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JEE MainMathematicsArea Under Curves

The area of the region S = (x, y) : 0 y 2 - |x^2 - 2| is

Options

  1. A32 3
  2. B8 3 (2- 2 )
  3. C16 3 (2- 2 )
  4. D8 3 (3 2 -4 )

Correct answer

C. 16 3 (2- 2 )

Step-by-step solution

The given region is S = (x, y) : 0 y 2 - |x^2 - 2| . The area is bounded by the x-axis ( y=0 ) and the curve y = 2 - |x^2 - 2| . To find the intersection with the x-axis, set y = 0 : 2 - |x^2 - 2| = 0 |x^2 - 2| = 2 x^2 - 2 = 2 or x^2 - 2 = -2 x^2 = 4 or x^2 = 0 x = 2, 0 . The region is symmetric about the y-axis, so the total area A is: A = 2 ₀² (2 - |x^2 - 2|) dx The expression inside the modulus changes sign at x = 2 . We split the integral at this critical point: For x [0, 2 ] , x^2 - 2 0 |x^2 - 2| = 2 - x^2 . F

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