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

Step 1: Analyze the inequalities

We are given a region defined by two inequalities. The lower bound for $y$ is a parabola $y_1 = x^2 + 4x + 2$, and the upper bound for $y$ is an absolute value function $y_2 = |x + 2|$. We need to find the area between these two curves.

Step 2: Simplify the parabola equation

Let's simplify the equation of the parabola by completing the square. This makes it easier to see its vertex and how it relates to the absolute value function. The parabola $y_1 = x^2 + 4x + 2$ can be rewritten as $y_1 = (x + 2)^2 - 4 + 2$, which simplifies to $y_1 = (x + 2)^2 - 2$.

Step 3: Find intersection points

To find the intersection points of the two curves, we set their equations equal to each other: $(x + 2)^2 - 2 = |x + 2|$. Let $u = x + 2$. Then the equation becomes $u^2 - 2 = |u|$. We consider two cases: $u \geq 0$ and $u < 0$.

Step 4: Solve for intersection points (Case 1: $u \geq 0$)

For $u \geq 0$, the equation is $u^2 - 2 = u$, which rearranges to $u^2 - u - 2 = 0$. Factoring this quadratic equation gives $(u - 2)(u + 1) = 0$. The solutions are $u = 2$ or $u = -1$. Since we assumed $u \geq 0$, we take $u = 2$. Substituting back $u = x + 2$, we get $x + 2 = 2$, so $x = 0$.

Step 5: Solve for intersection points (Case 2: $u < 0$)

For $u < 0$, the equation is $u^2 - 2 = -u$, which rearranges to $u^2 + u - 2 = 0$. Factoring this quadratic equation gives $(u + 2)(u - 1) = 0$. The solutions are $u = -2$ or $u = 1$. Since we assumed $u < 0$, we take $u = -2$. Substituting back $u = x + 2$, we get $x + 2 = -2$, so $x = -4$.

Step 6: Set up the integral for the area

The intersection points are $x = -4$ and $x = 0$. In the interval $[-4, 0]$, the absolute value function $|x + 2|$ is above the parabola $(x + 2)^2 - 2$. Therefore, the area is given by the definite integral of the upper function minus the lower function from $x = -4$ to $x = 0$. We can simplify the integral by using the substitution $u = x + 2$, so $du = dx$. When $x = -4$, $u = -2$. When $x = 0$, $u = 2$.

Step 7: Evaluate the integral

Due to the symmetry of the integrand $(|u| - u^2 + 2)$ around $u=0$, we can integrate from $0$ to $2$ and multiply the result by $2$. For $u \geq 0$, $|u| = u$. So the integral becomes $2 \int_{0}^{2} (u - u^2 + 2) du$. Evaluating the integral, we get $2 \left[ \frac{u^2}{2} - \frac{u^3}{3} + 2u \right]_{0}^{2}$. Plugging in the limits, we find the area to be $2 \left( \frac{2^2}{2} - \frac{2^3}{3} + 2(2) \right) = 2 \left( 2 - \frac{8}{3} + 4 \right) = 2 \left( 6 - \frac{8}{3} \right) = 2 \left( \frac{18 - 8}{3} \right) = 2 \left( \frac{10}{3} \right) = \frac{20}{3}$.

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