1. Key Concepts and Formulas

• Property of Definite Integrals (King Property): For a continuous function $h(x)$ on $[a, b]$, $\int_{a}^{b} h(x) \,dx = \int_{a}^{b} h(a+b-x) \,dx$
• Function Property: If $f(x) + f(c-x) = k$ for some constant $c$ and $k$, this often simplifies integrals with limits related to $0$ and $c$.

2. Step-by-Step Solution

Step 1: Analyze the function $f(x)$ and determine the sum of the limits of integration. We are given $f(x) = \frac{e^x}{1+e^x}$. The limits of integration for both $I_1$ and $I_2$ are $f(-a)$ and $f(a)$. Let's find $f(-x)$: $f(-x) = \frac{e^{-x}}{1+e^{-x}}$ To simplify, multiply the numerator and denominator by $e^x$: $f(-x) = \frac{e^{-x} \cdot e^x}{(1+e^{-x}) \cdot e^x} = \frac{1}{e^x+1}$ Now, let's find the sum $f(x) + f(-x)$: $f(x) + f(-x) = \frac{e^x}{1+e^x} + \frac{1}{1+e^x} = \frac{e^x+1}{1+e^x} = 1$ This means that for any value of $a$, we have $f(a) + f(-a) = 1$. Let the lower limit of integration be $L = f(-a)$ and the upper limit be $U = f(a)$. From our finding, $L+U = 1$. This is a crucial observation as it simplifies the application of the King Property.

Step 2: Apply the King Property to $I_1$. We are given: $I_1 = \int_{f(-a)}^{f(a)} xg\{x(1-x)\} \,dx$ Let $A = f(-a)$ and $B = f(a)$. Then $A+B = 1$. Using the King Property, $\int_{A}^{B} h(x) \,dx = \int_{A}^{B} h(A+B-x) \,dx$, we replace $x$ with $(A+B-x) = (1-x)$ in the integrand: The term $x$ outside the function $g$ becomes $(1-x)$. The argument of $g$, which is $x(1-x)$, becomes $(1-x)(1-(1-x))$. Simplifying the argument: $(1-x)(1-1+x) = (1-x)x = x(1-x)$. So, the integrand transforms as follows: $xg\{x(1-x)\} \longrightarrow (1-x)g\{x(1-x)\}$ Thus, $I_1$ can be rewritten as: $I_1 = \int_{f(-a)}^{f(a)} (1-x)g\{x(1-x)\} \,dx$

Step 3: Split the transformed integral and relate it to $I_1$ and $I_2$. The transformed expression for $I_1$ can be split into two integrals: $I_1 = \int_{f(-a)}^{f(a)} g\{x(1-x)\} \,dx - \int_{f(-a)}^{f(a)} xg\{x(1-x)\} \,dx$ Now, let's compare these with the definitions of $I_1$ and $I_2$: $I_2 = \int_{f(-a)}^{f(a)} g\{x(1-x)\} \,dx$ $I_1 = \int_{f(-a)}^{f(a)} xg\{x(1-x)\} \,dx$ Substituting these into the split equation for $I_1$: $I_1 = I_2 - I_1$

Step 4: Solve for the ratio $\frac{I_2}{I_1}$. Rearrange the equation $I_1 = I_2 - I_1$: $I_1 + I_1 = I_2$ $2I_1 = I_2$ To find the ratio $\frac{I_2}{I_1}$, divide both sides by $I_1$ (assuming $I_1 \neq 0$): $\frac{I_2}{I_1} = 2$

3. Common Mistakes & Tips

• Failure to recognize $f(x) + f(-x) = 1$: This property of $f(x)$ is key to establishing that the sum of the integration limits is 1, which is essential for applying the King Property effectively.
• Incorrect application of the King Property: Ensure that every instance of $x$ in the integrand, including those within nested functions like $x(1-x)$, is replaced by $(a+b-x)$.
• Algebraic errors in simplifying the argument of $g$: The argument $x(1-x)$ transforming to $(1-x)(1-(1-x))$ and simplifying back to $x(1-x)$ is a critical step. Any mistake here will lead to an incorrect result.

4. Summary

The problem requires the application of a fundamental property of definite integrals, often called the King Property, in conjunction with an analysis of the given function $f(x)$. We first established that $f(x) + f(-x) = 1$, which implies that the sum of the integration limits for $I_1$ and $I_2$ is 1. By applying the King Property to $I_1$, where we substitute $x$ with $(1-x)$ (since the sum of limits is 1), we transformed the integral. This transformed integral could then be expressed as $I_2 - I_1$. Equating this to the original $I_1$ led to the equation $2I_1 = I_2$, from which the ratio $\frac{I_2}{I_1}$ was found to be 2.

The final answer is $\boxed{2}$.