Answer
$$\eqalign{
& {\text{Trapezoidal Rule}} \approx 20.2 \cr
& {\text{Simpson's Rule}} \approx 19.47 \cr
& {\text{Graphing utility}} \approx 19.3132 \cr} $$
Work Step by Step
$$\eqalign{
& \int_1^5 {\frac{{12}}{x}} dx \cr
& {\text{For }}n = 4,{\text{ }}\Delta x = \frac{{b - a}}{n} = \frac{{5 - 1}}{4} = 1,{\text{ then,}} \cr
& {x_0} = 1,{\text{ }}{x_1} = 2,{\text{ }}{x_2} = 3,{\text{ }}{x_3} = 4,{\text{ }}{x_4} = 5 \cr
& \cr
& {\text{*Using the trapezoidal Rule }}\left( {{\text{THEOREM 4}}{\text{.17}}} \right) \cr
& \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{2n}}\left[ {f\left( {{x_0}} \right) + 2f\left( {{x_1}} \right) + \cdots 2f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr
& \int_1^5 {\frac{{12}}{x}} dx \approx \frac{1}{2}\left[ {\frac{{12}}{1} + \frac{{2\left( {12} \right)}}{2} + \frac{{2\left( {12} \right)}}{3} + \frac{{2\left( {12} \right)}}{4} + \frac{{12}}{5}} \right] \cr
& {\text{Simplifying}} \cr
& \int_1^5 {\frac{{12}}{x}} dx \approx \frac{1}{2}\left( {\frac{{202}}{5}} \right) \cr
& \int_1^5 {\frac{{12}}{x}} dx \approx 20.2 \cr
& \cr
& {\text{*Using Simpson's Rule }}\left( {{\text{THEOREM 4}}{\text{.19}}} \right) \cr
& \int_a^b {f\left( x \right)} dx \approx \frac{{b - a}}{{3n}}\left[ {f\left( {{x_0}} \right) + 4f\left( {{x_1}} \right) + 2f\left( {{x_2}} \right) + 4f\left( {{x_3}} \right) + \cdots } \right. \cr
& \left. {{\text{ }} + 4f\left( {{x_{n - 1}}} \right) + f\left( {{x_n}} \right)} \right] \cr
& \int_1^5 {\frac{{12}}{x}} dx \approx \frac{1}{3}\left[ {\frac{{12}}{1} + \frac{{4\left( {12} \right)}}{2} + \frac{{2\left( {12} \right)}}{3} + \frac{{4\left( {12} \right)}}{4} + \frac{{12}}{5}} \right] \cr
& {\text{Simplifying}} \cr
& \int_1^5 {\frac{{12}}{x}} dx \approx 19.47 \cr
& \cr
& {\text{Using a graphing utility we obtain}} \cr
& \int_1^5 {\frac{{12}}{x}} dx \approx 19.3132 \cr
& \cr
& {\text{Trapezoidal Rule}} \approx 20.2 \cr
& {\text{Simpson's Rule}} \approx 19.47 \cr
& {\text{Graphing utility}} \approx 19.3132 \cr} $$
