Answer :

Given the boundaries of the area O befound are,

• Curve is y^{2} = 2x +1

• Line x-y = 1

Consider the curve

y^{2} = 2x +1

This clearly shows, the curve is a parabola with vertex

Consider the curve, y^{2} = 2x +1 and substitute the line x = y +1 in the curve

y^{2} = 2(y+1) +1

y^{2} = 2y +2 +1

y^{2} = 2y +3

y^{2} -2y -3 = 0

y = 3 (or) y = -1

substituting y in x-y = 1

x = 4 (or) x = 0

So , the parabola meets the line x-y =1 at 2 points, B (4,3) and C (0,-1)

As per the given boundaries,

• The parabola y^{2} = 2x +1, with vertex at A(-0.5,0) and symmetric about the x-axis as y has even powers.

• Line x-y = 1

The boundaries of the region to be found are,

•Point A, where the curve y^{2} = 2x +1 has the extreme end the vertex i.e. A (-0.5,0)

•Point B, where the curve y^{2} = 2x +1 and the line x-y = 1 meet i.e. B (4,3)

•Point C, where the curve y^{2} = 2x +1 and the line x-y = 1 meet i.e. B (0,-1) on the negative y

•Point D, where the line x-y = 1 meets the x-axis i.e. D(1,0)

Consider the curve,

y^{2} = 2x +1

2x = y^{2} – 1

Consider the line x – y = 1

x = y +1

Area of the required region = Area of ABDC

Area of ABDC = Area above CDB – Area above CAB

[Using the formula ]

The Area of the required region

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