To solve for the area between the curves $y = |x|$ and $y = x^2 - 2$, we need to follow these steps:

Step 1: Find the points of intersection.

The curves $y = |x|$ and $y = x^2 - 2$ intersect at points where they have the same $y$-value. This means we need to solve the equation $|x| = x^2 - 2$.

For $x \geq 0$:

In this case, $|x| = x$, so the equation becomes: $x = x^2 - 2$ $x^2 - x - 2 = 0$ Factoring this gives: $(x - 2)(x + 1) = 0$ Thus, $x = 2$ or $x = -1$. Since we are considering $x \geq 0$, we have $x = 2$.

For $x < 0$:

In this case, $|x| = -x$, so the equation becomes: $-x = x^2 - 2$ $x^2 + x - 2 = 0$ Factoring gives: $(x - 1)(x + 2) = 0$ Thus, $x = 1$ or $x = -2$. Since $x < 0$, we take $x = -2$.

Step 2: Set up the integral.

The area between the curves is the integral of the difference between the two functions over the interval where they overlap, which is $[-2, 2]$. We split the integral into two parts based on the symmetry of $|x|$:

1. For $x \in [-2, 0]$, $|x| = -x$.
2. For $x \in [0, 2]$, $|x| = x$.

Thus, the total area is given by: $\text{Area} = \int_{-2}^{0} \left( (-x) - (x^2 - 2) \right) \, dx + \int_{0}^{2} \left( x - (x^2 - 2) \right) \, dx$

Step 3: Compute the integrals.

First integral ($x \in [-2, 0]$):

$\int_{-2}^{0} \left( -x - x^2 + 2 \right) \, dx$ $= \int_{-2}^{0} (-x) \, dx - \int_{-2}^{0} x^2 \, dx + \int_{-2}^{0} 2 \, dx$ Evaluating these integrals:

1. $\int_{-2}^{0} (-x) \, dx = \left[ \frac{-x^2}{2} \right]_{-2}^{0} = 0 - \frac{-4}{2} = 2$
2. $\int_{-2}^{0} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{-2}^{0} = 0 - \frac{(-2)^3}{3} = \frac{8}{3}$
3. $\int_{-2}^{0} 2 \, dx = \left[ 2x \right]_{-2}^{0} = 0 - (-4) = 4$

Thus, the first integral becomes: $2 - \frac{8}{3} + 4 = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$

Second integral ($x \in [0, 2]$):

$\int_{0}^{2} \left( x - x^2 + 2 \right) \, dx$ $= \int_{0}^{2} x \, dx - \int_{0}^{2} x^2 \, dx + \int_{0}^{2} 2 \, dx$ Evaluating these integrals:

1. $\int_{0}^{2} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{4}{2} - 0 = 2$
2. $\int_{0}^{2} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{2} = \frac{8}{3} - 0 = \frac{8}{3}$
3. $\int_{0}^{2} 2 \, dx = \left[ 2x \right]_{0}^{2} = 4 - 0 = 4$

Thus, the second integral becomes: $2 - \frac{8}{3} + 4 = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$

Step 4: Add the results.

The total area is: $\frac{10}{3} + \frac{10}{3} = \frac{20}{3}$

Thus, the area between the curves is $\frac{20}{3}$.

Would you like more details on any step?

Here are five related questions:

1. How do you solve the system of equations to find points of intersection?
2. What is the significance of symmetry in integrals involving absolute values?
3. How can you check if two functions always intersect at certain points?
4. How does the absolute value affect the integration process in such problems?
5. How would you find the area if the limits of integration were different?

Tip: When dealing with absolute values in integrals, split the integral into regions where the function behaves linearly for easier calculations.