Let f:R to R and g:R to R be continuous funct | vedclass

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(D) Let $h(x) = [f(x) + f(-x)][g(x) - g(-x)]$. Now,calculate $h(-x)$: $h(-x) = [f(-x) + f(-(-x))][g(-x) - g(-(-x))] = [f(-x) + f(x)][g(-x) - g(x)]$. W

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(D) Let $h(x) = [f(x) + f(-x)][g(x) - g(-x)]$. Now,calculate $h(-x)$: $h(-x) = [f(-x) + f(-(-x))][g(-x) - g(-(-x))] = [f(-x) + f(x)][g(-x) - g(x)]$. We can rewrite this as: $h(-x) = -[f(x) + f(-x)][g(x) - g(-x)] = -h(x)$. Since $h(-x) = -h(x)$,the function $h(x)$ is an odd function. For any odd function $h(x)$,the integral over a symmetric interval $[-a, a]$ is zero: $\int_{-a}^{a} h(x) \, dx = 0$. Therefore,$\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx = 0$.

Let $f:R \to R$ and $g:R \to R$ be continuous functions,then the value of the integral $\int_{-\pi/2}^{\pi/2} [f(x) + f(-x)][g(x) - g(-x)] \, dx$ is:

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