From the ellipse equation, y = (3/4)√(16-x²) for the upper half. By symmetry, the total area is 4 times the area in the first quadrant: Area = 4∫[0,4] (3/4)√(16-x²) dx = 3∫[0,4] √(16-x²) dx. Using the standard integral ∫√(a²-x²)dx = (x/2)√(a²-x²) + (a²/2)sin⁻¹(x/a) + C with a=4: ∫[0,4] √(16-x²) dx = [(x/2)√(16-x²) + 8 sin⁻¹(x/4)] from 0 to 4 = (0 + 8×(π/2)) − 0 = 4π. So the total area = 3 × 4π = 12π square units, which matches the standard formula πab = π(4)(3) = 12π.
Read the following and answer the questions that follow: The cross-section of a small ornamental pond, as seen from above along one axis, is modelled by the curve y = 4 - x² and the x-axis, for -2 ≤ x ≤ 2 (measurements in metres). (a) Find the x-intercepts of the curve y = 4 - x². (b) Set up the definite integral for the area of this cross-section. (c) Evaluate the integral to find the cross-sectional area. (d) The pond's full water-surface area is twice this cross-sectional area, by symmetry across a perpendicular axis. Find the total surface area.
Answer
(a) Setting 4 - x² = 0 gives x² = 4, so x = -2 and x = 2. (b) The area is A = ∫[-2,2] (4-x²) dx. (c) Since 4-x² is an even function, A = 2∫[0,2] (4-x²) dx = 2[4x - x³/3] from 0 to 2 = 2[8 - 8/3] = 2[24/3-8/3] = 2(16/3) = 32/3 square metres. (d) Total surface area = 2 × 32/3 = 64/3 square metres.
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