解答题
22.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∫abf(x)dx∫xbf(y)dy=∫abf(x)[F(b)－F(x)]dx
=F(b)∫abf(x)dx－∫abf(x)F(x)dx=F2(b)－∫abF(x)dF(x)
=F2(b)－
则∫abf(x)dx∫xbf(y)dy=∫abf(x)[F(b)－F(x)]dx
=F(b)∫abf(x)dx－∫abf(x)F(x)dx=F2(b)－∫abF(x)dF(x)
=F2(b)－
【答案解析】