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
- A125 3
- B128 3
- C124 3
- 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