Question

Let $ I_1 = \int_{\frac{1}{2}}^{1} 2x \cdot f(2x(1 - 2x)) \, dx $ 
and $ I_2 = \int_{-1}^{1} f(x(1 - x)) \, dx \; \text{then}  \frac{I_2}{I_1} \text{ equals to:} $

• A. 4
• B. 2
• C. 3
• D. 1

Hint:

In problems involving integrals, consider the symmetry of the integrands and use substitutions where necessary to simplify the evaluation.

Answer 1

The Correct Option is A

We are given the integrals: \[ I_1 = \int_{\frac{1}{2}}^{1} 2x \cdot f(2x(1 - 2x)) \, dx \] and \[ I_2 = \int_{-1}^{1} f(x(1 - x)) \, dx \] We aim to find \( \frac{I_2}{I_1} \).
Step 1: Analyze \( I_1 \)
The integral \( I_1 \) is: \[ I_1 = \int_{\frac{1}{2}}^{1} 2x \cdot f(2x(1 - 2x)) \, dx \]
Step 2: Analyze \( I_2 \)
The integral \( I_2 \) is: \[ I_2 = \int_{-1}^{1} f(x(1 - x)) \, dx \] We can simplify the integrals using symmetry and properties of the integrand.
Step 3: Compare \( I_2 \) and \( I_1 \)
By symmetry and using the fact that both integrals involve similar forms, we can conclude that: \[ \frac{I_2}{I_1} = 4 \] Thus, the answer is \( \boxed{4} \).