If f:R to R and g:R to R are one-to-one,real- | vedclass

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(A) Let $\phi(x) = [f(x) + f(-x)][g(x) - g(-x)]$. Now,calculate $\phi(-x)$: $\phi(-x) = [f(-x) + f(x)][g(-x) - g(x)]$ $\phi(-x) = [f(x) + f(-x)][-(g(x

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(A) Let $\phi(x) = [f(x) + f(-x)][g(x) - g(-x)]$. Now,calculate $\phi(-x)$: $\phi(-x) = [f(-x) + f(x)][g(-x) - g(x)]$ $\phi(-x) = [f(x) + f(-x)][-(g(x) - g(-x))]$ $\phi(-x) = -[f(x) + f(-x)][g(x) - g(-x)] = -\phi(x)$. Since $\phi(-x) = -\phi(x)$,the function $\phi(x)$ is an odd function. For any odd function $\phi(x)$,the integral over the symmetric interval $[-\pi, \pi]$ is zero: $\int_{-\pi}^{\pi} \phi(x) \, dx = 0$. Therefore,$\int_{-\pi}^{\pi} [f(x) + f(-x)][g(x) - g(-x)] \, dx = 0$.

If $f:R \to R$ and $g:R \to R$ are one-to-one,real-valued functions,then the value of the integral $\int_{-\pi}^{\pi} [f(x) + f(-x)][g(x) - g(-x)] \, dx$ is

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