Let f(x) be a positive continuous function. Let I₁ = ₀^ 2 x f( ^2 x) dx and I₂ = ₀^ /2 f( ^2 x) dx . If I₁ = k I₂ , then the value of the constant k is equal to

Step-by-step solution

We are given I₁ = ₀^ 2 x f( ^2 x) dx . Applying the property _a^b g(x) dx = _a^b g(a+b-x) dx , we get: I₁ = ₀^ 2 (2 - x) f( ^2(2 - x)) dx I₁ = ₀^ 2 (2 - x) f( ^2 x) dx Adding this to the original integral I₁ : 2I₁ = ₀^ 2 2 f( ^2 x) dx I₁ = ₀^ 2 f( ^2 x) dx Since the function f( ^2 x) is periodic with period , we can write: ₀^ 2 f( ^2 x) dx = 2 ₀^ f( ^2 x) dx Now, applying the property ₀^ 2a g(x) dx = 2 ₀^a g(x) dx (since g(2a-x) = g(x) ): Here, f( ^2( - x)) = f( ^2 x) , so: 2 ₀^ f( ^2 x) dx = 4 ₀^ /2 f( ^2 x) dx Th