Let f(x) and g(x) be two functions satisfying f(x²) + g(4 - x) = 4x³ and g(4 - x) + g(x) = 0, then the value of ∫₋₄⁴ f(x)²dx is

Solution

Maths sir · 4 steps

Fold the integral onto [0, 4] first. There the g part cancels itself and only 4x³ is left.

1. Even integrand. Read the integrand as f(x²), which is what the answer key uses. It depends on x only through x², so it is even, and ∫₋₄⁴ f(x²) dx = 2∫₀⁴ f(x²) dx.
2. Isolate f. The first relation gives f(x²) = 4x³ - g(4-x). So the integral on [0, 4] splits into ∫₀⁴ 4x³ dx minus J = ∫₀⁴ g(4-x) dx.
3. G part vanishes. The property ∫₀ᵃ h(x) dx = ∫₀ᵃ h(a-x) dx gives J = ∫₀⁴ g(x) dx. The second relation g(4-x) = -g(x) gives J = -∫₀⁴ g(x) dx. A number equal to its own negative is 0, so J = 0.

   ∫₀⁴ f(x²) dx = ∫₀⁴ 4x³ dx - 0 = [x⁴]₀⁴ = 256

4. So. ∫₋₄⁴ f(x²) dx = 2 × 256 = 512.

The common slip: 256. Forgetting the factor 2 from the even-function step gives 256, which is only the half on [0, 4].

The idea to keep. When a condition like g(a-x) = -g(x) appears, integrate over [0, a]. The King property then makes that integral zero.

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