Question

Find the area bounded by the curves $y=\sqrt{x},2y+3=x$ and x-axis.

Answer 1

We have $y=\sqrt{x}...\left(i\right),2y+3=x....\left(ii\right)$
Solving (i) and (ii), $2\sqrt{x}+3=x\Rightarrow (x-3{)}^2=4x$
$\Rightarrow x^2-10x+9=0\Rightarrow (x-9)(x-1)=0$
$\therefore x=1,9\Rightarrow y=-1,3$
But note that $y=\sqrt{x}$ so, y > 0
Therefore, the point of intersection is (9,3).
Now, required area $={\int }_0^3(2y+3)dy={\int }_0^3{y}^{2}dy$
$\Rightarrow ={[\frac{(2y+3{)}^2}{2\times 2}-\frac{y^3}{3}]}_0^3$
$\Rightarrow =[\frac{81}{2\times 2}-\frac{27}{3}]-[\frac{9}{2\times 2}-0]=\frac{72}{2\times 2}-\frac{27}{3}=18-9$
$\Rightarrow =9$ Sq. units.