Evaluate $\int\limits_{-a}^{a} f(x) dx$, where $f(x) = \frac{9^x}{1 + 9^x}$.
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $a$
$I = \int\limits_{-a}^{a} f(x) = \int\limits_{-a}^{a} \frac{9^x}{1 + 9^x} dx$
$I = \int\limits_{-a}^{a} \frac{9^{(a-a-x)}}{1 + 9^{(a-a-x)}} dx$
$I = \int\limits_{-a}^{a} \frac{9^{-x}}{1 + 9^{-x}} dx \dots(i)$
$I = \int\limits_{-a}^{a} \frac{1}{1 + 9^x} dx \dots(ii)$
On adding eqs. (i) and (ii)
$2I = \int\limits_{-a}^{a} \frac{9^x}{1 + 9^x} dx + \int\limits_{-a}^{a} \frac{1}{1 + 9^x} dx$
$2I = \int\limits_{-a}^{a} \frac{1 + 9^x}{1 + 9^x} dx = \int\limits_{-a}^{a} 1.dx$
$2I = [x]_{-a}^{a}$
$2I = 2a$
$∴I = a$