设f(x)连续,∫0xtf(x-t)dt=1一cosx,求f(x)dx.

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问题 设f(x)连续,∫0xtf(x-t)dt=1一cosx,求f(x)dx.

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答案由∫0xtf(x—t)dt[*]∫x0(x一u)f(u)(一du)=∫0x(x—u)f(u)du =x∫0xf(u)du—∫0xuf(u)du,得x∫0xf(u)du—∫0xuf(u)du=1一cosx, 两边求导得∫0xf(u)du=sinx,令x=[*]f(x)dx=1.

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