Question

Area of the region $(x,y)ϵR^2:y\ge R^2:y\ge \sqrt{|x+3|},5y\le (x+9)\le 15$ is equal to

• A. $\frac{1}{6}$
  
• B. $\frac{4}{3}$
  
• C. $\frac{3}{2}$
  
• D. $\frac{5}{3}$

Answer 1

The correct option is C $\frac{3}{2}$


Here, $(x,y)ϵR^2:y\ge R^2:y\ge \sqrt{|x+3|},5y\le (x+9)\le 15$
$\therefore y\ge \sqrt{|x+3|}\Rightarrow y\ge \{\begin{array}{c}\sqrt{x+3},whenx\ge -3\\ \sqrt{-x-3},whenx\le -3\end{array}or{y}^2\ge \{\begin{array}{cc}x+3,whenx\ge -3& \\ -3-x,whenx\le -3& \end{array}$
$Shownas$


Also, $5y\le (x+9)\le 15\Rightarrow (x+9)\ge 5yandx\le 6Shownas$

$\therefore (x,y)ϵR^2:y\ge \sqrt{|x+3|},5y\le (x+9)\le 15$

$\therefore$ Required area =Area of trapezium ABCD - Area ABE under the parabola -Area of CDE under the parabola
$=\frac{1}{2}(1+2)\left(5\right)-{\int }_{-4}^{-3}\sqrt{-(x+3)}dx-{\int }_{-a}^1\sqrt{(x+3)dx}=\frac{15}{2}-{\left[\frac{(-3-x{)}^{\frac{3}{2}}}{-\frac{3}{2}}\right]}_{-4}^{-3}-{\left[\frac{(x+3{)}^{\frac{3}{2}}}{\frac{3}{2}}\right]}_{-3}^1=\frac{15}{2}+\frac{2}{3}[0-1]-\frac{2}{3}[8-0]=\frac{15}{2}-\frac{2}{3}-\frac{16}{3}=\frac{15}{2}-\frac{18}{3}=\frac{3}{2}$