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JEE Main20266 April 2026Morning ShiftMathematicsArea Under CurvesActual

The area of the region (x, y) : 0 y 6 - x, y^2 4x - 3, x 0 is:

Options

  1. A8
  2. B9
  3. C12
  4. D15

Correct answer

B. 9

Step-by-step solution

The given region is defined by the inequalities: x 0 0 y 6 - x y 0 and x 6 - y y^2 4x - 3 x y^2 + 3 4 From these inequalities, for a given y 0 , the value of x ranges from 0 to (6 - y, y^2 + 3 4 ) . To find the point where the two bounding curves intersect, we equate them: 6 - y = y^2 + 3 4 24 - 4y = y^2 + 3 y^2 + 4y - 21 = 0 (y + 7)(y - 3) = 0 Since y 0 , the intersection occurs at y = 3 . For 0 y 3 , the right boundary is the parabola x = y^2 + 3 4 . For 3 y 6 , the right boundary is the line x = 6 - y . The tota

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