1. We are given the derivative of a function: $f'(x) = x^2 \sin(3x)$ and the initial condition $f(0) = 5$. We want to find the original function $f(x)$ on the interval $[-2, 2]$. 2. To find $f(x)$, we need to integrate $f'(x)$: $$f(x) = \int f'(x) \, dx = \int x^2 \sin(3x) \, dx + C$$ where $C$ is the constant of integration. 3. Use integration by parts to solve $\int x^2 \sin(3x) \, dx$. Let $u = x^2$, so $du = 2x \, dx$. Let $dv = \sin(3x) \, dx$, so $v = -\frac{1}{3} \cos(3x)$. 4. Applying integration by parts: $$\int x^2 \sin(3x) \, dx = -\frac{1}{3} x^2 \cos(3x) + \frac{2}{3} \int x \cos(3x) \, dx$$ 5. Now, integrate $\int x \cos(3x) \, dx$ by parts again. Let $u = x$, so $du = dx$. Let $dv = \cos(3x) \, dx$, so $v = \frac{1}{3} \sin(3x)$. 6. Applying integration by parts again: $$\int x \cos(3x) \, dx = \frac{1}{3} x \sin(3x) - \frac{1}{3} \int \sin(3x) \, dx$$ 7. Integrate $\int \sin(3x) \, dx$: $$\int \sin(3x) \, dx = -\frac{1}{3} \cos(3x) + C$$ 8. Substitute back: $$\int x \cos(3x) \, dx = \frac{1}{3} x \sin(3x) + \frac{1}{9} \cos(3x) + C$$ 9. Substitute this into step 4: $$\int x^2 \sin(3x) \, dx = -\frac{1}{3} x^2 \cos(3x) + \frac{2}{3} \left( \frac{1}{3} x \sin(3x) + \frac{1}{9} \cos(3x) \right) + C$$ 10. Simplify: $$f(x) = -\frac{1}{3} x^2 \cos(3x) + \frac{2}{9} x \sin(3x) + \frac{2}{27} \cos(3x) + C$$ 11. Use the initial condition $f(0) = 5$ to find $C$: $$f(0) = -\frac{1}{3} \cdot 0^2 \cdot \cos(0) + \frac{2}{9} \cdot 0 \cdot \sin(0) + \frac{2}{27} \cos(0) + C = \frac{2}{27} + C = 5$$ 12. Solve for $C$: $$C = 5 - \frac{2}{27} = \frac{135}{27} - \frac{2}{27} = \frac{133}{27}$$ 13. Final solution: $$f(x) = -\frac{1}{3} x^2 \cos(3x) + \frac{2}{9} x \sin(3x) + \frac{2}{27} \cos(3x) + \frac{133}{27}$$ This function satisfies the given derivative and initial condition on $[-2, 2]$.