设f(x),g(x)在区间[-a,a](a>0)上连续,g(x)为偶函数,且f(x)满足条件 f(x)+f(-x)=A(A为常数) 证明:∫-aaf(x)g(x)dx=A∫0ag(x)dx

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问题 设f(x),g(x)在区间[-a,a](a>0)上连续,g(x)为偶函数,且f(x)满足条件
f(x)+f(-x)=A(A为常数)

证明:∫-aaf(x)g(x)dx=A∫0ag(x)dx

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答案-aaf(x)g(x)dx=∫-a0f(x)g(x)dx+∫0af(x)g(x)dx而 ∫-a0f(x)g(x)dx[*]-∫a0f(-t)g(-t)dt=∫0af(-x)g(x)dx 于是 ∫-aaf(x)g(x)dx=∫0af(-x)g(x)dx+∫0af(x)g(x)dx =∫0a[f(x)+f(-x)]g(x)dx=A∫0ag(x)dx

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