The value of ∫_-a^a f(x)dx where f(x) =(7^x)/(1+7^x) , is:

The correct answer is Option (1) → \(a\)

On a symmetric interval, the property

    \( \displaystyle \int_{-a}^{a} f(x)\,dx = \int_{0}^{a} \big[f(x) + f(-x)\big]\,dx \)

is useful here. Multiplying the numerator and denominator by \(7^x\):

    \( \displaystyle f(-x) = \frac{7^{-x}}{1 + 7^{-x}} = \frac{1}{7^x + 1} \)

so

    \( \displaystyle f(x) + f(-x) = \frac{7^x + 1}{1 + 7^x} = 1 \)

Hence

    \( \displaystyle \int_{-a}^{a} f(x)\,dx = \int_{0}^{a} 1\,dx = a \)

Tip: For any base \(b \gt 0\), \(\frac{b^x}{1 + b^x} + \frac{b^{-x}}{1 + b^{-x}} = 1\), so \(\int_{-a}^{a} \frac{b^x}{1 + b^x}\,dx = a\), whatever the base.