Chapter 4 - Integration - 4.6 Exercises - Page 310: 12

$$\eqalign{ & {\text{Trapezoidal Rule}} \approx 1.3973 \cr & {\text{Simpson's Rule}} \approx 1.4051 \cr & {\text{Graphing utility}} \approx 1.4021 \cr} $$

$$\eqalign{ & \int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}} dx \cr & {\text{*Using the trapezoidal Rule }}\left( {{\text{THEOREM 4}}{\text{.17}}} \right) \cr & \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{2n}}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + \cdots 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr & {\text{For }}n = 4,{\text{ }}\Delta x = \frac{{b - a}}{n} = \frac{{2 - 0}}{4} = \frac{1}{2},{\text{ then,}} \cr & {x_0} = 0,{\text{ }}{x_1} = \frac{1}{2},{\text{ }}{x_2}{\text{ = 1, }}{x_3} = \frac{3}{2},{\text{ }}{x_4} = 2 \cr & \int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}} dx \approx \frac{{2 - 0}}{{2\left( 4 \right)}}\left[ {f\left( 0 \right) + 2f\left( {\frac{1}{2}} \right) + 2f\left( {\frac{3}{2}} \right) + f\left( 2 \right)} \right] \cr & \int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}} dx \approx \frac{1}{4}\left[ {1 + \frac{2}{{\sqrt {1 + 1/8} }} + \frac{2}{{\sqrt {1 + 1} }} + \frac{2}{{\sqrt {1 + 27/8} }} + \frac{1}{3}} \right] \cr & {\text{Simplifying}} \cr & \int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}} dx \approx 1.3973 \cr & \cr & {\text{*Using the Simpson's Rule }}\left( {{\text{THEOREM 4}}{\text{.19}}} \right) \cr & \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{3n}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + \cdots } \right. \cr & \left. { + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr & \int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}} dx \approx \frac{{2 - 0}}{{3\left( 4 \right)}}\left[ {1 + \frac{4}{{\sqrt {1 + 1/8} }} + \frac{2}{{\sqrt {1 + 1} }} + \frac{4}{{\sqrt {1 + 27/8} }} + \frac{1}{3}} \right] \cr & {\text{Simplifying}} \cr & \int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}} dx \approx 1.40519 \cr & \cr & {\text{Using a graphing utility we obtain}} \cr & \int_0^2 {\frac{1}{{\sqrt {1 + {x^3}} }}} dx \approx 1.40218 \cr} $$