MHT CET202617 April 2026Morning ShiftMathematicsArea Under CurvesActual
Area enclosed by curve y = 2x^2 and lines x 1, y 4 is ..... sq. units
Options
- A8 2 -10 3
- B8( 2 -1) 3
- C4 2 -5 3
- D4( 2 -1) 3
Correct answer
A. 8 2 -10 3
Step-by-step solution
The region is bounded by the curve y = 2x^2 , the vertical line x = 1 , and the horizontal line y = 4 . The intersection of y = 2x^2 and y = 4 in the region x 1 is given by: 2x^2 = 4 x^2 = 2 x = 2 The required area A is the integral of the upper boundary minus the lower boundary from x = 1 to x = 2 : A = ₁^ 2 (4 - 2x^2) dx A = [ 4x - 2x^3 3 ]₁^ 2 A = ( 4 2 - 4 2 3 ) - ( 4 - 2 3 ) A = 8 2 3 - 10 3 = 8 2 - 10 3 Answer: 8 2 -10 3