Answer
$$\eqalign{
& {\text{Trapezoidal Rule}} \approx 3.283 \cr
& {\text{Simpson's Rule}} \approx 3.240 \cr
& {\text{Graphing utility}} \approx 3.2413 \cr} $$
Work Step by Step
$$\eqalign{
& \int_0^2 {\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 {\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 {\sqrt {1 + {x^3}} } dx \approx \frac{1}{4}\left[ {1 + 2\sqrt {1 + \frac{1}{8}} + 2\sqrt {1 + 1} + 2\sqrt {1 + \frac{{27}}{8}} + 3} \right] \cr
& \int_0^2 {\sqrt {1 + {x^3}} } dx \approx \frac{1}{4}\left[ {1 + 2\sqrt {\frac{9}{8}} + 2\sqrt 2 + 2\sqrt {\frac{{35}}{8}} + 3} \right] \cr
& {\text{Simplifying}} \cr
& \int_0^2 {\sqrt {1 + {x^3}} } dx \approx 3.283 \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 {\sqrt {1 + {x^3}} } dx \approx \frac{{2 - 0}}{{3\left( 4 \right)}}\left[ {1 + 4\sqrt {1 + \frac{1}{8}} + 2\sqrt {1 + 1} + 4\sqrt {1 + \frac{{27}}{8}} + 3} \right] \cr
& \int_0^2 {\sqrt {1 + {x^3}} } dx \approx \frac{1}{{16}}\left[ {1 + 4\sqrt {\frac{9}{8}} + 2\sqrt 2 + 4\sqrt {\frac{{35}}{8}} + 3} \right] \cr
& \int_0^2 {\sqrt {1 + {x^3}} } dx \approx 3.240 \cr
& \cr
& {\text{Using a graphing utility we obtain}} \cr
& \int_0^2 {\sqrt {1 + {x^3}} } dx \approx 3.2413 \cr} $$