Question

The evaluation of the double integral \( \int_{-a}^{a} \int_{0}^{x} \frac{e^y}{(1+e^y)^2} \, dy \, dx \) is: 

• A. \(a\)
• B. \(e^a\)
• C. \(e^{a^2}\)
• D. \(0\)

Hint:

Always check if the final single-variable integrand is odd when the outer limits are symmetric (\([-a, a]\)).
This avoids unnecessary computation of complex anti-derivatives.

Answer 1

The Correct Option is D

Concept:
• A function \(f(x)\) is odd if \(f(-x) = -f(x)\).
• The definite integral of an odd function over a symmetric interval \([-a, a]\) is zero: \(\int_{-a}^{a} f(x) \, dx = 0\).

Step 1: Evaluate the inner integral with respect to \(y\)
Let \(I_1 = \int_{0}^{x} \frac{e^y}{(1+e^y)^2} \, dy\).
Using substitution \(u = 1+e^y\), \(du = e^y dy\): \[ I_1 = \left[ -\frac{1}{1+e^y} \right]_{0}^{x} = -\frac{1}{1+e^x} + \frac{1}{1+e^0} = \frac{1}{2} - \frac{1}{1+e^x} \]

Step 2: Simplify the resulting single variable function \(f(x)\)
\[ f(x) = \frac{1}{2} - \frac{1}{1+e^x} = \frac{1+e^x - 2}{2(1+e^x)} = \frac{e^x - 1}{2(1+e^x)} \] Check if \(f(x)\) is odd: \[ f(-x) = \frac{e^{-x} - 1}{2(1+e^{-x})} = \frac{\frac{1}{e^x} - 1}{2\left(1+\frac{1}{e^x}\right)} = \frac{1 - e^x}{2(e^x + 1)} = -f(x) \] Thus, \(f(x)\) is an odd function.

Step 3: Integrate the odd function over the symmetric limits
\[ \int_{-a}^{a} f(x) \, dx = 0 \]