Let f(x) be a positive function and I₁ = ∫_(-1/2)¹ 2x f(2x(1-2x)) dx and I₂ = ∫₋₁² f(x(1-x)) dx. Then the value of I₂/I₁ is equal to ________

Solution

Maths sir · 3 steps

Put t = 2x so both integrals run from -1 to 2, then use t → 1 - t to remove the extra t.

1. Match the limits. In I₁, put t = 2x, so dx = dt/2 and the limits -1/2 and 1 become -1 and 2. Then I₁ = 1/2∫₋₁² t f(t(1-t)) dt. Call this integral J, so I₁ = J/2.
2. Reflect the interval. Here a + b = -1 + 2 = 1, so replace t by 1 - t. The product t(1-t) stays the same, and the factor t becomes 1 - t.

   J = ∫₋₁² (1-t) f(t(1-t)) dt = I₂ - J ⇒ J = I₂/2

3. So. I₁ = J/2 = I₂/4. Since f is positive, I₁ ≠ 0, so I₂/I₁ = 4. The answer is (d).

The idea to keep. When the function inside stays the same under x → a + b - x, the extra factor x can be replaced by (a+b)/2. First substitute so that both integrals have the same limits.

Answer from the NTA final answer key. Solution by PrepWiser.