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Application of Integrals – Class 12 Mathematics notes, important questions and MCQs
NCERT Class 12 Mathematics, chapter 8. Notes, key terms, important questions with answers and MCQs.
Chapter overview
Definite integrals find areas under curves and between curves.
Area under a curve
- Area between y = f(x), the x-axis, x = a and x = b: ∫ₐᵇ |f(x)| dx.
- Area with the y-axis: ∫ x dy between y-limits.
- If the curve goes below the x-axis, take the absolute value of that part.
Standard results
- Area of the circle x² + y² = a² is πa² (4 × quadrant area).
- Area of the ellipse x²/a² + y²/b² = 1 is πab.
- Area between two curves = ∫(upper − lower) dx between intersection points.
Key terms
- Area under a curve
- region bounded by the curve, the axis and two ordinates
- Area between two curves
- ∫ (upper curve − lower curve) dx between their points of intersection
- Area along the y-axis
- ∫ x dy between y-limits, used when the region is easier to describe horizontally
Formulas
- Area = ∫ₐᵇ (f(x) − g(x)) dx, f ≥ g
- Ellipse area = πab
Important questions with answers
Find the area under y = x³ from x = 0 to x = 2.2 marks
∫₀² x³ dx = [x⁴/4]₀² = 4 square units.
Find the area of the region bounded by y² = 4x and the line x = 3.3 marks
The region is symmetric about the x-axis. Area = 2∫₀³ 2√x dx = 4 × [⅔x^(3/2)]₀³ = (8/3) × 3√3 = 8√3 square units.
Find the area of the region enclosed between the parabola y = x² and the line y = x + 2.5 marks
Intersection: x² = x + 2, x² − x − 2 = 0, x = −1 or 2. On (−1, 2), the line is above the parabola. Area = ∫₋₁² (x + 2 − x²) dx = [x²/2 + 2x − x³/3]₋₁² = (2 + 4 − 8/3) − (1/2 − 2 + 1/3) = 10/3 − (−7/6) = 27/6 = 9/2 square units.
Quick revision
- Use symmetry to reduce work.
- Area is always positive.
- Circle area πa², ellipse πab.
Exam tips
- Sketch the region first to find limits and which curve is on top.
- Find points of intersection first; if a curve crosses the x-axis, split the integral and add absolute values.
Practice MCQs
Area of the ellipse x²/9 + y²/4 = 1 is
- 13π
- 6π
- 36π
- 12π
Show answer
Answer: (b) 6π. π × 3 × 2.
Area under y = x from x = 0 to x = 2 is
- 1
- 2
- 4
- 8
Show answer
Answer: (b) 2. ½ × 2 × 2.
Area bounded by y = x² , the x-axis and x = 3 is
- 27
- 9
- 3
- 6
Show answer
Answer: (b) 9. [x³/3]₀³.
Area of a quadrant of x² + y² = 16 is
- 8π
- 16π
- 4π
- 2π
Show answer
Answer: (c) 4π. ¼ × 16π.
Area bounded by y = 2x, the x-axis and x = 3 is
- 6
- 9
- 18
- 3
Show answer
Answer: (b) 9. ½ × 3 × 6.
Area between y = x² and y = x is
- 1/6
- 1/3
- 1/2
- 1/12
Show answer
Answer: (a) 1/6. ∫₀¹ (x − x²) dx.
Area under y = sin x from 0 to π is
- 2
- 1
- 0
- π
Show answer
Answer: (a) 2. [−cos x]₀^π = 2.
Area enclosed by the circle x² + y² = 4 is
- 2π
- 4π
- 8π
- 16π
Show answer
Answer: (b) 4π. πr² with r = 2.
All chapters of Class 12 Mathematics
Written by Kundali IQ from standard, well-established facts to help with revision. This is not an official NCERT publication; always read your textbook too.