Chapter 8 – Application of Integrals is a direct scoring chapter for AAI ATC. It applies integral calculus to calculate areas bounded by curves, lines, circles, and ellipses. Typically 1–2 questions appear in the AAI ATC paper from this chapter, often as direct formula applications.
✦ Area under a curve
✦ Area w.r.t. y-axis
✦ Area of circle
✦ Area of ellipse
✦ Area above & below x-axis
✦ Area bounded by line
✦ Area of y = cos x, sin x
✦ Parabola regions
8.1
Introduction – Why Do We Need Integrals for Area?
Elementary geometry vs. Integration · Area as a definite integral
🔍 The Problem with Geometry Alone
Simple geometry formulae (triangles, circles) cannot handle areas enclosed by arbitrary curves. Integral calculus bridges this gap by treating area as the sum of infinitely thin strips.
Core Idea: Think of area under a curve y = f(x) between x = a and x = b as the sum of very thin vertical rectangles of height y and width dx. Each such strip has area dA = y dx. The total area = ∫ₐᵇ y dx.
Area with x-axis: A = ∫ₐᵇ f(x) dx (vertical strips, width dx)
Area with y-axis: A = ∫ᶜᵈ g(y) dy (horizontal strips, height dy)
If f(x) < 0, the integral is negative; we take the absolute value for area
If the curve crosses the x-axis, split the integral and add absolute values: A = |A₁| + A₂
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Insert Image: Fig 8.1 – Vertical strip dA = y dx between x = a and x = b (Add your diagram here)
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Insert Image: Fig 8.2 – Horizontal strip dA = x dy between y = c and y = d (Add your diagram here)
Two Fundamental Area Formulae
A = ∫ₐᵇ f(x) dx [curve y = f(x), bounded by x-axis, x=a, x=b]
A = ∫ᶜᵈ g(y) dy [curve x = g(y), bounded by y-axis, y=c, y=d]
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Insert Image: Fig 8.3 & Fig 8.4 – Curve below x-axis (negative area) and mixed case (A₁ + A₂) (Add your diagram showing both above and below x-axis regions)
📝 Practice MCQs – Section 8.1
Q1. The area of the region bounded by the curve y = f(x), x-axis and ordinates x = a, x = b is given by:
A) ∫ₐᵇ x dy
B) ∫ₐᵇ y dx
C) ∫ₐᵇ xy dx
D) ∫ₐᵇ f'(x) dx
✅ Answer: (B) | Each thin vertical strip has area dA = y dx. Total area A = ∫ₐᵇ y dx = ∫ₐᵇ f(x) dx
Q2. If the curve y = f(x) lies entirely below the x-axis between x = 1 and x = 3, then the area bounded is:
A) ∫₁³ f(x) dx
B) |∫₁³ f(x) dx|
C) –∫₁³ f(x) dx only if f(x) > 0
D) ∫₁³ –f(x) dx only if f(x) > 0
✅ Answer: (B) | When the curve is below x-axis, integral gives a negative value. We take absolute value since area is always positive.
Q3. Area of region bounded by x = g(y), y-axis, y = 2 and y = 5 is:
A) ∫₂⁵ y dy
B) ∫₂⁵ g(y) dy
C) ∫₂⁵ g(x) dx
D) ∫₂⁵ x dy
✅ Answer: (B) | For a curve x = g(y), we use horizontal strips. Area = ∫ᶜᵈ x dy = ∫₂⁵ g(y) dy
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8.2
Area of a Circle
x² + y² = a² · Using symmetry · Vertical and horizontal strips
⭕ Standard Result: Area of Circle x² + y² = a²
By symmetry (circle is symmetric about both axes), we compute the area in the first quadrant and multiply by 4.
Derivation Key Steps
Area = 4 × ∫₀ᵃ y dx = 4 × ∫₀ᵃ √(a²–x²) dx
Using formula: ∫√(a²–x²) dx = (x/2)√(a²–x²) + (a²/2)sin⁻¹(x/a) + C
AAI ATC Tip: The result Area of circle = πa² where a is the radius is a direct formula you can quote in the exam. The derivation using symmetry (×4) and the integral ∫√(a²–x²) dx is asked frequently as a worked example type question.
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Insert Image: Fig 8.5 – Circle x²+y²=a² with first quadrant region AOBA shaded, vertical strip shown (Add your diagram here)
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Insert Image: Fig 8.6 – Circle x²+y²=a² with horizontal strips (alternative method) (Add your diagram here)
Circle
Area Formula
Method
x² + y² = a²
πa²
4 × ∫₀ᵃ √(a²–x²) dx
x² + y² = 4 (radius 2)
4π
a = 2 → π(2²) = 4π
x² + y² = 9 (radius 3)
9π
a = 3 → π(3²) = 9π
First quadrant only (x²+y²=a²)
πa²/4
∫₀ᵃ √(a²–x²) dx
📝 Practice MCQs – Area of Circle
Q1. The area enclosed by the circle x² + y² = 16 is:
A) 8π sq units
B) 16π sq units
C) 4π sq units
D) 32π sq units
✅ Answer: (B) | a² = 16, so a = 4. Area = πa² = π × 16 = 16π sq units
Q2. Area lying in the first quadrant bounded by circle x² + y² = 4 and lines x = 0, x = 2 is:
A) π sq units
B) π/2 sq units
C) 2π sq units
D) 4π sq units
✅ Answer: (A) | a=2, first quadrant area = πa²/4 = π(4)/4 = π sq units (NCERT Exercise 8.1 Q3)
Q3. To find area of circle x²+y²=a² in first quadrant, after symmetry we evaluate:
A) 4∫₀ᵃ x dx
B) 4∫₀ᵃ √(a²–x²) dx
C) 2∫₀ᵃ √(a²–x²) dx
D) ∫₋ₐᵃ √(a²–x²) dx
✅ Answer: (B) | y = √(a²–x²) in first quadrant. Full circle = 4 × first quadrant area = 4∫₀ᵃ √(a²–x²) dx
Special Case: When a = b, the ellipse becomes a circle. Area = πa² (consistent with circle formula). For AAI ATC, remember: ellipse area = πab where a, b are semi-major and semi-minor axes.
