Application of Integrals

⭐ Important Questions

1. Find area bounded by y = x² and y = 6 - x.
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   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.
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   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