Suppose \( f \) is an odd function, \( \int_{0}^{6} f(x) d x=4 \), and \( \int_{0}^{12} f(x) d x=6 \). \( \begin{array}{l}\text { a. Evaluate } \int_{-6}^{12} f(x) d x \\ \text { b. Evaluate } \int_{-12}^{6} f(x) d x \\ \begin{array}{l}\text { a. } \int_{-6}^{12} f(x) d x=\square \text { (Simplify your answer.) }\end{array}\end{array} \)

To solve the problem, we will use the properties of odd functions and the properties of definite integrals. ### Known Conditions: 1. \( f \) is an odd function. 2. \( \int_{0}^{6} f(x) \, dx = 4 \) 3. \( \int_{0}^{12} f(x) \, dx = 6 \) ### Step-by-Step Solution: #### Part a: Evaluate \( \int_{-6}^{12} f(x) \, dx \) 1. **Using the property of odd functions**: - For an odd function \( f \), we have: \[ \int_{-a}^{a} f(x) \, dx = 0 \] - Therefore, \( \int_{-6}^{0} f(x) \, dx = -\int_{0}^{6} f(x) \, dx \). 2. **Calculate \( \int_{-6}^{0} f(x) \, dx \)**: \[ \int_{-6}^{0} f(x) \, dx = -\int_{0}^{6} f(x) \, dx = -4 \] 3. **Now, we can express \( \int_{-6}^{12} f(x) \, dx \)**: \[ \int_{-6}^{12} f(x) \, dx = \int_{-6}^{0} f(x) \, dx + \int_{0}^{12} f(x) \, dx \] Substituting the known values: \[ \int_{-6}^{12} f(x) \, dx = -4 + 6 = 2 \] Thus, the answer for part a is: \[ \int_{-6}^{12} f(x) \, dx = 2 \] #### Part b: Evaluate \( \int_{-12}^{6} f(x) \, dx \) 1. **Calculate \( \int_{-12}^{0} f(x) \, dx \)**: \[ \int_{-12}^{0} f(x) \, dx = -\int_{0}^{12} f(x) \, dx = -6 \] 2. **Now, we can express \( \int_{-12}^{6} f(x) \, dx \)**: \[ \int_{-12}^{6} f(x) \, dx = \int_{-12}^{0} f(x) \, dx + \int_{0}^{6} f(x) \, dx \] Substituting the known values: \[ \int_{-12}^{6} f(x) \, dx = -6 + 4 = -2 \] Thus, the answer for part b is: \[ \int_{-12}^{6} f(x) \, dx = -2 \] ### Final Answers: - a. \( \int_{-6}^{12} f(x) \, dx = 2 \) - b. \( \int_{-12}^{6} f(x) \, dx = -2 \)