解答题
8.设f(x)在[a，b]上连续，证明：∫abf(x)dx∫xbf(y)dy=
【正确答案】令F(x)=∫axf(t)dt，
则
∫f(x)dx∫f(y)dy
=∫f(x)[F(b)-F(x)]dx
=F(b)∫abf(x)dx-∫abf(x)F(x)
=F(b)-∫abF(x)dF(x)
=F2(b)-
F2(x)｜ab
=
F2(b)
=
则
∫f(x)dx∫f(y)dy
=∫f(x)[F(b)-F(x)]dx
=F(b)∫abf(x)dx-∫abf(x)F(x)
=F(b)-∫abF(x)dF(x)
=F2(b)-
F2(x)｜ab=
F2(b)=
【答案解析】