The value of \(\int_0^6{\frac{dx}{1+x^2}}\) by Simpson's \(\frac{1}{3}\) rule is

Concept:

Simpson’s one-third rule:

\(\mathop \smallint \limits_{{x_0}}^{{x_0} + nh} f\left( x \right)dx = \frac{h}{3}\left[ {\left( {{y_0} + {y_n}} \right) + 4\left( {{y_1} + {y_3} + - - - - {y_{n - 1}}} \right)} \right] + 2\left( {{y_2} + {y_4} + - - - - {y_{n - 2}}} \right)\)

The given interval must be divided into even number of equal sub intervals.

\(h=\frac{b-a}{n}=\frac{6-0}{6}=1\)

Analysis:

\(\displaystyle \int^6_0 \frac{dx}{1+x^2}=\frac{h}{3}\left[(y_0+y_6)+4\times(y_1+y_3+y_5)+2\times(y_2+y_4)\right]\)

\(=\frac{1}{3}\left[\left(1+\frac{1}{37}\right)+4\left(\frac{1}{2}+\frac{1}{10}+\frac{1}{26}\right)+2\left(\frac{1}{5}+\frac{1}{17}\right)\right]\)

\(=\frac{1}{3}\left(\frac{38}{37}+4\times\frac{83}{130}+2\times\frac{22}{85}\right)\)

\(=\frac{1}{3}\left(\frac{38}{37}\times\frac{166}{65}\times\frac{44}{85}\right)\)

= 1.36617 ≈ 1.3662