1. **State the problem:** We are given a continuous function $f(x)$ on the interval $[0,2]$ with two integral conditions: $$\int_0^2 (f(x) + x) \, dx = 8$$ and $$\int_0^1 f(x) \, dx = 2$$ We need to find the value of $$\int_1^2 f(x) \, dx$$ 2. **Recall the properties of integrals:** The integral of a sum is the sum of the integrals: $$\int_a^b (f(x) + g(x)) \, dx = \int_a^b f(x) \, dx + \int_a^b g(x) \, dx$$ 3. **Apply this to the given integral:** $$\int_0^2 (f(x) + x) \, dx = \int_0^2 f(x) \, dx + \int_0^2 x \, dx = 8$$ 4. **Calculate the integral of $x$ from 0 to 2$:** $$\int_0^2 x \, dx = \left[ \frac{x^2}{2} \right]_0^2 = \frac{2^2}{2} - 0 = \frac{4}{2} = 2$$ 5. **Substitute back:** $$\int_0^2 f(x) \, dx + 2 = 8 \implies \int_0^2 f(x) \, dx = 6$$ 6. **Use the property of integrals over adjacent intervals:** $$\int_0^2 f(x) \, dx = \int_0^1 f(x) \, dx + \int_1^2 f(x) \, dx$$ 7. **Substitute known values:** $$6 = 2 + \int_1^2 f(x) \, dx$$ 8. **Solve for the unknown integral:** $$\int_1^2 f(x) \, dx = 6 - 2 = 4$$ **Final answer:** $$\boxed{4}$$