令F()=0/()-f(x)k-30(x)-f(o 则F()=(0()J(xk-J(x)+3/b-0) F(a)=-3f(x)-f(a)dx<0 F(b)=If(b)-f(x)ldx>0 故由零点定理知3∈(,6)上()=0又 F()=f(t)(t-a+3b-3t)=f(t)(b-a+2(b-1)>0 故ξ唯一  = − − − t a b t F(t) [ f (t) f (x)]dx 3 [ f (x) f (t)]dx   = − − − + − t a b t 则F(t) f (t)(t a) f (x)dx 3 f (x)dx 3 f (t)(b t)  = − −  b a F(a) 3 [ f (x) f (a)]dx 0  = −  b a F(b) [ f (b) f (x)]dx 0 故由零点定理知  (a,b) 使F( ) = 0 又 F(t) = f (t)(t − a + 3b − 3t) = f (t)(b − a + 2(b − t))  0 故 唯一 令