The area (in square units) bounded by the curves y = x ,   2 y - x + 3 = 0 , X -axis and lying in the…

First draw the given curves and represent the required area.

Eliminate $\mathit{x}$ to find point of intersection of $y=\sqrt{x}$ and $\mathrm{2}y-x+\mathrm{3}=\mathrm{0}$

$\Rightarrow$$2y-y^2+3=0$

$\Rightarrow$$y^2-2y-3=0$

$\Rightarrow \left(y-3\right)\left(y+1\right)=0$

$\Rightarrow y=3,-1$
$\therefore$Required area $A=\underset{0}{\overset{9}{\int }}\sqrt{x}dx-\frac{1}{2}\times 6\times 3$

$A={\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}} \right]}_0^9-9$

$\therefore \mathrm{A}=-9$

Hence, area is $9 \mathrm{sq}. \mathrm{units}$