设f(x)连续,x∈[0,a]且∫0af(x)dx=,求∫0adx∫xaf(x)f(y)dy.

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问题 设f(x)连续,x∈[0,a]且∫0af(x)dx=,求∫0adx∫xaf(x)f(y)dy.

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答案方法一: ∫0adx∫xaf(x)f(y)dy=∫0af(x)dx∫xaf(y)dy=∫0af(y)dy∫0yf(x)dx =∫0af(x)dx∫0xf(y)dy=[*][∫0af(x)dx∫xaf(y)dy+∫0af(x)dx∫0xf(y)dy] =[*]∫0af(x)dx∫0af(y)dy=[*][∫0af(x)dx]2=[*]([*])2=1 方法二: 令F(x)=∫0af(y)dy,就有F’(x)=-f(x),于是 ∫0adx∫xaf(x)f(y)dy=∫0af(x)dx∫xaf(y)dy=-∫0aF(x)d[F(x)]=[*]|0a=1

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