Question

The area bounded by the curves $y=\sqrt{x},2y+3=x$ and $x$-axis in the $1^{st}$ quadrant is

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

Answer 1

The correct option is A $9sq.units$

Given answer $y=\sqrt{x}\Rightarrow y^2=x$ .....(i)
and $x=2y+3$ ......(ii)
$4y^2={(x-3)}^2\Rightarrow 4x=x^2-6x+9$
From Eqs (i) and (ii) we get
$\Rightarrow x^2-10x+9=0$
$\Rightarrow (x-1)(x-9)=0$
Therefore, required Area $=A+B$
$={\int }_0^3\sqrt{x}dx+{\int }_3^9[\sqrt{x}-\left(\frac{x-3}{2}\right)]dx=\frac{2}{3}{\left[x^{3/2}\right]}_0^3+\frac{2}{3}{\left[x^{3/2}\right]}_3^9-\frac{1}{2}{[\frac{x^2}{2}-3x]}_3^9$
$=\frac{2}{3}.3\sqrt{3}+\frac{2}{3}[9\times 3-3\sqrt{3}]-\frac{1}{2}[(\frac{18}{2}-27)-(\frac{9}{2}-9)]=9$ sq.units