The area (in square units) bounded by the curves $y=\sqrt{x}$, $2y-x+3=0$, $x$ -axis, and lying in the first quadrant is:

1. Intersection: $\sqrt{x}=\frac{x-3}{2}\Rightarrow 2\sqrt{x}=x-3\Rightarrow 4x=x^2-6x+9\Rightarrow x^2-10x+9=0$.$x=9$ or $x=1$. Since $x\ge 3$ for $y\ge 0$ in the line, $x=9$.2. The area is bounded by $y=\sqrt{x}$ from $0$ to $9$ minus the area under the line from $3$ to $9$.
$$\text{Area}={\int }_0^9\sqrt{x}dx-{\int }_3^9\frac{x-3}{2}dx$$
$$\text{Area}=[\frac{2}{3}{x}^{3/2}{]}_0^9-\frac{1}{2}[\frac{x^2}{2}-3x{]}_3^9$$
$$\text{Area}=\frac{2}{3}(27)-\frac{1}{2}[(\frac{81}{2}-27)-(\frac{9}{2}-9)]$$
$$\text{Area}=18-\frac{1}{2}[\frac{27}{2}-(-\frac{9}{2})]=18-\frac{1}{2}(18)=18-9=9$$
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