Let f(x) and g(x) be two functions satisfying | vedclass

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(D) Given $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) = -g(x)$.Substituting $g(4-x) = -g(x)$ into the first equation,we get $f(x^{2}) - g(x) = 4x^{3}$,so

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(D) Given $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) = -g(x)$.Substituting $g(4-x) = -g(x)$ into the first equation,we get $f(x^{2}) - g(x) = 4x^{3}$,so $f(x^{2}) = 4x^{3} + g(x)$.We want to evaluate $I = \int_{-4}^{4} f(x) dx$. Let $x = t^{2}$,then $dx = 2t dt$. However,since $f(x)$ is defined via $f(x^{2})$,we consider the integral $\int_{-4}^{4} f(x^{2}) dx$.Since $f(x^{2}) = 4x^{3} + g(x)$,we have $\int_{-4}^{4} f(x^{2}) dx = \int_{-4}^{4} (4x^{3} + g(x)) dx$.$= \int_{-4}^{4} 4x^{3} dx + \int_{-4}^{4} g(x) dx$.Since $4x^{3}$ is an odd function,$\int_{-4}^{4} 4x^{3} dx = 0$.Given $g(4-x) = -g(x)$,let $u = 4-x$,then $du = -dx$. The integral $\int_{-4}^{4} g(x) dx$ becomes $\int_{8}^{0} g(4-u) (-du) = \int_{0}^{8} g(4-u) du = \int_{0}^{8} -g(u) du$. This implies the integral of $g(x)$ over symmetric intervals is $0$.Thus,the integral evaluates to $0$.

Let $f(x)$ and $g(x)$ be two functions satisfying $f(x^{2}) + g(4-x) = 4x^{3}$ and $g(4-x) + g(x) = 0$. Then the value of $\int_{-4}^{4} f(x) dx$ is

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