(2) (t-jddt 2×2(2-2)!J0 j=0,≠2 故 atb n=(b-a)(f(a)+f( )+∫(b)) 6 6 b (f(a)+4(0b )+f(b)=S 6 2 即为辛甫生( Simpson)公式 同样的,当n=4时, 4-0 4) 4 4×0!×(4-0)!J0 ∏Ⅰ(t-/)a j=0,≠=0 90 4)_(-1) ∏Ⅰ(t-/)dt 32 4×1×(4-1)!1001 90 (4) (-1)4-2 SoII(t-jxlt=9o 4×2×(4-2)!00 4-3 (4) ∏I(-d 32 4×3×(4-3)!00,x3 90 4 (4) 7 4-4×4.(4-4)!)xx (t-idt 90 故2 2 2 2 (2) 2 0 0, 2 ( 1) 1 ( ) 2 2! (2 2)! 6 j c t j dt − =  − = − =   −   故： 1 4 1 ( )( ( ) ( ) ( )) 6 6 2 6 ( ( ) 4 ( ) ( )) 6 2 n a b I b a f a f f b b a a b f a f f b S + = − + + − + = + + = 即为辛甫生（ Simpson ）公式。 同样的，当 n = 4 时， 4 0 4 4 (4) 0 0 0, 0 ( 1) 7 ( ) 4 0! (4 0)! 90 j c t j dt − =  − = − =   −   4 1 4 4 (4) 1 0 0, 1 ( 1) 32 ( ) 4 1! (4 1)! 90 j c t j dt − =  − = − =   −   90 12 ( ) 4 2! (4 2)! ( 1) 4 0 4 0, 2 4 2 (4) 2 − =   − − =  =  − c t j dt j 90 32 ( ) 4 3! (4 3)! ( 1) 4 0 4 0, 3 4 3 (4) 3 − =   − − =  =  − c t j dt j 90 7 ( ) 4 4! (4 4)! ( 1) 4 0 4 0, 4 4 4 (4) 4 − =   − − =  =  − c t j dt j 故：