1. **State the problem:** Find the value of the integral $$\int_0^1 x \cdot f''(x) \, dx$$ given the values of $f(x)$, $f'(x)$, and $f''(x)$ at $x=0$ and $x=1$. 2. **Recall integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ We choose: - $u = x \implies du = dx$ - $dv = f''(x) dx \implies v = f'(x)$ 3. **Apply integration by parts:** $$\int_0^1 x f''(x) dx = \left. x f'(x) \right|_0^1 - \int_0^1 f'(x) dx$$ 4. **Evaluate the boundary term:** $$\left. x f'(x) \right|_0^1 = 1 \cdot f'(1) - 0 \cdot f'(0) = f'(1) = 6$$ 5. **Evaluate the remaining integral:** $$\int_0^1 f'(x) dx = f(1) - f(0) = 2 - 1 = 1$$ 6. **Combine results:** $$\int_0^1 x f''(x) dx = 6 - 1 = 5$$ **Final answer:** $$\boxed{5}$$