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
- A+4
- B+ 2 3
- C2 +2
- 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