Area of the region ≤ft(x, y) ∈ R^2 : y ≥ √(|x+3|), 5y ≤x+9≤15

We have, $y ≥\sqrt{|x+3|}$

$⇒y^2 ≥|x+3|$

$⇒y^2≥(x+3)$ for $x ≥-3$ and $y^2 ≥-(x+3)$ for $x <- 3$

Required area A is given by

$A=\int\limits_{-4}^{-3}\left(\frac{x+9}{5}-\sqrt{-x-3}\right)dx+\int\limits_{-3}^{1}\left(\frac{x+9}{5}-\sqrt{x+3}\right)dx$

$A=\frac{1}{5}\int\limits_{-4}^{-1}(x+9)dx-\int\limits_{-4}^{-3}\sqrt{-x-3}dx-\int\limits_{-3}^{1}\sqrt{x+3}dx$

$A=\frac{1}{5}\left[\frac{(x+9)^2}{2}\right]_{-4}^{-1}+\frac{2}{3}\left[(-x-3)^{3/2}\right]_{-4}^{-3}-\frac{2}{3}\left[(x+3)^{3/2}\right]_{-3}^{1}$

$A=\frac{1}{5}\left(50-\frac{25}{2}\right)+\frac{2}{3}(0-1)-\frac{2}{3}(8-0)=\frac{3}{2}$