Question

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 ∫$_{-4}^{4}$ f(x²) dx is __________.

Hint:

The integral of an odd function over a symmetric interval $[-a, a]$ is always zero.

Answer 1

Correct Answer: 0

Step 1: Given $g(4-x) = -g(x)$. Substitute this into the first equation: $f(x^2) - g(x) = 4x^3$.
Step 2: We need to evaluate $I = \int_{-4}^{4} f(x^2) dx$.
Step 3: From Step 1, $f(x^2) = 4x^3 + g(x)$.
Step 4: $I = \int_{-4}^{4} (4x^3 + g(x)) dx = \int_{-4}^{4} 4x^3 dx + \int_{-4}^{4} g(x) dx$.
Step 5: $\int_{-4}^{4} 4x^3 dx = 0$ because $4x^3$ is an odd function.
Step 6: For $\int_{-4}^{4} g(x) dx$, use property $\int_a^b g(x) dx = \int_a^b g(a+b-x) dx$. Here, $a+b = -4+4 = 0$. So $\int_{-4}^{4} g(x) dx = \int_{-4}^{4} g(-x) dx$. Since $g(4-x) = -g(x)$, and the symmetry suggests $g(x)$ is odd relative to the interval, the integral results in 0.