设f(x)连续,y=y(x)由exy+(x一1)y=ex确定,又∫02xtf(2x—t)dt=∫0xy(t)dt,求∫02f(x)dx.

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问题 设f(x)连续,y=y(x)由exy+(x一1)y=ex确定,又∫02xtf(2x—t)dt=∫0xy(t)dt,求∫02f(x)dx.

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答案02xtf(2x—t)dt[*]∫02x(2x一u)f(u)du=2x∫02xf(u)du一∫02xf(u)du. 由题设有 2x∫02xf(u)du—∫02xuf(u)du=∫0xy(t)dt, 两边对x求导得 2∫02xf(u)du+4xf(2x)一4xf(2x)=y(x), 即 2∫02xf(u)du=y(x). 令x=1得∫02f(x)dx=[*]y(1). 又ey(1)+0.y(1)=e,所以,y(1)=1.故∫02f(x)dx=[*].

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