Area of the region $\{(x,y)\in\mathbb{R}^{2}: y\geq\sqrt{|x+3|},\,5y\leq x+9\leq 15\}$ is equal to (1) $\frac{1}{6}$ (2) $\frac{4}{3}$ (3) $\frac{3}{2}$ (4) $\frac{5}{3}$

Area of the region {(x,y)∈R²: y≥√|x+3|, 5y≤ x+9≤ 15} is equal to

$y \ge \sqrt{|x+3|}$

$5y \le x+9 \le 15$

$y \le \dfrac{x+9}{5},\qquad x \le 6$

$\text{area between } y=\sqrt{|x+3|} \text{ and } y=\dfrac{x+9}{5}$

$1.\ \text{intersections of } \sqrt{|x+3|} \text{ and } \dfrac{x+9}{5}$

$2.\ \text{the interval where the line is above}$

$3.\ \text{integrate and evaluate}$

$\text{lower } y=\sqrt{|x+3|},\quad \text{upper } y=\dfrac{x+9}{5}$

$\text{region needs } \sqrt{|x+3|}\le\dfrac{x+9}{5}$

$\sqrt{|x+3|} = \dfrac{x+9}{5},\quad u=x+3$

$25|u| = (u+6)^2$

$u\ge0:\ u^2-13u+36=0 \Rightarrow u=4,\ 9$

$u<0:\ u^2+37u+36=0 \Rightarrow u=-1\ (u>-6)$

$\Rightarrow x=-4,\ 1,\ 6$

$\text{region only where } \sqrt{|u|}\le\dfrac{u+6}{5}$

$u\in[-1,4] \Rightarrow x\in[-4,1]$

$A = \int_{-4}^{1}\left[\dfrac{x+9}{5}-\sqrt{|x+3|}\right]dx$

$= \int_{-1}^{4}\left[\dfrac{u+6}{5}-\sqrt{|u|}\right]du$

$= \int_{-1}^{0}\left[\dfrac{u+6}{5}-\sqrt{-u}\right]du + \int_{0}^{4}\left[\dfrac{u+6}{5}-\sqrt{u}\right]du$

$\int_{-1}^{4}\dfrac{u+6}{5}\,du = \dfrac{15}{2},\qquad \int_{-1}^{4}\sqrt{|u|}\,du = 6$

$A = \dfrac{15}{2}-6 = \dfrac{3}{2}$

$\text{Area} = \dfrac{3}{2}$

$\boxed{\dfrac{3}{2}}\quad \text{option (3)}$