Since a = 0 {\displaystyle a=0} and b = 2 {\displaystyle b=2} , we have Δ x = 2 − 0 6 = 1 3 {\displaystyle \Delta x={\frac {2-0}{6}}={\frac {1}{3}}} , and so
x 0 = 0 {\displaystyle x_{0}=0} , x 1 = 1 3 {\displaystyle x_{1}={\frac {1}{3}}} , x 2 = 2 3 {\displaystyle x_{2}={\frac {2}{3}}} , x 3 = 1 {\displaystyle x_{3}=1} , x 4 = 4 3 {\displaystyle x_{4}={\frac {4}{3}}} , x 5 = 5 3 {\displaystyle x_{5}={\frac {5}{3}}} , and x 6 = 2 {\displaystyle x_{6}=2} .
Since Simpson’s Rule with n = 6 {\displaystyle n=6} in general is
Δ x 3 ( f ( x 0 ) + 4 f ( x 1 ) + 2 f ( x 2 ) + 4 f ( x 3 ) + 2 f ( x 4 ) + 4 f ( x 5 ) + f ( x 6 ) ) , {\displaystyle {\frac {\Delta x}{3}}{\big (}f(x_{0})+4f(x_{1})+2f(x_{2})+4f(x_{3})+2f(x_{4})+4f(x_{5})+f(x_{6}){\big )},}
with f ( x ) = ( x − 3 ) 5 {\displaystyle f(x)=(x-3)^{5}} , the desired approximation is
1 / 3 3 ( ( − 3 ) 5 + 4 ( 1 3 − 3 ) 5 + 2 ( 2 3 − 3 ) 5 + 4 ( − 2 ) 5 + 2 ( 4 3 − 3 ) 5 + 4 ( 5 3 − 3 ) 5 + ( − 1 ) 5 ) . {\displaystyle {\frac {1/3}{3}}{\bigg (}(-3)^{5}+4{\Big (}{\frac {1}{3}}-3{\Big )}^{5}+2{\Big (}{\frac {2}{3}}-3{\Big )}^{5}+4(-2)^{5}+2{\Big (}{\frac {4}{3}}-3{\Big )}^{5}+4{\Big (}{\frac {5}{3}}-3{\Big )}^{5}+(-1)^{5}{\bigg )}.}
and 
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in general is
, the desired approximation is
