Chapter 4 - Integration - 4.6 Exercises - Page 311: 44

$$ \approx 3.2304$$

$$\eqalign{ & \int_0^1 {\frac{4}{{1 + {x^2}}}} dx \cr & {\text{For }}n = 6,{\text{ }}\Delta x = \frac{{b - a}}{n} = \frac{{1 - 0}}{6} = \frac{1}{6},{\text{ then,}} \cr & {x_0} = 0,{\text{ }}{x_1} = \frac{1}{6},{\text{ }}{x_2} = \frac{1}{3},{\text{ }}{x_3} = \frac{1}{2},{\text{ }}{x_4} = \frac{2}{3},{\text{ }}{x_5} = \frac{5}{6},{\text{ }}{x_6} = 1 \cr & {\text{*Using the Simpson's Rule }}\left( {{\text{THEOREM 4}}{\text{.19}}} \right) \cr & \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{3n}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + \cdots } \right. \cr & \left. { + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr & f\left( {{x_0}} \right) = \frac{4}{{1 + {{\left( 0 \right)}^2}}} = 4 \cr & 4f\left( {{x_1}} \right) = \frac{{16}}{{1 + {{\left( {1/6} \right)}^2}}} = \frac{{576}}{{37}} \cr & 2f\left( {{x_2}} \right) = \frac{8}{{1 + {{\left( {1/3} \right)}^2}}} = \frac{{36}}{5} \cr & 4f\left( {{x_3}} \right) = \frac{{16}}{{1 + {{\left( {1/3} \right)}^2}}} = \frac{{72}}{5} \cr & 2f\left( {{x_4}} \right) = \frac{8}{{1 + {{\left( {2/3} \right)}^2}}} = \frac{{72}}{{13}} \cr & 4f\left( {{x_5}} \right) = \frac{{16}}{{1 + {{\left( {5/6} \right)}^2}}} = \frac{{576}}{{61}} \cr & f\left( {{x_6}} \right) = \frac{4}{{1 + {{\left( 1 \right)}^2}}} = 2 \cr & {\text{Therefore,}} \cr & \int_0^1 {\frac{4}{{1 + {x^2}}}} dx \approx \frac{1}{{3\left( 6 \right)}}\left[ {4 + \frac{{576}}{{37}} + \frac{{36}}{5} + \frac{{72}}{5} + \frac{{72}}{{13}} + \frac{{576}}{{61}} + 2} \right] \cr & \int_0^1 {\frac{4}{{1 + {x^2}}}} dx \approx 3.2304 \cr} $$