例6设f(x)连续证明 (x-tf(t)di f(u)du dt 证一记F(x)=∫(x-)( G(x)=f(a)则 (x)=C(x)=「f()dt→F(x)-G(x)=C 而F(0)=G(0)=0故F(x)=G(x)设 f ( x ) 连续 证明 x t f t dt f u du dt x x t          − = 0 0 0 ( ) ( ) ( ) 证一记  = − x F x x t f t dt 0 ( ) ( ) ( ) G x f u du dt x t         = 0 0 ( ) ( ) 则   =  = x F x G x f t dt 0 ( ) ( ) ( )  F(x) − G(x) = C 而 F(0) = G(0) = 0 故 F(x) = G(x) 例6