12th Class
Mathematics
Definite Integrals
Question Bank
Critical Thinking
question_answer
If \[f(x)=\frac{{{e}^{x}}}{1+{{e}^{x}}},\,\,\,\ {{I}_{1}}=\int_{f(-a)}^{f(a)}{xg\{x(1-x)\}dx}\], and \[{{I}_{2}}=\int_{f(-a)}^{f(a)}{g\{x(1-x))\}dx}\], then the value of \[\frac{{{I}_{2}}}{{{I}_{1}}}\] is [AIEEE 2004]