Given $$ \int_{1}^{6} F(x) d x=7, \int_{1}^{6} g(x) d x=9 $$ and $$ \int_{6}^{3} F(x) d x=-4 $$ Firld, A, $$ \int_{3}^{1} F(x) d x $$ $$ B_{1} \int_{1}^{6}(2 F(x)-3 g(x)) d x $$ $$ C_{1} \int_{1}^{6}(3 F(x)+4 g(x)) d x $$

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Let's evaluate each of the integrals step by step using the given information:

1. **Given Integrals:**
   - $$\int_{1}^{6} F(x) \, dx = 7$$
   - $$\int_{1}^{6} g(x) \, dx = 9$$
   - $$\int_{6}^{3} F(x) \, dx = -4$$

2. **Understanding $$\int_{6}^{3} F(x) \, dx = -4$$:**
   
   Recall that reversing the limits of integration changes the sign of the integral:
   $$
   \int_{a}^{b} f(x) \, dx = -\int_{b}^{a} f(x) \, dx
   $$
   
   Therefore,
   $$
   \int_{3}^{6} F(x) \, dx = 4
   $$
   
3. **Finding $$\int_{1}^{3} F(x) \, dx$$:**
   
   We can express $$\int_{1}^{6} F(x) \, dx$$ as the sum of $$\int_{1}^{3} F(x) \, dx$$ and $$\int_{3}^{6} F(x) \, dx$$:
   $$
   \int_{1}^{6} F(x) \, dx = \int_{1}^{3} F(x) \, dx + \int_{3}^{6} F(x) \, dx
   $$
   
   Plugging in the known values:
   $$
   7 = \int_{1}^{3} F(x) \, dx + 4 \implies \int_{1}^{3} F(x) \, dx = 3
   $$
   
4. **Evaluating Each Part:**

**A. $$\int_{3}^{1} F(x) \, dx$$:**
   
   Reversing the limits changes the sign:
   $$
   \int_{3}^{1} F(x) \, dx = -\int_{1}^{3} F(x) \, dx = -3
   $$
   
   **B. $$\int_{1}^{6} (2F(x) - 3g(x)) \, dx$$:**
   
   Use the linearity of integrals:
   $$
   \int_{1}^{6} (2F(x) - 3g(x)) \, dx = 2\int_{1}^{6} F(x) \, dx - 3\int_{1}^{6} g(x) \, dx = 2(7) - 3(9) = 14 - 27 = -13
   $$
   
   **C. $$\int_{1}^{6} (3F(x) + 4g(x)) \, dx$$:**
   
   Again, apply the linearity of integrals:
   $$
   \int_{1}^{6} (3F(x) + 4g(x)) \, dx = 3\int_{1}^{6} F(x) \, dx + 4\int_{1}^{6} g(x) \, dx = 3(7) + 4(9) = 21 + 36 = 57
   $$
   
5. **Final Answers:**
   - **A.** $$\int_{3}^{1} F(x) \, dx = -3$$
   - **B.** $$\int_{1}^{6} (2F(x) - 3g(x)) \, dx = -13$$
   - **C.** $$\int_{1}^{6} (3F(x) + 4g(x)) \, dx = 57$$
A. $$\int_{3}^{1} F(x) \, dx = -3$$ B. $$\int_{1}^{6} (2F(x) - 3g(x)) \, dx = -13$$ C. $$\int_{1}^{6} (3F(x) + 4g(x)) \, dx = 57$$

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