The shaded area OBAO represents the area bounded by the curve x2 = 4y and line x = 4y – 2.

Let A and B be the points of intersection of the line and parabola.
Co-ordinates of point A are (−1,

1

4

). Co-ordinates of point B are (2, 1).
Area OBAO = Area OBCO + Area OACO … (1)
Area OBCO =

2

∫

0

x+2

4

dx -

2

∫

0

x2

4

dx
=

1

4

[

x2

2

+2x]02 -

1

4

[

x3

3

]02
=

1

2

(2 + 4) -

1

4

[

8

3

]
=

3

2

−

2

3

=

5

6

Area OACO =

0

∫

−1

x+2

4

dx -

0

∫

−1

x2

4

dx
=

1

4

|x2+2+2x|−10| -

1

4

|

x3

3

|−10
=

1

4

|

−12

2

−2−1| -

1

4

|−(−

1

3

)3|
=

1

4

|−

1

2

+2| -

1

4

(

1

3

)
=

3

8

−

1

12

=

7

24

Therefore, required area = (

5

6

+

7

24

) =

9

8

sq. units
OR
Given equations of the circles are
x2+y2 = 4 ... (1)
(x−2)2+y2 = 4 ... (2)
Equation (1) is a circle with centre O at the origin and radius 2. Equation (2) is a circle with centre C (2, 0) and radius 2.
Solving (1) and (2), we have:
(x−2)2+y2 = x1+y2
x2 - 4x + 4 + y2 = x2+y2
x = 1
This gives y = ±√3
Thus, the points of intersection of the given circles are A (1 , √3) and A' (1 , - √3) as shown in the figure.

Required area
= Area of the region OACA'O
= 2 [area of the region ODCAO]
= 2 [area of the region ODAO + area of the region DCAD]
= 2 [

1

∫

0

ydx+

2

∫

1

ydx]
= 2 [

1

∫

0

√4−(x−2)2 dx +

2

∫

1

√4−x2 dx]
= 2 + [x√4−x2+4sin−1

x

2

]12
= + [4sin−11−√3−4sin−1

1

2

]
= [(−√3−4×

π

6

)+4×

π

2

] + [4×

π

2

−√3−4×

π

6

]
=

8π

3

- 2 √3