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

If x + 3 < 0 ⇒ y \ge \sqrt{- ( x + 3 )} & y \ge 0
If x + 3 \ge 0 ⇒ y \ge \sqrt{( x + 3 )} & y \ge 0
Also
$5y \le x + 9 \le 15$
⇒ x - 5y + 9 \ge 0 & x \le 6
Intersection point of line & parabolas
5y = x + 9 & {y}^{2} = | x + 3 | ( y \ge 0 )
${( \dfrac{x + 9}{5} )}^{2} = | x + 3 |$
${x}^{2} + 18x + 81 = 25x + 75or - 25x - 75$
${x}^{2} - 7x + 6 = 0or{x}^{2} + 43x + 156 = 0$
$x = 1, 6orx = - 39, - 4$
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) $= \dfrac{1}{2}\left(1 + 2\right)\left(5\right) = \dfrac{15}{2}$
${A}_{1} = \mathop{∫}_{- 4}^{- 3}\sqrt{- \left(x + 3\right)}dx$
$= \dfrac{2}{3}$
and ${A}_{1} = \mathop{∫}_{- 3}^{1}{\left(x + 3\right)}^{\dfrac{1}{2}}dx = \dfrac{16}{3}$
$∴$ From equation (i), we get required area $= \dfrac{15}{2} - \left(\dfrac{2}{3} + \dfrac{16}{3}\right) = \dfrac{3}{2}$