Let $f(x)$ be a positive function and $$I_1 = \int_{-\frac{1}{2}}^{1} 2x f(2x(1 - 2x)) \, dx \quad \text{and} \quad I_2 = \int_{-1}^{\frac{1}{2}} f(x(1 - x)) \, dx.$$ Then the value of $\frac{I_2}{I_1}$ is equal to:

Step 1: Evaluate $$I_1$$ using substitution

We begin by evaluating the integral $$I_1$$ . We will use a substitution to simplify the integrand. Let $$t = 2x$$ . Then $$dt = 2 \, dx$$ . We also need to change the limits of integration. When $$x = -\frac{1}{2}$$ , $$t = 2(-\frac{1}{2}) = -1$$ . When $$x = 1$$ , $$t = 2(1) = 2$$ .

Step 2: Substitute into $$I_1$$

After substituting $$t = 2x$$ and $$dx = \frac{dt}{2}$$ , the integral $$I_1$$ becomes $$\int_{-1}^{2} t f(t(1 - t)) \frac{dt}{2}$$ . We can pull the constant $$\frac{1}{2}$$ outside the integral.

Step 3: Evaluate $$I_2$$ using substitution

Now, let's evaluate the integral $$I_2$$ . We will use a substitution to simplify the integrand. Let $$u = 1 - x$$ . Then $$du = -dx$$ , so $$dx = -du$$ . We also need to change the limits of integration. When $$x = -1$$ , $$u = 1 - (-1) = 2$$ . When $$x = \frac{1}{2}$$ , $$u = 1 - \frac{1}{2} = \frac{1}{2}$$ . Also, $$x = 1 - u$$ .

Step 4: Substitute into $$I_2$$

After substituting $$u = 1 - x$$ , $$dx = -du$$ , and $$x = 1 - u$$ , the integral $$I_2$$ becomes $$\int_{2}^{\frac{1}{2}} f((1 - u)u) (-du)$$ . We can pull the negative sign outside the integral and then reverse the limits of integration, which changes the sign back to positive. This gives us $$\int_{\frac{1}{2}}^{2} f(u(1 - u)) \, du$$ .

Step 5: Apply property of definite integrals to $$I_2$$

Since the variable of integration is a dummy variable, we can replace $$u$$ with $$x$$ in the expression for $$I_2$$ . This gives us $$I_2 = \int_{\frac{1}{2}}^{2} f(x(1 - x)) \, dx$$ .

Step 6: Combine $$I_1$$ and $$I_2$$

We can split the integral $$I_1$$ into two parts using the property $$\int_a^b g(x) dx = \int_a^c g(x) dx + \int_c^b g(x) dx$$ . We choose $$c = \frac{1}{2}$$ to match the limits of $$I_2$$ .

Step 7: Evaluate the ratio $$\frac{I_2}{I_1}$$

Let $$J = \int_{\frac{1}{2}}^{2} f(x(1 - x)) \, dx$$ . Then $$I_2 = J$$ . For the first part of $$I_1$$ , let $$t = 1 - y$$ . Then $$dt = -dy$$ . When $$t = -1$$ , $$y = 2$$ . When $$t = \frac{1}{2}$$ , $$y = \frac{1}{2}$$ . So, $$\int_{-1}^{\frac{1}{2}} t f(t(1 - t)) \, dt = \int_{2}^{\frac{1}{2}} (1 - y) f((1 - y)y) (-dy) = \int_{\frac{1}{2}}^{2} (1 - y) f(y(1 - y)) \, dy = \int_{\frac{1}{2}}^{2} (1 - x) f(x(1 - x)) \, dx$$ . Therefore, $$I_1 = \frac{1}{2} \left( \int_{\frac{1}{2}}^{2} (1 - x) f(x(1 - x)) \, dx + \int_{\frac{1}{2}}^{2} x f(x(1 - x)) \, dx \right) = \frac{1}{2} \int_{\frac{1}{2}}^{2} (1 - x + x) f(x(1 - x)) \, dx = \frac{1}{2} \int_{\frac{1}{2}}^{2} f(x(1 - x)) \, dx = \frac{1}{2} J$$ . Thus, $$\frac{I_2}{I_1} = \frac{J}{\frac{1}{2} J} = 2$$ .

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