Find the area bounded by the curve $y = \sqrt{x}$, $x = 2y + 3$ in the first quadrant and X-axis.

$9$ square units

The correct answer is Option (2) → $9$ square units

Given equation of the curves are $y = \sqrt{x}$ and $x = 2y + 3$ in the first quadrant.

On solving both the equations, we get

$y = \sqrt{2y + 3}$

$\Rightarrow y^2 = 2y + 3$

$\Rightarrow y^2 - 2y - 3 = 0$

$\Rightarrow y^2 - 3y + y - 3 = 0$

$\Rightarrow y(y - 3) + 1(y - 3) = 0$

$\Rightarrow (y + 1)(y - 3) = 0$

$\Rightarrow y = -1, 3$

Let $A_1 = \text{Area above the straight line } x = 2y + 3$

$= \text{Area of OCADO}$

and $A_2 = \text{Area above the curve } y = \sqrt{x}$

$= \text{Area of OADO}$

$∴$ Required area of shaded region $= A_1 - A_2$

$A = \int\limits_{0}^{3} (2y + 3 - y^2) \, dy = \left[ \frac{2y^2}{2} + 3y - \frac{y^3}{3} \right]_{0}^{3}$

$= [9 + 9 - 9 - 0] = 9 \text{ sq. units}$