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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.

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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

  1. Area of the ellipse x²/9 + y²/4 = 1 is

    1. 13π
    2. 6π
    3. 36π
    4. 12π
    Show answer

    Answer: (b) 6π. π × 3 × 2.

  2. Area under y = x from x = 0 to x = 2 is

    1. 1
    2. 2
    3. 4
    4. 8
    Show answer

    Answer: (b) 2. ½ × 2 × 2.

  3. Area bounded by y = x² , the x-axis and x = 3 is

    1. 27
    2. 9
    3. 3
    4. 6
    Show answer

    Answer: (b) 9. [x³/3]₀³.

  4. Area of a quadrant of x² + y² = 16 is

    1. 8π
    2. 16π
    3. 4π
    4. 2π
    Show answer

    Answer: (c) 4π. ¼ × 16π.

  5. Area bounded by y = 2x, the x-axis and x = 3 is

    1. 6
    2. 9
    3. 18
    4. 3
    Show answer

    Answer: (b) 9. ½ × 3 × 6.

  6. Area between y = x² and y = x is

    1. 1/6
    2. 1/3
    3. 1/2
    4. 1/12
    Show answer

    Answer: (a) 1/6. ∫₀¹ (x − x²) dx.

  7. Area under y = sin x from 0 to π is

    1. 2
    2. 1
    3. 0
    4. π
    Show answer

    Answer: (a) 2. [−cos x]₀^π = 2.

  8. Area enclosed by the circle x² + y² = 4 is

    1. 2π
    2. 4π
    3. 8π
    4. 16π
    Show answer

    Answer: (b) 4π. πr² with r = 2.

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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.