Question

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

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

Answer 1

Answer: (A)

$9$

Given that

$y=\sqrt{x}$; $2y-x+3=0$

First we find the intersection point of these

$\Rightarrow 2y=x-3$

$\Rightarrow 2\sqrt{x}=x-3$

Squaring on both sides and rearranging, we get

$\Rightarrow (x-1)(x-9)=0$

$x=1;9\Rightarrow y=1,3$

But only $(9,3) lies on the curve

Hence area $A={\int }_0^9\sqrt{x}dx-{\int }_3^9\frac{x-3}{2}dx$

$\Rightarrow A=[\frac{x^{3/2}}{3/2}{]}_0^9-[\frac{\frac{x^2}{2}-3x}{2}{]}_3^9$

On simplifying we get,

$\Rightarrow A=9$ sq.units