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 evaluate the integrals involving the odd function \( f \), we can utilize the properties of odd functions and the given integrals. ### Part a: Evaluate \( \int_{-6}^{12} f(x) \, dx \) Since \( f \) is an odd function, we have: \[ \int_{-a}^{a} f(x) \, dx = 0 \] Thus, we can break down the integral: \[ \int_{-6}^{12} f(x) \, dx = \int_{-6}^{0} f(x) \, dx + \int_{0}^{12} f(x) \, dx \] Using the property of odd functions: \[ \int_{-6}^{0} f(x) \, dx = -\int_{0}^{6} f(x) \, dx \] Given that \( \int_{0}^{6} f(x) \, dx = 4 \), we have: \[ \int_{-6}^{0} f(x) \, dx = -4 \] Next, we substitute \( \int_{0}^{12} f(x) \, dx \): \[ \int_{-6}^{12} f(x) \, dx = -4 + 6 = 2 \] So, we conclude that: \[ \int_{-6}^{12} f(x) \, dx = 2 \] ### Part b: Evaluate \( \int_{-12}^{6} f(x) \, dx \) Again, using the property of odd functions: \[ \int_{-12}^{6} f(x) \, dx = \int_{-12}^{0} f(x) \, dx + \int_{0}^{6} f(x) \, dx \] And: \[ \int_{-12}^{0} f(x) \, dx = -\int_{0}^{12} f(x) \, dx \] Given that \( \int_{0}^{12} f(x) \, dx = 6 \), we have: \[ \int_{-12}^{0} f(x) \, dx = -6 \] Now we substitute this into our integral: \[ \int_{-12}^{6} f(x) \, dx = -6 + 4 = -2 \] Thus, we reach the final results: \[ \int_{-12}^{6} f(x) \, dx = -2 \] In summary: - a. \( \int_{-6}^{12} f(x) \, dx = 2 \) - b. \( \int_{-12}^{6} f(x) \, dx = -2 \)