1. **State the problem:** Find the area of the region bounded by the curves $y = x^2 - 2$ and $y = 2$ by finding their intersection points and integrating. 2. **Find the intersection points:** Set the two functions equal to find $x$ values where they intersect: $$x^2 - 2 = 2$$ Add 2 to both sides: $$x^2 = 4$$ Take the square root: $$x = \pm 2$$ So, the curves intersect at $x = -2$ and $x = 2$. 3. **Set up the integral:** The area between the curves from $x = -2$ to $x = 2$ is given by: $$\text{Area} = \int_{-2}^{2} \left(2 - (x^2 - 2)\right) dx$$ Simplify the integrand: $$2 - x^2 + 2 = 4 - x^2$$ 4. **Integrate:** $$\int_{-2}^{2} (4 - x^2) dx = \int_{-2}^{2} 4 dx - \int_{-2}^{2} x^2 dx$$ Calculate each integral: $$4x \Big|_{-2}^{2} - \frac{x^3}{3} \Big|_{-2}^{2}$$ Evaluate at the bounds: $$4(2) - 4(-2) - \left(\frac{2^3}{3} - \frac{(-2)^3}{3}\right) = 8 + 8 - \left(\frac{8}{3} - \frac{-8}{3}\right) = 16 - \frac{16}{3}$$ 5. **Simplify the result:** $$16 - \frac{16}{3} = \frac{48}{3} - \frac{16}{3} = \frac{32}{3}$$ **Final answer:** The area of the region bounded by the curves is $\boxed{\frac{32}{3}}$.