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

We have, $y = \sqrt x ,$ $x = 2y + 3$

Solving we get
$y = \sqrt {2y + 3} ,y \ge 0$

$\Rightarrow$ ${y^2} = 2y + 3,y \ge 0$

$\Rightarrow$ ${y^2} - 2y - 3 = 0,y \ge 0$

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

$\Rightarrow$ $y = 3$

The graph of function $y = \sqrt x$ is part of parabola ${y^2} = x$ lying above x axis.

The graph is as shown in the following figure.

From the figure, area of shaded region,

$A = \int_0^3 {\left( {2y + 3 - {y^2}} \right)} dy$

$= \left[ {\frac{{2{y^2}}}{2} + 3y - \frac{{{y^3}}}{3}} \right]_0^3 = \left[ {\frac{{18}}{2} + 9 - 9 - 0} \right] = 9$ sq. units

Long Answer Questions (L.A.)