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JEE Main202622 January 2026Evening ShiftMathematicsArea Under CurvesActual

The area of the region A = (x, y): 4 x²+y² 8 . and .y² 4 x is:

Options

  1. A+4
  2. B+ 2 3
  3. C2 +2
  4. D2 + 1 3

Correct answer

B. + 2 3

Step-by-step solution

A = ₀² 2 x , dx + 2 ₁^ 2 8 - 4x^2 , dx = 8 3 ( x^ 3 2 ) |₀¹ + 4 ₁^ 2 2 - x^2 , dx = 8 3 + 4 1 2 [ x 2 - x^2 + 2 ⁻¹ ( x 2 ) ] |₁^ 2 = 8 3 + 2 [ 2 2 - 1 - 2 4 ] = 8 3 + 2 - 2 - = + 2 3 sq. units

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