We are tasked with finding the area between the curves $y = |x|$ and $y = x^2 - 2$.

Steps:

1. Find the points of intersection: To determine the limits of integration, we need to find where the curves intersect by solving $|x| = x^2 - 2$.

   This splits into two cases based on the definition of the absolute value:

   • Case 1: $y = x$ (for $x \geq 0$)
   • Case 2: $y = -x$ (for $x < 0$)

   Let's solve these two cases.

Case 1: $x = x^2 - 2$

$x^2 - x - 2 = 0$ Factoring: $(x - 2)(x + 1) = 0$ So, $x = 2$ or $x = -1$. Since this is the positive case, we take $x = 2$.

Case 2: $-x = x^2 - 2$

$x^2 + x - 2 = 0$ Factoring: $(x - 1)(x + 2) = 0$ So, $x = 1$ or $x = -2$. Since this is the negative case, we take $x = -2$.

Thus, the curves intersect at $x = -2$ and $x = 2$.

2. Set up the integral:

The area is given by the difference between the upper curve and the lower curve. The curve $y = |x|$ is split into two parts:

• From $x = -2$ to $x = 0$, the upper curve is $-x$ and the lower curve is $x^2 - 2$.
• From $x = 0$ to $x = 2$, the upper curve is $x$ and the lower curve is $x^2 - 2$.

Thus, the total area is:

[ A = \int_{-2}^{0} [(-x) - (x^2 - 2)] , dx + \int_{0}^{2} [x - (x^2 - 2)] , dx ]

3. Solve the integrals:

1. First integral: [ \int_{-2}^{0} [(-x) - (x^2 - 2)] , dx = \int_{-2}^{0} [-x - x^2 + 2] , dx ] The antiderivative is: $-\frac{x^2}{2} - \frac{x^3}{3} + 2x$ Evaluating from $-2$ to $0$: $\left( 0 - 0 + 0 \right) - \left( -2 - \frac{-8}{3} - 4 \right) = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$

2. Second integral: [ \int_{0}^{2} [x - (x^2 - 2)] , dx = \int_{0}^{2} [x - x^2 + 2] , dx ] The antiderivative is: $\frac{x^2}{2} - \frac{x^3}{3} + 2x$ Evaluating from $0$ to $2$: $\left( \frac{4}{2} - \frac{8}{3} + 4 \right) - \left( 0 - 0 + 0 \right) = 2 - \frac{8}{3} + 4 = 6 - \frac{8}{3} = \frac{18}{3} - \frac{8}{3} = \frac{10}{3}$

4. Total area:

Adding both parts: $A = \frac{10}{3} + \frac{10}{3} = \frac{20}{3}$

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

Would you like further clarification or additional details?

Related Questions:

1. How do we calculate areas between curves for functions with absolute values?
2. What happens if the curves intersect at more than two points?
3. How does the symmetry of the curves help in simplifying area calculations?
4. What are other methods to solve integrals involving absolute values?
5. Can we use numerical methods to approximate areas between curves?

Tip:

When dealing with absolute value functions, always split the domain where the argument inside the absolute value changes its sign.