1. The problem gives the derivative of a function as $f'(x) = (x^2 - 2) \sin(x)$ and an initial condition $f(-1) = 8$. We are asked to find the original function $f(x)$ on the interval $[-2, 2]$. 2. To find $f(x)$, we need to integrate $f'(x)$: $$f(x) = \int (x^2 - 2) \sin(x) \, dx + C$$ where $C$ is a constant determined by the initial condition. 3. Use integration by parts to solve $\int (x^2 - 2) \sin(x) \, dx$. Let $u = x^2 - 2$, so $du = 2x \, dx$. Let $dv = \sin(x) \, dx$, so $v = -\cos(x)$. 4. Applying integration by parts: $$\int (x^2 - 2) \sin(x) \, dx = - (x^2 - 2) \cos(x) + \int 2x \cos(x) \, dx$$ 5. Now integrate $\int 2x \cos(x) \, dx$ by parts again. Let $u = 2x$, so $du = 2 \, dx$. Let $dv = \cos(x) \, dx$, so $v = \sin(x)$. 6. Applying integration by parts again: $$\int 2x \cos(x) \, dx = 2x \sin(x) - \int 2 \sin(x) \, dx = 2x \sin(x) + 2 \cos(x) + C$$ 7. Substitute back: $$f(x) = - (x^2 - 2) \cos(x) + 2x \sin(x) + 2 \cos(x) + C$$ 8. Simplify: $$f(x) = -x^2 \cos(x) + 2 \cos(x) + 2x \sin(x) + 2 \cos(x) + C = -x^2 \cos(x) + 2x \sin(x) + 4 \cos(x) + C$$ 9. Use the initial condition $f(-1) = 8$ to find $C$: $$8 = -(-1)^2 \cos(-1) + 2(-1) \sin(-1) + 4 \cos(-1) + C$$ Since $\cos(-1) = \cos(1)$ and $\sin(-1) = -\sin(1)$, $$8 = -1 \cdot \cos(1) - 2 \cdot (-\sin(1)) + 4 \cos(1) + C = (-\cos(1) + 2 \sin(1) + 4 \cos(1)) + C = (3 \cos(1) + 2 \sin(1)) + C$$ 10. Solve for $C$: $$C = 8 - 3 \cos(1) - 2 \sin(1)$$ 11. Final answer: $$f(x) = -x^2 \cos(x) + 2x \sin(x) + 4 \cos(x) + 8 - 3 \cos(1) - 2 \sin(1)$$