The area of the region $\{(x,y): x^2+4x+2\leq y\leq |x+2|\}$ is equal to
Step-by-Step Solution
Key Concept: Note $x^2+4x+2 = (x+2)^2-2$; setting $u=x+2$ the region becomes $u^2-2\leq|u|$, i.e., $|u|^2-|u|-2\leq 0$, giving $|u|\leq 2$, so $x\in[-4,0]$. Use symmetry in $u$ to integrate from $0$ to $2$.
With $u = x+2$: region is $u^2-2 \leq |u|$, valid for $|u|\leq 2$, i.e., $u\in[-2,2]$.
$$\text{Area} = 2\int_0^2\!(u-(u^2-2))du = 2\int_0^2(u-u^2+2)du = 2\left[\frac{u^2}{2}-\frac{u^3}{3}+2u\right]_0^2$$
$$= 2\left(2-\frac{8}{3}+4\right) = 2\cdot\frac{10}{3} = \frac{20}{3}.$$
Correct Answer: 4