MHT CET202122 Sep 2021Evening ShiftMathematicsArea Under CurvesActual
The area bounded by the parabola y^2=x and the line x+y=2 in the first quadrant is
Options
- A7 6 aq. units
- B1 6 sq. units
- C2 3 sq. units
- D6 7 aq. units
Correct answer
A. 7 6 aq. units
Step-by-step solution
The point of intersection of y^2=x and x+y=2 is, (2-x)^2=x x^2-5 x+4=0 (x-4)(x-1)=0 Let A =(1,1) in first quadrant and B =(4,-2) in fourth quadrant The line x + y =2 cuts X axis at P (2,0) Refer figure Required area is shaded aligned & A = ₀^1 x dx + ₁^2(2- x ) dx & = [ x ^ 2 3 ( 3 2 ) ]₀^1+[2 x ]₁^2- [ x ^2 2 ]₁^2 & = [ ( 2 3 )(1) ]+[2(2-1)]- [ ( 4-1 2 ) ]= 2 3 +2- 3 2 & = 7 6 sq. units aligned