y² = 2x is a parabola

In y² = 2x parabola it is not defined for negative values of x hence the parabola will be to the right of Y-axis passing through (0, 0)

x² + y² = 4x is equation of circle

The general equation of circle is given by x² + y² + 2gx + 2fy + c = 0

Centre of circle is (-g, -f) and radius is

In x² + y² – 4x = 0, 2g = -4 ⇒ g = -2 and f = c = 0

Hence center is (-(-2), 0) that is (2, 0) and radius is which is 2

Hence plot the circle and parabola roughly and to mark the intersection points solve the parabola equation and circle equation simultaneously

Put y² = 2x in x² + y² = 4x

⇒ x² + 2x = 4x

⇒ x² – 2x = 0

⇒ x(x – 2) = 0

⇒ x = 0 and x = 2

Put x = 2 in y² = 2x

⇒ y² = 2(2)

⇒ y = ±2

Hence the circle and parabola intersects at (0, 0), (2, 2) and (2, -2)

By integrating we will get the area only in the 1^st quadrant but given parabola and circle are symmetric about X-axis hence area above and below X-axis will be equal

Hence area of shaded region will be twice the area we will get by integration in 1^st quadrant …(a)

Observe that

⇒ area of shaded in 1^st quadrant = area under circle – area under parabola …(i)

Let us find area under circle

x² + y² = 4x

⇒ y² = 4x – x²

Integrate from 0 to 2

Using

Now let us find area under parabola

⇒ y² = 2x

⇒ y = √2√x

Integrate from 0 to 2

Using (i)

⇒ area of shaded in 1^st quadrant = unit²

Using (a)

The area required of shaded region = unit²