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

Find points of intersection.

We need to see where the parabola $$y = x^2 + 4x + 2$$ intersects the absolute value function $$y = |x + 2|$$.

• Case 1: $$x \geq -2$$ (where $$|x+2| = x+2$$)

$$x^2 + 4x + 2 = x + 2 \implies x^2 + 3x = 0 \implies x = 0$$ or $$x = -3$$ (reject $$-3$$).

• Case 2: $$x < -2$$ (where $$|x+2| = -(x+2)$$)

$$x^2 + 4x + 2 = -x - 2 \implies x^2 + 5x + 4 = 0 \implies (x+1)(x+4) = 0 \implies x = -4$$ or $$x = -1$$ (reject $$-1$$).

Set up the integral.

The area is bounded between $$x = -4$$ and $$x = 0$$.

$$Area = \int_{-4}^{0} (|x+2| - (x^2+4x+2)) dx$$

Since the functions are symmetric about $$x = -2$$, we can calculate the area from $$-2$$ to $$0$$ and double it:

$$Area = 2 \int_{-2}^{0} ((x+2) - (x^2+4x+2)) dx = 2 \int_{-2}^{0} (-x^2 - 3x) dx$$

$$= 2 \left[ -\frac{x^3}{3} - \frac{3x^2}{2} \right]_{-2}^{0} = 2 \left[ 0 - \left( \frac{8}{3} - 6 \right) \right] = 2 \left( \frac{10}{3} \right) = \frac{20}{3}$$