设f(x)连续,∫0xtf(x-t)dt=1-cosx,求

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问题 设f(x)连续,∫0xtf(x-t)dt=1-cosx,求

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答案由∫0xtf(x-t)dt[*]∫x0(x-u)f(u)(-du)=∫0x(x-u)f(u)du =x∫0xf(u)du-∫0xuf(u)du, 得x∫0xf(u)du-∫0xf(u)du=1-cosx, 两边求导得∫0xf(u)du=sinx,令[*]

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