1. **State the problem:** We want to find the value of the integral $$\int_2^4 f(x) g''(x) \, dx$$ given the values of $f(x)$, $g(x)$, $g'(x)$, and $g''(x)$ at $x=2$ and $x=4$, and that $f'(x) = 3$ for all $x$. 2. **Recall the integration by parts formula:** $$\int u \, dv = uv - \int v \, du$$ We choose: - $u = f(x)$ so that $du = f'(x) dx = 3 dx$ - $dv = g''(x) dx$ so that $v = g'(x)$ 3. **Apply integration by parts:** $$\int_2^4 f(x) g''(x) \, dx = \left[ f(x) g'(x) \right]_2^4 - \int_2^4 g'(x) f'(x) \, dx$$ 4. **Evaluate the boundary term:** $$\left[ f(x) g'(x) \right]_2^4 = f(4) g'(4) - f(2) g'(2) = 13 \times 7 - 7 \times 1 = 91 - 7 = 84$$ 5. **Evaluate the remaining integral:** Since $f'(x) = 3$ is constant, $$\int_2^4 g'(x) f'(x) \, dx = 3 \int_2^4 g'(x) \, dx$$ 6. **Use the Fundamental Theorem of Calculus:** $$\int_2^4 g'(x) \, dx = g(4) - g(2) = 9 - 2 = 7$$ 7. **Calculate the integral:** $$3 \times 7 = 21$$ 8. **Combine results:** $$\int_2^4 f(x) g''(x) \, dx = 84 - 21 = 63$$ **Final answer:** $$\boxed{63}$$