Question

Let \( a \) be a positive real number. If \( f \) is a continuous and even function defined on the interval \( [-a, a] \), then \( \int_{-a}^{a} \dfrac{f(x)}{1+e^x} dx \) is equal to 
 

• A. \( \int_{0}^{a} f(x) dx \)
• B. \( 2 \int_{0}^{a} \dfrac{f(x)}{1+e^x} dx \)
• C. \( 2\int_{0}^{a} f(x) dx \)
• D. \( 2a \int_{0}^{a} \dfrac{f(x)}{1+e^x} dx \)

Hint:

For even functions, the integral over a symmetric interval can be simplified by doubling the integral from 0 to the positive limit.

Answer 1

The Correct Option is A

Step 1: Understanding even functions.
An even function satisfies \( f(-x) = f(x) \). Thus, the integral of an even function over a symmetric interval can be written as twice the integral from 0 to \( a \).

Step 2: Analyzing the options.
(A) \( \int_{0}^{a} f(x) dx \): This is the integral from 0 to \( a \), but we need the full range \( [-a, a] \).
(B) \( 2 \int_{0}^{a} f(x) dx \): Correct — Since \( f(x) \) is even, the integral from \( -a \) to \( a \) is twice the integral from 0 to \( a \).
(C) \( \int_{-a}^{a} f(x) dx \): This is the correct integral but does not account for the even nature of \( f(x) \).
(D) \( 2a \int_{0}^{a} \frac{f(x){1+x^2} dx \):} This is incorrect as it introduces an unnecessary factor.

Step 3: Conclusion.
The correct answer is (B) \( 2 \int_{0}^{a} f(x) dx \).