Given the equation of the curves are:

$y=x^2+2$

$y=x$

$x=0\;and\;x=3.$

Here clearly $y=x^2+2$ represents a parabola with vertex at (0,2) and axis along positive direction of y-axis.The parabola is open upwards.

y=x is a line passing through the origin .

x=0 is y-axis and x=3 is a line parallel to y-axis at a distance of 3units from it.

Area of the required region is bounded by these curves and it is the shaded region as shown in the fig.

Required area is $A=\int_0^3(y_2-y_1)dx$

$A=\int_0^3[(x^2+2)-x]dx$

on integrating we get,

$A=\begin{bmatrix}\frac{x^3}{3}+2x-\frac{x^2}{2}\end{bmatrix}_0^3$

on applying the limits we get,

$A=\begin{bmatrix}9+6-\frac{9}{2}\end{bmatrix}=\frac{21}{2}$sq.units.

Hence the required area is $\frac{21}{2}$ sq.units.