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JEE Main20207 Jan 2020Evening ShiftMathematicsArea Under CurvesActual

The area (in sq. units) of the region x , y ∈ R 2 4 x 2 ≤ y ≤ 8 x + 12 is

Options

  1. A125 3
  2. B128 3
  3. C124 3
  4. D127 3

Correct answer

B. 128 3

Step-by-step solution

4 x 2 = y y = 8 x + 12 4 x 2 = 8 x + 12 x 2 - 2 x - 3 = 0 x = - 1 , 3 A = ∫ - 1 3 8 x + 12 - 4 x 2 d x A = 8 x 2 2 + 12 x - 4 x 3 3 3 - 1 = 4 9 + 36 - 36 - 4 - 12 + 4 3 = 36 + 8 - 4 3 = 44 - 4 3 = 132 - 4 3 = 128 3

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