1. **State the problem:** Calculate the value of $$x = \int_0^\pi \left(3 \sin(t) + \frac{12}{\pi^2} t\right) dt + \det(\ln(e^2))$$ 2. **Recall formulas and rules:** - The integral of a sum is the sum of the integrals. - Integral of $\sin(t)$ is $-\cos(t)$. - Integral of $t$ is $\frac{t^2}{2}$. - $\ln(e^2) = 2$ because $\ln(e^x) = x$. - The determinant of a scalar (single number) is the number itself. 3. **Calculate the integral:** $$\int_0^\pi 3 \sin(t) dt + \int_0^\pi \frac{12}{\pi^2} t dt$$ 4. **Evaluate each integral separately:** - For $3 \sin(t)$: $$3 \int_0^\pi \sin(t) dt = 3[-\cos(t)]_0^\pi = 3[-\cos(\pi) + \cos(0)] = 3[-(-1) + 1] = 3[1 + 1] = 6$$ - For $\frac{12}{\pi^2} t$: $$\frac{12}{\pi^2} \int_0^\pi t dt = \frac{12}{\pi^2} \left[\frac{t^2}{2}\right]_0^\pi = \frac{12}{\pi^2} \cdot \frac{\pi^2}{2} = \frac{12}{\pi^2} \cdot \frac{\pi^2}{2}$$ 5. **Simplify the fraction with cancellation:** $$\frac{12}{\cancel{\pi^2}} \cdot \frac{\cancel{\pi^2}}{2} = \frac{12}{2} = 6$$ 6. **Sum the integrals:** $$6 + 6 = 12$$ 7. **Calculate the determinant term:** $$\det(\ln(e^2)) = \det(2) = 2$$ 8. **Add all parts to find $x$:** $$x = 12 + 2 = 14$$ **Final answer:** $$\boxed{14}$$