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 = 6 - x.
    Show Solution
    Intersection: x2 = 6-x => x2+x-6=0 => (x+3)(x-2)=0 => x=-3,2\nArea = ∫_{-3}^2 (6-x-x2)dx\n= [6x - x2/2 - x3/3]_{-3}^2\n= (12-2-8/3) - (-18-9/2+9) = 125/6 sq units
  2. Find area between y2 = 4x and x2 = 4y.
    Show Solution
    Intersection: x = 0,4\nArea = ∫_04 (2√x - x2/4)dx\n= [4x^{3/2}/3 - x3/12]_04\n= 32/3 - 64/12 = 32/3 - 16/3 = 16/3 sq units

📅 Year 2023

  1. Find area bounded by y = x2 and y = 6 - x.
    Show Solution
    Intersection: x2 = 6-x => x2+x-6=0 => (x+3)(x-2)=0 => x=-3,2\nArea = ∫_{-3}^2 (6-x-x2)dx\n= [6x - x2/2 - x3/3]_{-3}^2\n= (12-2-8/3) - (-18-9/2+9) = 125/6 sq units
  2. Find area between y2 = 4x and x2 = 4y.
    Show Solution
    Intersection: x = 0,4\nArea = ∫_04 (2√x - x2/4)dx\n= [4x^{3/2}/3 - x3/12]_04\n= 32/3 - 64/12 = 32/3 - 16/3 = 16/3 sq units

📅 Year 2024

  1. Find area bounded by y = x2 and y = 6 - x.
    Show Solution
    Intersection: x2 = 6-x => x2+x-6=0 => (x+3)(x-2)=0 => x=-3,2\nArea = ∫_{-3}^2 (6-x-x2)dx\n= [6x - x2/2 - x3/3]_{-3}^2\n= (12-2-8/3) - (-18-9/2+9) = 125/6 sq units
  2. Find area between y2 = 4x and x2 = 4y.
    Show Solution
    Intersection: x = 0,4\nArea = ∫_04 (2√x - x2/4)dx\n= [4x^{3/2}/3 - x3/12]_04\n= 32/3 - 64/12 = 32/3 - 16/3 = 16/3 sq units