Answer :

Given the boundaries of the area O befound are,


• Curve is y2 = 2x +1


• Line x-y = 1


Consider the curve


y2 = 2x +1



This clearly shows, the curve is a parabola with vertex


Consider the curve, y2 = 2x +1 and substitute the line x = y +1 in the curve


y2 = 2(y+1) +1


y2 = 2y +2 +1


y2 = 2y +3


y2 -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 y2 = 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 y2 = 2x +1 has the extreme end the vertex i.e. A (-0.5,0)


Point B, where the curve y2 = 2x +1 and the line x-y = 1 meet i.e. B (4,3)


Point C, where the curve y2 = 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,


y2 = 2x +1


2x = y2 – 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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