1. We need to calculate the definite integral $$\int_{-2}^{2} (4 - x^2)(4 + x^2) \, dx$$. 2. First, expand the integrand using the distributive property: $$ (4 - x^2)(4 + x^2) = 4 \cdot 4 + 4 \cdot x^2 - x^2 \cdot 4 - x^2 \cdot x^2 = 16 + 4x^2 - 4x^2 - x^4 $$ 3. Simplify the expression inside the integral: $$ 16 + 4x^2 - 4x^2 - x^4 = 16 - x^4 $$ 4. So the integral becomes: $$ \int_{-2}^{2} (16 - x^4) \, dx $$ 5. Since the function is even (because $x^4$ is even and 16 is constant), we can use the property: $$ \int_{-a}^{a} f(x) \, dx = 2 \int_0^{a} f(x) \, dx $$ 6. Therefore: $$ \int_{-2}^{2} (16 - x^4) \, dx = 2 \int_0^{2} (16 - x^4) \, dx $$ 7. Integrate term by term: $$ \int_0^{2} 16 \, dx = 16x \Big|_0^{2} = 16 \times 2 = 32 $$ $$ \int_0^{2} x^4 \, dx = \frac{x^5}{5} \Big|_0^{2} = \frac{2^5}{5} = \frac{32}{5} $$ 8. Combine the results: $$ 2 \left( 32 - \frac{32}{5} \right) = 2 \left( \frac{160}{5} - \frac{32}{5} \right) = 2 \times \frac{128}{5} = \frac{256}{5} $$ 9. Final answer: $$ \boxed{\frac{256}{5}} $$