I noticed that I get the exact error, using midpoint rule error bound formula, but with f ″ ( b − a 2 ) for K, i.e. : K ( b − a ) 3 24 n 2 f ″ ( b − a 2 ) ( b − a ) 3 24 n 2

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I noticed that I get the exact error, using midpoint rule error bound formula, but with       f    ″    (            b      −      a        2    ) for K, i.e. :            K      (      b      −      a              )        3                    24              n        2                                f        ″            (                        b          −          a                2            )      (      b      −      a              )        3                    24              n        2

Zayden Andrade · Accepted Answer

Step 1What you observe happens because your test example is cubic, so that its second derivative is linear. This gives a certain symmetry about the midpoint.To highlight this, consider that the integral value over   [  0  ,  a  ] is the same as the one for the symmetrized function      f    e    (  x  )  =      1    2    (  f  (  x  )  +  f  (  a  −  x  )  )  . .For this modified function the second derivative is a constant, the linear contributions of both terms in it are complementary. This constant has the value      f    e    ″    (  x  )  =      f    e    ″    (  a      /    2  )  =      f    ″    (  a      /    2  )  , ,so that indeed all the error formulas are correct when inserting this value.

I noticed that I get the exact error, using midpoint rule error bound formula, but with f

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