The area of the region $\left\{(\mathrm{x},\mathrm{y}):{\mathrm{x}}^2+4\mathrm{x}+2\le \mathrm{y}\le |\mathrm{x}+2|\right\}$ is equal to 

${\mathrm{x}}^2+4\mathrm{x}+2\le \mathrm{y}\le |\mathrm{x}+2|$
The area bounded between 
y = x² + 4x + 2 = (x + 2)² –2 and y = |x + 2| is same as area bounded 
between y = x²–2 and y = |x| For P.O.I |x|² – 2 = |x|
$\begin{array}{l}\Rightarrow \left|\mathrm{x}\right|=2\Rightarrow \mathrm{x}=\pm 2\\ \therefore \text{ Required area }=-{\int }_{-2}^2 \left({\mathrm{x}}^2-2\right)\mathrm{dx}+{\int }_{-2}^2 \left|\mathrm{x}\right|\mathrm{dx}\\ =-2{\int }_0^2 \left({\mathrm{x}}^2-2\right)\mathrm{dx}+2{\int }_0^2 \mathrm{x}\cdot \mathrm{dx}\\ =-2{\left[\frac{{\mathrm{x}}^3}{3}-2\mathrm{x}\right]}_0^2+2{\left[\frac{{\mathrm{x}}^2}{2}\right]}_0^2\\ =-2\left[\frac{8}{3}-4\right]+2\left[\frac{4}{2}\right]\\ =-2\times \left(\frac{-4}{3}\right)+4\\ =\frac{20}{3}\end{array}$