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MHT CET202020 Oct 2020Morning ShiftMathematicsArea Under CurvesActual

The area of the region included between the parabola y²=x and the line x+y=2 in the first quardrant is

Options

  1. A1 6 sq. units
  2. B2 7 6 sq . units
  3. C1 2 sq . units
  4. D2 3 sq. units

Correct answer

B. 2 7 6 sq . units

Step-by-step solution

Point of intersection of y²=x and x+y=2 is (2-x)²=x x²-4 x-x+4=0 x²-5 x+4=0 (x-4)(x-1)=0 x=1,4 But since we want area in 1^ st quadrant only, we take x=1 y²=1 y= 1 y=1 in 1^ st quadrant. A (1,1) and P (1,0) Point of intersection of x+y=2 with X axis is B=(2,0) Hence area required is array l = ₀¹ x d x+ ₁²(2-x) d x = 2 3 [x x ]₀¹+2[x]₁²- 1 2 [x² ]₁² = 2 3 +2- ( 1 2 3 )= 2 3 +2- 3 2 = 4+12-9 6 = 7 6 array

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