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Insert Image: Fig 8.7 – Ellipse x²/a² + y²/b² = 1 with first quadrant region AOBA and vertical strip dx (Add your diagram here)
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Insert Image: Fig 8.8 – Ellipse with horizontal strips (alternative method giving same result πab) (Add your diagram here)
Ellipse
a
b
Area = πab
x²/16 + y²/9 = 1
4
3
12π sq units
x²/4 + y²/9 = 1
2
3
6π sq units
x²/25 + y²/16 = 1
5
4
20π sq units
x² + y² = a² (circle)
a
a
πa² sq units
📝 Practice MCQs – Area of Ellipse
Q1. Area of the ellipse x²/16 + y²/9 = 1 is:
A) 9π sq units
B) 16π sq units
C) 12π sq units
D) 144π sq units
✅ Answer: (C) | a² = 16 → a = 4, b² = 9 → b = 3. Area = πab = π × 4 × 3 = 12π sq units (NCERT Ex 8.1 Q1)
Q2. Area of the ellipse x²/4 + y²/9 = 1 is:
A) 4π sq units
B) 9π sq units
C) 6π sq units
D) 18π sq units
✅ Answer: (C) | a = 2, b = 3. Area = πab = π × 2 × 3 = 6π sq units (NCERT Ex 8.1 Q2)
Q3. For the ellipse x²/a² + y²/b² = 1, the area in the first quadrant is:
A) πab
B) πab/4
C) πab/2
D) 2πab
✅ Answer: (B) | Total ellipse area = πab. By symmetry (4 quadrants equal), first quadrant area = πab/4
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8.4
Miscellaneous Area Problems
Area bounded by lines · Area of y = cos x · Parabola y² = 4x
📏 Area Bounded by a Line and x-axis
When the line crosses the x-axis within the given interval, split the integral at the crossing point and add absolute values.
Example 3 (NCERT): y = 3x + 2, x-axis, x = –1 to x = 1
Line crosses x-axis at x = –2/3 (set y=0: 3x+2=0 → x=–2/3)
Area = |∫₋₁^(–2/3) (3x+2) dx| + ∫_(–2/3)^1 (3x+2) dx
= 1/6 + 25/6 = 13/3 sq units
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Insert Image: Fig 8.9 – Line y = 3x+2 crossing x-axis at (–2/3, 0), showing regions ACBA and ADEA (Add your graph here)
📉 Area Bounded by y = cos x (0 to 2π)
cos x ≥ 0 on [0, π/2] → Area₁ = ∫₀^(π/2) cos x dx = [sin x]₀^(π/2) = 1
cos x ≤ 0 on [π/2, 3π/2] → Area₂ = |∫_(π/2)^(3π/2) cos x dx| = |–2| = 2
cos x ≥ 0 on [3π/2, 2π] → Area₃ = ∫_(3π/2)^(2π) cos x dx = 1
Total Area = 1 + 2 + 1 = 4 sq units
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Insert Image: Fig 8.10 – y = cos x from 0 to 2π showing three regions (above-below-above x-axis) (Add your graph here)
📐 Area of Parabola y² = 4x, y-axis, y = 3
NCERT Exercise 8.1 Q4
x = y²/4. Area = ∫₀³ x dy = ∫₀³ y²/4 dy = [y³/12]₀³ = 27/12 = 9/4 sq units
📝 Practice MCQs – Miscellaneous Area Problems
Q1. Area bounded by y = cos x from x = 0 to x = 2π is:
Exam-ready summary · Common results · Direct application
📋 Master Formula Table for Chapter 8
Curve / Region
Formula
Result
y = f(x), x-axis, x=a to x=b
∫ₐᵇ f(x) dx
Area under curve
x = g(y), y-axis, y=c to y=d
∫ᶜᵈ g(y) dy
Area (horizontal strips)
Circle: x²+y²=a²
4∫₀ᵃ √(a²–x²) dx
πa²
Ellipse: x²/a²+y²/b²=1
(4b/a)∫₀ᵃ √(a²–x²) dx
πab
y = sin x, 0 to 2π
Split at 0, π, 2π
4 sq units
y = cos x, 0 to 2π
Split at π/2, 3π/2
4 sq units
y² = 4x, y-axis, y=0 to y=k
∫₀ᵏ y²/4 dy
k³/12
y = x³, x-axis, x=–2 to x=1
|∫₋₂⁰ x³ dx| + ∫₀¹ x³ dx
17/4 sq units
🔑 Memory Trick for AAI ATC:
Circle area = πr² (r = radius = a) | Ellipse area = πab (a = semi-major, b = semi-minor)
For sin x / cos x area over full cycle [0, 2π] → always = 4 sq units
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Insert Image: Summary diagram showing all curve types — circle, ellipse, parabola, trig functions — with their areas labelled (Add your summary chart or mind map here)
📝 Final Practice MCQs – Mixed Chapter 8
Q1. Area bounded by y = x³, x-axis and ordinates x = –2, x = 1 is: