Step 1: Use the property of definite integrals
The integral \( \int_{-a}^a f(x) \, dx = 0 \) if the function \( f(x) \) is odd. A function \( f(x) \) is odd if: \[ f(-x) = -f(x). \]
Step 2: Simplify the integral for odd functions
For odd functions: \[ \int_{-a}^a f(x) \, dx = \int_{-a}^0 f(x) \, dx + \int_{0}^a f(x) \, dx. \]
Since \( f(-x) = -f(x) \), the integral over \( [-a, 0] \) cancels with the integral over \( [0, a] \), resulting in: \[ \int_{-a}^a f(x) \, dx = 0. \]
Step 3: {Conclude the result}
The given integral equals 0 when \( f(-x) = -f(x) \).
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.