Question

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:

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

Answer 1

Answer: (D)

$9$

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

from above we have,

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

$2\sqrt{x}=x-3$
$4x=x^2-6x+9$
$x^2-10x+9=0$
$x=9,x=1$

now,

${\int }_0^3\left[\right(2y+3)-y^2]dy$
$={[y^2+3y-\frac{y^3}{3}]}_0^3=9+9-9=9$