Prove that the simple Simpson’s rule ∫ a b f ( x ) d x ≈ ( b − a ) 6 [ f ( a ) + 4 f ( a + b 2 ) + f ( b ) ]is exact for all cubic polynomials.

Question

Prove that the simple Simpson’s rule      ∫    a    b    f  (  x  )  d  x  ≈            (      b      −      a      )        6    [  f  (  a  )  +  4  f      (                  a        +        b            2        )    +  f  (  b  )  ]is exact for all cubic polynomials.

Dustin Durham · Accepted Answer

The change of variable allows to replace the range   [  a  ,  b  ] by   [  0  ,  1  ], to simplify the notation. Then setting  g  (  t  )  :=  f  (  a  +  t  (  b  −  a  )  )  ,we have      ∫    a    b    f  (  x  )    d  x  =  (  b  −  a  )      ∫    0    1    g  (  t  )    d  t  ≈  (  b  −  a  )      1    6        [    g    (    0    )    +    4    g          (              1        2            )        +    g    (    1    )    ]    .Now we can show that the relation holds exactly for all powers of   x up to third:      ∫    0    1    d  t  =  1  =      1    6        [    1    +    4    ⋅    1    +    1    ]    ,      ∫    0    1    t    d  t  =      1    2    =      1    6        [    0    +    4    ⋅          1      2        +    1    ]    ,      ∫    0    1        t    2    d  t  =      1    3    =      1    6        [    0    +    4    ⋅          1      4        +    1    ]    ,      ∫    0    1        t    3    d  t  =      1    4    =      1    6        [    0    +    4    ⋅          1      8        +    1    ]    .By linearity, it holds for all cubic polynomials.

Prove that the simple Simpson’s rule <msubsup> ∫ a b </msubs

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