$$x^2+4 x+2 \leq y \leq|x+2|$$

The area bounded between

$$y=x^2+4 x+2=(x+2)^2-2$$

and $y=|x+2|$ is same as area bounded between $y=x^2-2$ and $y=|x|$

For P.O.I $|x|^2-2=|x|$

$$\begin{aligned} & \Rightarrow|x|=2 \Rightarrow x= \pm 2 \\ & \therefore \text { Required area }=-\int_{-2}^2\left(x^2-2\right) d x+\int_{-2}^2|x| d x \\ & =-2 \int_0^2\left(x^2-2\right) d x+2 \int_0^2 x \cdot d x \\ & =-2\left[\frac{x^3}{3}-2 x\right]_0^2+2\left[\frac{x^2}{2}\right]_0^2 \\ & =--2\left[\frac{8}{3}-4\right]+2\left[\frac{4}{2}\right] \\ & =-2 \times\left(\frac{-4}{3}\right)+4 \\ & =\frac{20}{3} \end{aligned}$$