The shaded area OBAO represents the area bounded by the curve $x^2$ = 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 = $\int\limits_{0}^{2} \frac{x+2}{4}$ dx - $\int\limits_{0}^{2} \frac{x^2}{4}$ dx = $\frac{1}{4}\left[ \frac{x^2}{2} + 2x \right]_{0}^{2}$ - $\frac{1}{4}\left[ \frac{x^3}{3} \right]_{0}^{2}$ = $\frac{1}{2}$ (2 + 4) - $\frac{1}{4} \left[ \frac{8}{3} \right]$ = $\frac{3}{2} - \frac{2}{3}$ = $\frac{5}{6}$ Area OACO = $\int\limits_{-1}^{0} \frac{x+2}{4}$ dx - $\int\limits_{-1}^{0} \frac{x^2}{4}$ dx = $\frac{1}{4}\left| x^2 + 2 + 2x \right|_{-1}^{0}$ - $\frac{1}{4}\left| \frac{x^3}{3} \right|_{-1}^{0}$ = $\frac{1}{4}\left| \frac{(-1)^2}{2} - 2 - 1 \right|$ - $\frac{1}{4}\left| -\left( -\frac{1}{3} \right)^3 \right|$ = $\frac{1}{4}\left| -\frac{1}{2} + 2 \right|$ - $\frac{1}{4} \left( \frac{1}{3} \right)$ = $\frac{3}{8} - \frac{1}{12}$ = $\frac{7}{24}$ Therefore, required area = $\left( \frac{5}{6} + \frac{7}{24} \right)$ = $\frac{9}{8}$ sq. units OR Given equations of the circles are $x^2 + y^2$ = 4 ... (1) $(x-2)^2 + y^2$ = 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: $(x-2)^2 + y^2$ = $x^1 + y^2$ $x^2$ - 4x + 4 + $y^2$ = $x^2 + y^2$ 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 $\left[ \int\limits_{0}^{1} y\,dx + \int\limits_{1}^{2} y\,dx \right]$ = 2 [$\int\limits_{0}^{1} \sqrt{4-(x-2)^2}$ dx + $\int\limits_{1}^{2} \sqrt{4-x^2}$ dx] = 2 + $\left[ x \sqrt{4-x^2} + 4 \sin^{-1}\left(\frac{x}{2}\right) \right]_{1}^{2}$ = + $\left[ 4 \sin^{-1} 1 - \sqrt{3} - 4 \sin^{-1}\left(\frac{1}{2}\right) \right]$ = $\left[ \left( -\sqrt{3} - 4 \times \frac{\pi}{6} \right) + 4 \times \frac{\pi}{2} \right]$ + $\left[ 4 \times \frac{\pi}{2} - \sqrt{3} - 4 \times \frac{\pi}{6} \right]$ = $\frac{8\pi}{3}$ - 2 $\sqrt{3}$