Mathematics - Area Under Curves Question with Solution | TestHub

If $x+3<0\Rightarrow y\ge \sqrt{-(x+3)}\&\mathrm{ y}\ge 0$
If $x+3\ge 0\Rightarrow y\ge \sqrt{(x+3)}\&\mathrm{ y}\ge 0$
Also
$5y\le x+9\le 15$
$\Rightarrow x-5y+9\ge 0\&x\le 6$
Intersection point of line & parabolas
$5y=x+9\&y^2=\left|x+3\right|(y\ge 0)$
${\left(\frac{x+9}{5}\right)}^2=\left|x+3\right|$
$x^2+18x+81=25x+75or-25x-75$
$x^2-7x+6=0or{x}^2+43x+156=0$
$x=1,6orx=-39,-4$
So Clearly,  $x=1,y=2$
and when  $x=-4,y=1$ are required points
Drawing all these, we get this graph.

Shaded portion is required area.
Clearly required area = area (trapezium ABCD) - $\left(A_1+A_2\right)$ .......(i)
area (trapezium ABCD) $=\frac{1}{2}\left(1+2\right)\left(5\right)=\frac{15}{2}$
$A_1= \underset{-4}{\overset{-3}{\int }}\sqrt{-\left(x+3\right)}dx$
$=\frac{2}{3}$
and $A_1= \underset{-3}{\overset{1}{\int }}{\left(x+3\right)}^{\frac{1}{2}}dx=\frac{16}{3}$
$\therefore$ From equation (i), we get required area $=\frac{15}{2}-\left(\frac{2}{3}+\frac{16}{3}\right)=\frac{3}{2}$