1. We are asked to calculate the definite integral of the function $f(x) = (4 - x^2)(4 + x^2)$ from $x = -2$ to $x = 2$. 2. First, expand the product inside the integral: $$ (4 - x^2)(4 + x^2) = 4 \cdot 4 + 4 \cdot x^2 - x^2 \cdot 4 - x^2 \cdot x^2 = 16 + 4x^2 - 4x^2 - x^4 = 16 - x^4 $$ 3. So the integral simplifies to: $$ \int_{-2}^{2} (16 - x^4) \, dx $$ 4. Since the function $16 - x^4$ is even (because $x^4$ is even), we can use the property of even functions: $$ \int_{-a}^{a} f(x) \, dx = 2 \int_0^{a} f(x) \, dx $$ 5. Therefore: $$ \int_{-2}^{2} (16 - x^4) \, dx = 2 \int_0^{2} (16 - x^4) \, dx $$ 6. Now compute the integral: $$ \int_0^{2} (16 - x^4) \, dx = \int_0^{2} 16 \, dx - \int_0^{2} x^4 \, dx $$ 7. Calculate each integral separately: $$ \int_0^{2} 16 \, dx = 16x \Big|_0^{2} = 16 \times 2 - 16 \times 0 = 32 $$ $$ \int_0^{2} x^4 \, dx = \frac{x^5}{5} \Big|_0^{2} = \frac{2^5}{5} - 0 = \frac{32}{5} $$ 8. Substitute back: $$ \int_0^{2} (16 - x^4) \, dx = 32 - \frac{32}{5} = \frac{160}{5} - \frac{32}{5} = \frac{128}{5} $$ 9. Multiply by 2 to get the original integral: $$ 2 \times \frac{128}{5} = \frac{256}{5} $$ 10. Final answer: $$ \boxed{\frac{256}{5}} $$