Monomath – CBSE | GSEB | IB | JEE Maths by Ajay Yadav (Math King of Katargam) – 15+ Years Experience. Study Notes, Video Lessons, Formula Sheets & PYQs.
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Application of Integrals

Previous Year Questions (2022-2024)

📅 Year 2022

  1. Find area bounded by y = x2 and y = x.
    Show Solution
    Intersection: x2 = x => x(x-1) = 0 => x=0,1\nArea = ∫_01 (x - x2)dx = [x2/2 - x3/3]_01\n= 1/2 - 1/3 = 1/6 sq units
  2. Find area bounded by y2 = 4x and y = 2x.
    Show Solution
    Intersection: 4x = 4x2 => 4x(1-x)=0 => x=0,1\nPoints: (0,0), (1,2)\nArea = ∫_01 (2√x - 2x)dx = [4x^{3/2}/3 - x2]_01\n= 4/3 - 1 = 1/3 sq units

📅 Year 2023

  1. Find area bounded by y = x2 and y = x.
    Show Solution
    Intersection: x2 = x => x(x-1) = 0 => x=0,1\nArea = ∫_01 (x - x2)dx = [x2/2 - x3/3]_01\n= 1/2 - 1/3 = 1/6 sq units
  2. Find area bounded by y2 = 4x and y = 2x.
    Show Solution
    Intersection: 4x = 4x2 => 4x(1-x)=0 => x=0,1\nPoints: (0,0), (1,2)\nArea = ∫_01 (2√x - 2x)dx = [4x^{3/2}/3 - x2]_01\n= 4/3 - 1 = 1/3 sq units

📅 Year 2024

  1. Find area bounded by y = x2 and y = x.
    Show Solution
    Intersection: x2 = x => x(x-1) = 0 => x=0,1\nArea = ∫_01 (x - x2)dx = [x2/2 - x3/3]_01\n= 1/2 - 1/3 = 1/6 sq units
  2. Find area bounded by y2 = 4x and y = 2x.
    Show Solution
    Intersection: 4x = 4x2 => 4x(1-x)=0 => x=0,1\nPoints: (0,0), (1,2)\nArea = ∫_01 (2√x - 2x)dx = [4x^{3/2}/3 - x2]_01\n= 4/3 - 1 = 1/3 sq units