Question

The area bounded by the curve $y=\sqrt{x}$, the line $2y+3=x$ and the $x$-axis in the first quadrant is

• A. $9$
• B. $\frac{27}{4}$
• C. $36$
• D. $18$

Answer 1

Answer: (A)

$9$

Given curves are $y=\sqrt{x}$ ...(1)
and $2y-x+3=0$ ...(2)
Solving (1) and (2), we get
$2\sqrt{x}-{\left(\sqrt{x}\right)}^2+3=0\Rightarrow {\left(\sqrt{x}\right)}^2-2\sqrt{x}-3=0\Rightarrow (\sqrt{x}-3)(\sqrt{x}+1)=0\Rightarrow \sqrt{x}=3$
$\because \sqrt{x}=-1$ is not possible
$\therefore y=3$
Hence required area
$={\int }_0^3(x_2-x_1)dy={\int }_0^3((2y+3)-y^2)dy$
$={[y^2+3y-\frac{y^3}{3}]}_0^3=9+9-9=9$