Question

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

• A. 9
• B. 6
• C. 18
• D. $\frac{27}{4}$

Answer 1

Answer: (A)

Given curves are $y=\sqrt{x}........(i)$
and $2y-x+3=0........(ii)$
On solving Eqs. (i) and (ii), we get
$2\sqrt{x}-(\sqrt{x}{)}^2+3=0$
$\Rightarrow (\sqrt{x}{)}^2-2\sqrt{x}-3=0$
$\Rightarrow (\sqrt{x}-3)(\sqrt{x}+1)=0$
$\Rightarrow \sqrt{x}=3$
$\Rightarrow [since,\sqrt{x}=-1isnotpossible]$
$\therefore y=3$
Hence, required area
$=\underset{0}{\overset{3}{\int }}(x_2-x_1)dy=\underset{0}{\overset{3}{\int }}(2y+3)-y^{2}dy$
$=[y^2+3y-\frac{y^3}{3}{]}_0^3=9+9-9=9squnits$