If f(x) = e^x1+e^x, I_1 = _f(-a)^f(a) xgx(1-x),dx and I_2 = _f(-a)^f(a) gx(1-x),dx, then t…

Question

If $f(x) = \dfrac{e^x}{1+e^x}$, $I_1 = \displaystyle\int_{f(-a)}^{f(a)} xg\{x(1-x)\}\,dx$ and $I_2 = \displaystyle\int_{f(-a)}^{f(a)} g\{x(1-x)\}\,dx$, then the value of $\dfrac{I_2}{I_1}$ is:

Accepted Answer

$$f\left( x \right) = {{{e^x}} \over {1 + {e^x}}}$$

$$ \Rightarrow f\left( { - x} \right) = {{{e^{ - x}}} \over {1 + {e^{ - x}}}} = {1 \over {{e^x} + 1}}$$

$$\therefore$$ $$f\left( x \right) + f\left( { - x} \right) = 1\forall x$$

Now $${I_1} = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {xg\left\{ {x\left( {1 - x} \right)} \right\}} dx$$

$$ = \int\limits_{f\left( { - a} \right)}^{f\left( a \right)} {\left( {1 - x} \right)} g\left\{ {x\left( {1 - x} \right)} \right\}dx$$

$$\left[ {} \right.$$ using $$\int\limits_a^b {f\left( x \right)} dx\,a$$

$$ = \int\limits_a^b {f\left( {a + b - x} \right)dx} $$ $$\left. \, \right]$$

$$ = {I_2} - {I_1} \Rightarrow 2{I_1} = {I_2}$$

$$\therefore$$ $${{{I_2}} \over {{I_1}}} = 2$$

Solutions · 1