The shaded area OBAO represents the area bounded by the curve ${\mathit{x}}^2$ = 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 $\left(-1,\frac{1}{4}\right)$. Co-ordinates of point B are (2, 1).
Area OBAO = Area OBCO + Area OACO … (1)
Area OBCO = $\underset{0}{\overset{2}{\int }}\frac{\mathit{x}+2}{4}$ dx - $\underset{0}{\overset{2}{\int }}\frac{{\mathit{x}}^2}{4}$ dx
= $\frac{1}{4}{\left[\frac{{\mathit{x}}^2}{2}+2\mathit{x}\right]}_0^2$ - $\frac{1}{4}{\left[\frac{{\mathit{x}}^3}{3}\right]}_0^2$
= $\frac{1}{2}$ (2 + 4) - $\frac{1}{4}\left[\frac{8}{3}\right]$
= $\frac{3}{2}-\frac{2}{3}$ = $\frac{5}{6}$
Area OACO = $\underset{-1}{\overset{0}{\int }}\frac{\mathit{x}+2}{4}$ dx - $\underset{-1}{\overset{0}{\int }}\frac{{\mathit{x}}^2}{4}$ dx
= $\frac{1}{4}|{\mathit{x}}^2+2+2\mathit{x}{|}_{-1}^0|$ - $\frac{1}{4}|\frac{{\mathit{x}}^3}{3}{|}_{-1}^0$
= $\frac{1}{4}|\frac{{-1}^2}{2}-2-1|$ - $\frac{1}{4}|-{\left(-\frac{1}{3}\right)}^3|$
= $\frac{1}{4}|-\frac{1}{2}+2|$ - $\frac{1}{4}\left(\frac{1}{3}\right)$
= $\frac{3}{8}-\frac{1}{12}$ = $\frac{7}{24}$
Therefore, required area = $\left(\frac{5}{6}+\frac{7}{24}\right)$ = $\frac{9}{8}$ sq. units
OR
Given equations of the circles are
${\mathit{x}}^2+{\mathit{y}}^2$ = 4 ... (1)
${\left(\mathit{x}-2\right)}^2+{\mathit{y}}^2$ = 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:
${\left(\mathit{x}-2\right)}^2+{\mathit{y}}^2$ = ${\mathit{x}}^1+{\mathit{y}}^2$
${\mathit{x}}^2$ - 4x + 4 + ${\mathit{y}}^2$ = ${\mathit{x}}^2+{\mathit{y}}^2$
x = 1
This gives y = $\pm \sqrt{3}$
Thus, the points of intersection of the given circles are A (1 , $\sqrt{3})$ and A' (1 , - $\sqrt{3}$) 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 $\left[\underset{0}{\overset{1}{\int }}\mathit{y}\mathit{d}\mathit{x}+\underset{1}{\overset{2}{\int }}\mathit{y}\mathit{d}\mathit{x}\right]$
= 2 [$\underset{0}{\overset{1}{\int }}\sqrt{4-{\left(\mathit{x}-2\right)}^2}$ dx + $\underset{1}{\overset{2}{\int }}\sqrt{4-{\mathit{x}}^2}$ dx]
= 2 + ${\left[\mathit{x}\sqrt{4-{\mathit{x}}^2}+4\mathit{s}\mathit{i}{\mathit{n}}^{-1}\frac{\mathit{x}}{2}\right]}_1^2$
= + $\left[4\mathit{s}\mathit{i}{\mathit{n}}^{-1}1-\sqrt{3}-4\mathit{s}\mathit{i}{\mathit{n}}^{-1}\frac{1}{2}\right]$
= $\left[\left(-\sqrt{3}-4\times \frac{\mathit{\pi }}{6}\right)+4\times \frac{\mathit{\pi }}{2}\right]$ + $\left[4\times \frac{\mathit{\pi }}{2}-\sqrt{3}-4\times \frac{\mathit{\pi }}{6}\right]$
= $\frac{8\mathit{\pi }}{3}$ - 2 $\sqrt{3}$