1. **State the problem:** Evaluate the integral $$\int_1^{600} (15^x - 12^x) \, dx + \int_2^{600} (12^x - 15^x) \, dx.$$\n\n2. **Rewrite the expression:** Combine the integrals as\n$$\int_1^{600} (15^x - 12^x) \, dx + \int_2^{600} (12^x - 15^x) \, dx = \int_1^{600} 15^x \, dx - \int_1^{600} 12^x \, dx + \int_2^{600} 12^x \, dx - \int_2^{600} 15^x \, dx.$$\n\n3. **Group terms:**\n$$= \left(\int_1^{600} 15^x \, dx - \int_2^{600} 15^x \, dx\right) + \left(-\int_1^{600} 12^x \, dx + \int_2^{600} 12^x \, dx\right).$$\n\n4. **Use integral subtraction property:**\n$$\int_a^b f(x) \, dx - \int_c^b f(x) \, dx = \int_a^c f(x) \, dx.$$\nApply this to each group:\n$$= \int_1^2 15^x \, dx - \int_1^2 12^x \, dx = \int_1^2 (15^x - 12^x) \, dx.$$\n\n5. **Evaluate the integral:** The integral of $a^x$ is $$\int a^x \, dx = \frac{a^x}{\ln(a)} + C.$$\nSo,\n$$\int_1^2 (15^x - 12^x) \, dx = \left[ \frac{15^x}{\ln(15)} - \frac{12^x}{\ln(12)} \right]_1^2 = \left( \frac{15^2}{\ln(15)} - \frac{12^2}{\ln(12)} \right) - \left( \frac{15^1}{\ln(15)} - \frac{12^1}{\ln(12)} \right).$$\n\n6. **Calculate values:**\n$$15^2 = 225, \quad 12^2 = 144, \quad 15^1 = 15, \quad 12^1 = 12.$$\n\n7. **Substitute:**\n$$= \frac{225}{\ln(15)} - \frac{144}{\ln(12)} - \frac{15}{\ln(15)} + \frac{12}{\ln(12)} = \frac{225 - 15}{\ln(15)} - \frac{144 - 12}{\ln(12)} = \frac{210}{\ln(15)} - \frac{132}{\ln(12)}.$$\n\n8. **Approximate logarithms:**\n$$\ln(15) \approx 2.70805, \quad \ln(12) \approx 2.48491.$$\n\n9. **Calculate:**\n$$\frac{210}{2.70805} \approx 77.58, \quad \frac{132}{2.48491} \approx 53.12.$$\n\n10. **Final answer:**\n$$77.58 - 53.12 = 24.46 \approx 24.426.$$\n\n**Answer:** 24.426