QJEEM25Ma7615: JEE Main 2025 Mathematics - Class 12 JEEM25Ma7615 Solution: The area of the region is equal to

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

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 $|x|^2-2=|x|$

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