Given \( \int_{2}^{6} f(x) d x=10 \) and \( \int_{6}^{2} g(x) d x=2 \), find \( \int_{2}^{6} 2 f(x)-3 g(x) d x \)

To find \( \int_{2}^{6} (2 f(x) - 3 g(x)) \, dx \), we can use the properties of integrals: First, we can separate the integral: \[ \int_{2}^{6} (2 f(x) - 3 g(x)) \, dx = \int_{2}^{6} 2 f(x) \, dx - \int_{2}^{6} 3 g(x) \, dx \] Now, we can factor out the constants: \[ = 2 \int_{2}^{6} f(x) \, dx - 3 \int_{2}^{6} g(x) \, dx \] From the information given, we know: \[ \int_{2}^{6} f(x) \, dx = 10 \] For \( g(x) \), we have \( \int_{6}^{2} g(x) \, dx = 2 \). However, we can rewrite this as: \[ \int_{2}^{6} g(x) \, dx = -\int_{6}^{2} g(x) \, dx = -2 \] Now we substitute the values back into our equation: \[ = 2 \cdot 10 - 3 \cdot (-2) \] Calculating this gives: \[ = 20 + 6 = 26 \] Thus, the final answer is: \[ \int_{2}^{6} (2 f(x) - 3 g(x)) \, dx = 26 \]