Step 1:

Given $x^2=y$ and the line $y=x+2$

Consider the curve $x^2=y$

This is a parabola which is open upwards.

The focus of the parabola is $(4a=1\Rightarrow a=\large\frac{1}{4})$$\Rightarrow (0,\large\frac{1}{4}$)

$y=x+2$ is a straight line.

Step 2:

The points of intersection can be obtained by solving the two equations.

$x^2=x+2\Rightarrow x^2-x-2=0$

On factorising we get,

$(x-2)(x+1)=0$

Hence $x=2$ or $-1$

The points of intersection are $(2,4)(-1,1)$

Step 3:

The required area is the shaded portion shown in the fig.

Area of the shaded portion can be evaluated

$A=\int_{-1}^2(y_2-y_1)dx.$

Where $y_2=x+2$ and $y_1=x^2$

Hence $A=\int_{-1}^2[(x+2)-x^2]dx.$

Step 4:

On integrating we get,

$A=\begin{bmatrix}(\large\frac{x^2}{2}\normalsize+2x)-\large\frac{x^3}{3}\end{bmatrix}_{-1}^2$

On applying limits we get,

$A=[(2+4)-\large\frac{1}{2}$$-2)]-[\large\frac{8}{3}+\frac{1}{3}]$$=6+\large\frac{3}{2}$$-3$

$\;\;=3+\large\frac{3}{2}=\frac{9}{2}$sq.units.