Question

If \( g(x) \) is a continuous function satisfying \( g(-x) = -g(x) \), then \( \int_{0}^{2a} g(x) \, dx \) is equal to: {5pt}

• A. \( 0 \)
• B. \( 2 \int_{0}^{a} g(x) \, dx \)
• C. \( \int_{-a}^{a} g(x) \, dx \)
• D. \( -\int_{-2a}^{0} g(x) \, dx \)
  {5pt}

Hint:

For odd functions \( g(x) \): - \( g(-x) = -g(x) \). - The integral of an odd function over symmetric intervals is zero. - Over asymmetric intervals, use the relationship \( \int_{0}^{a} g(x) \, dx = -\int_{-a}^{0} g(x) \).

Answer 1

The Correct Option is D

Step 1: Understanding the property of \( g(x) \). 
The function \( g(x) \) satisfies \( g(-x) = -g(x) \), which means \( g(x) \) is an odd function. 
Step 2: Integral over an asymmetric interval. 
The integral \( \int_{0}^{2a} g(x) \, dx \) can be related to the integral over the negative interval using the property of odd functions: \[ \int_{0}^{2a} g(x) \, dx = -\int_{-2a}^{0} g(x) \, dx. \] 
Step 3: Analyzing the given options. 
Option (D) correctly expresses the relationship between the integral over \( [0, 2a] \) and \( [-2a, 0] \) for odd functions. 
Step 4: Conclusion. 
The integral \( \int_{0}^{2a} g(x) \, dx \) equals \( -\int_{-2a}^{0} g(x) \, dx \). Thus, the correct answer is (D). {10pt}