4 t(t-2)dt 2J0 4J0 于是 f(x)x≈(b-a∑C)f(x) bar(a)+4/(a+b)+/(bC t t dt  − − = 2 0 (2) 1 ( 2) 2 1 6 4 = C t tdt  = − 2 0 (2) 2 ( 1) 4 1 6 1 =  =  − 2 0 (2) ( ) ( ) ( ) k k k b a f x dx b a C f x ) ( )] 2 [ ( ) 4 ( 6 f b a b f a f b a + + + − = 于是