Discussion Forum

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 

Answer:

Explanation:

Given curves are   $y=\sqrt{x}$   ........(i)

   and 2y-x+3=0      .......(ii)

 On solving Eqs(i) and (ii) , we get

   $2\sqrt{x}-(\sqrt{x})^{2}+3=0$

  $\Rightarrow$    $(\sqrt{x})^{2}-2\sqrt{x}-3=0$

$\Rightarrow$   $(\sqrt{x}-3)(\sqrt{x}+1)=0$

  $\Rightarrow$    $\sqrt{x}=3$

  ( $\therefore$   $\sqrt{x}=-1$   is not possible)

 $\therefore$  y=3

   $\therefore$  Required area =   $\int_{0}^{3} (x_{2}-x_{1})dy$

  $\int_{0}^{3} ((2y+3)-y^{2})dy$

      = $\left[y^{2}+3y-\frac{y^{3}}{3}\right]_{0}^{3}$

   =9+9-9=9