If f(x) = frac{e^x}{1 + e^x},I_1 = int_{f(-a) | vedclass

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(D) Given $f(x) = \frac{e^x}{1 + e^x}$. Note that $f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1}$.Thus,$f(a) + f(-a) = \frac{e^a}{1 + e^a} + \

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$2$

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(D) Given $f(x) = \frac{e^x}{1 + e^x}$. Note that $f(-a) = \frac{e^{-a}}{1 + e^{-a}} = \frac{1}{e^a + 1}$.Thus,$f(a) + f(-a) = \frac{e^a}{1 + e^a} + \frac{1}{1 + e^a} = 1$.Let $I_1 = \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$. Using the property $\int_{A}^{B} h(x) dx = \int_{A}^{B} h(A + B - x) dx$,where $A + B = 1$:$I_1 = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)(1 - (1 - x))\} dx = \int_{f(-a)}^{f(a)} (1 - x) g\{(1 - x)x\} dx$.$I_1 = \int_{f(-a)}^{f(a)} g\{x(1 - x)\} dx - \int_{f(-a)}^{f(a)} x g\{x(1 - x)\} dx$.$I_1 = I_2 - I_1$.$2I_1 = I_2$.Therefore,$\frac{I_2}{I_1} = 2$.

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

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