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# Find the area of the region enclosed by the parabola $x^2 = y$, the line $y = x + 2$ and the $x$ - axis.

Suppose we are given two curves represented by y=f(x) and y=g(x), where $f(x)\geq g(x)$ in [a,b], the point of intersection of there curves are given by x=a and x=b taking the common values of y from the given equation of the two values.
Hence the required area is given by $A=\int_a^b[f(x)-g(x)]dx$
The area of the region enclosed by the parabola $x^2=y$ and the line y=x+2 is represented by the shaded region in the fig
The point of intersection of the parabola $x^2=y$ and the line y=x+2 can be obtained by equally both the equations.
$x^2=x+2$.
$\Rightarrow\;x^2-x-2=0.$
On factorising we get (x-2)(x+1)=0
$\Rightarrow\;x=2,-1.$ Hence y=4,1.
The point of intersection are (2,4) and (-1,1) respectively.
But the area of the region which is required is between the points (-2,0) and (-1,1).
Hence the limits are -2 to 0.
The required area can be integrated as.
$A=\int _{-2}^{-1} (x+2) dx+\int_{-1}^0 x^2dx.$
On integrating we get
$A=\begin{bmatrix}\frac{x^2}{2}+2x\end{bmatrix}_{-2}^{-1}+\begin{bmatrix}\frac{x^3}{3}\end{bmatrix}_{-1}^{0}$
on applying limits we get,
$A=\begin{bmatrix}\frac{(-1)^2}{2}+2(-1)-\frac{(-2)^2}{2}-2(-2)\end{bmatrix}+\begin{bmatrix}\frac{(-1)^3}{3}\end{bmatrix}$

$\;\;\;=\begin{bmatrix}\frac{1}{2}-2-2+4+\frac{1}{3}\end{bmatrix}$

$\;\;\;=\frac{5}{6}sq.units.$

Hence the required area is $\frac{5}{6}sq. units.$