设f(x)为非负连续函数,且满足f(x)∫0xf(x-t)dt=sin4x,求f(x)在[0,π/2]上的平均值.

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问题 设f(x)为非负连续函数,且满足f(x)∫0xf(x-t)dt=sin4x,求f(x)在[0,π/2]上的平均值.

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答案令x-t=u,则∫0xf(x-t)dt=∫0xf(u)du.于是 f(x)∫0xf(u)du=sin4x,d[∫0xf(u)du]2=2sin4xdx. 两边积分(∫0π/2)得[∫0π/2f(u)du]2=2∫0π/2sin4xdx [*] 故f(x)在[0,π/2]上的平均值为 [*]

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