Concept:

Simpson’s \(\frac{1}{3}\) Rule: It is numerical integration method which follows the three-point Newton quadrature rule and is given by,

\(\mathop \smallint \limits_{\rm{a}}^{\rm{b}} {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = {\rm{\;}}\frac{{\rm{h}}}{3}\left[ {\left( {{{\rm{y}}_0} + {{\rm{y}}_{\rm{n}}}} \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3} + \ldots } \right) + 2\left( {{{\rm{y}}_2} + {{\rm{y}}_4} + \ldots } \right)} \right]{\rm{\;}}\)

Where \({{\rm{x}}_0}{\rm{\;to\;}}{{\rm{x}}_{\rm{n}}}\) are equally spaced n number (must be even number) of integration points, \({{\rm{y}}_0}{\rm{\;to\;}}{{\rm{y}}_{\rm{n}}}\) are functional value at those points and h is the interval length or step length.

Calculation:

Given, a = 0, b = 4, step length, h = 1 and \({\rm{f}}\left( {\rm{x}} \right) = {\rm{\;}}\left( {{{\rm{x}}^4} + 10} \right)\)

Hence, the integration points and functional values at those points are listed in the following table:

∴ Integration by Simpson’s 1/3 Rule = \({\rm{I'}} = \frac{{\rm{h}}}{3}\left[ {\left( {{{\rm{y}}_0} + {{\rm{y}}_4}} \right) + 4\left( {{{\rm{y}}_1} + {{\rm{y}}_3}} \right) + 2\left( {{{\rm{y}}_2}} \right)} \right]\)

\(\therefore {\rm{I'}} = \frac{1}{3}\left[ {\left( {10 + 266} \right) + 4\left( {11 + 91} \right) + 2\left( {26} \right)} \right] = \frac{{736}}{3} = 245.33\)

∴ The Exact integration, \({\rm{I}} = \mathop \smallint \limits_0^4 \left( {{{\rm{x}}^4} + 10} \right){\rm{dx}} = \left[ {\frac{{{{\rm{x}}^5}}}{5} + 10{\rm{x}}} \right]_0^4 = 244.8\)

∴ The Error = \({\rm{\xi }} = {\rm{I'}} - {\rm{I}} = 245.33 - 244.8 = 0.53\)

Trick:

The absolute error by Simpson’s 1/3 Rule is given by,

\({\rm{\xi }} = \frac{{{{\rm{h}}^4}}}{{180}}\left( {{\rm{b}} - {\rm{a}}} \right){\rm{max}}\left[ {{{\rm{f}}^4}\left( {\rm{x}} \right)} \right]\)

where, x is some number between a and b and \({{\rm{f}}^4}\left( {\rm{x}} \right)\) is the fourth derivative of the function.

\({\rm{Given}},{\rm{\;\;f}}\left( {\rm{x}} \right) = {\rm{\;}}\left( {{{\rm{x}}^4} + 10} \right){\rm{\;\;}}\therefore {\rm{\;}}{{\rm{f}}^4}\left( {\rm{x}} \right) = 24{\rm{\;\;}}\therefore \max \left[ {{{\rm{f}}^4}\left( {\rm{x}} \right)} \right] = 24{\rm{\;\;}}\)

\(\therefore {\rm{\xi }} = \frac{{{1^4}}}{{180}}\left( {4 - 0} \right) \times 24 = 0.53\)