Answer
$$ \approx 3.141614$$
Work Step by Step
$$\eqalign{
& \int_0^{1/2} {\frac{6}{{\sqrt {1 - {x^2}} }}} dx \cr
& {\text{For }}n = 6,{\text{ }}\Delta x = \frac{{b - a}}{n} = \frac{{1/2 - 0}}{6} = \frac{1}{{12}},{\text{ then,}} \cr
& {x_0} = 0,{\text{ }}{x_1} = \frac{1}{{12}},{\text{ }}{x_2} = \frac{1}{6},{\text{ }}{x_3} = \frac{1}{4},{\text{ }}{x_4} = \frac{1}{3},{\text{ }}{x_5} = \frac{5}{{12}},{\text{ }}{x_6} = \frac{1}{2} \cr
& {\text{*Using the Simpson's Rule }}\left( {{\text{THEOREM 4}}{\text{.19}}} \right) \cr
& \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{3n}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + \cdots } \right. \cr
& \left. { + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr
& f\left( {{x_0}} \right) = \frac{6}{{\sqrt {1 - {{\left( 0 \right)}^2}} }} = 6 \cr
& 4f\left( {{x_1}} \right) = \frac{{24}}{{\sqrt {1 - {{\left( {1/12} \right)}^2}} }} = \frac{{288}}{{\sqrt {143} }} \cr
& 2f\left( {{x_2}} \right) = \frac{{12}}{{\sqrt {1 - {{\left( {1/6} \right)}^2}} }} = \frac{{72}}{{\sqrt {35} }} \cr
& 4f\left( {{x_3}} \right) = \frac{{24}}{{\sqrt {1 - {{\left( {1/4} \right)}^2}} }} = \frac{{32\sqrt 3 }}{{\sqrt 5 }} \cr
& 2f\left( {{x_4}} \right) = \frac{{12}}{{\sqrt {1 - {{\left( {1/3} \right)}^2}} }} = 9\sqrt 2 \cr
& 4f\left( {{x_5}} \right) = \frac{{24}}{{\sqrt {1 - {{\left( {5/12} \right)}^2}} }} = \frac{{288}}{{\sqrt {119} }} \cr
& f\left( {{x_6}} \right) = \frac{6}{{\sqrt {1 - {{\left( {1/2} \right)}^2}} }} = 4\sqrt 3 \cr
& {\text{Therefore,}} \cr
& \int_0^{1/2} {\frac{6}{{\sqrt {1 - {x^2}} }}} dx \approx \cr
& \approx \frac{1}{{36}}\left[ {6 + \frac{{288}}{{\sqrt {143} }} + \frac{{72}}{{\sqrt {35} }} + \frac{{32\sqrt 3 }}{{\sqrt 5 }} + 9\sqrt 2 + \frac{{288}}{{\sqrt {119} }} + 4\sqrt 3 } \right] \cr
& \approx 3.141614 \cr} $$
