Chapter 7 - Integration - Chapter Review - Review Exercises - Page 418: 68

$$ \int_{2}^{10} \frac{x d x}{x-1} $$ Here $a=2, b=10,$ and $n=4,$ with $(b-a) / n=(10-2) / 4=2 $ as the altitude of each trapezoid. Then $x_{0}=2, x_{1}=4 , x_{2}=6 , x_{3}=8 ,$ and $x_{4}=10 .$ Now find the corresponding function values. The work can be organized into a table, as follows. $$ \begin{aligned} &n=4, b=10, a=2, f(x)=\frac{x}{x-1}\\ &\begin{array}{c|c|c} \hline i & x_{i} & f\left(x_{i}\right) \\ \hline 0 & 2 & 2 \\ 1 & 4 & \frac{4}{3} \\ 2 & 6 & \frac{6}{5} \\ 3 & 8 & \frac{8}{7} \\ 4 & 10 & \frac{10}{9} \\ \hline \end{array} \end{aligned} $$ The trapezoidal rule: $$ \begin{aligned} \int_{2}^{10} \frac{x}{x-1} d x& \\ &\approx \frac{10-2}{4}\left[\frac{1}{2}(2)+\frac{4}{3}+\frac{6}{5}+\frac{8}{7}+\frac{1}{2}\left(\frac{10}{9}\right)\right]. \\ &\approx 10.46 \end{aligned} $$ Exact value: Let $u=x-1,$ so that $d u=d x$ and $x=u+1$ Then $$ \begin{aligned} \int_{2}^{10} \frac{x}{x-1} d x&=\int_{1}^{9} \frac{u+1}{u} d u\\ &=\int_{1}^{9}\left(1+\frac{1}{u}\right) d u \\ &=\int_{1}^{9} d u+\int_{1}^{9} \frac{1}{u} d u\\ &=\left.u\right|_{1} ^{9}+\left.\ln |u|\right|_{1} ^{9}\\ &=(9-1)+(\ln 9-\ln 1)\\ &=8+\ln 9\\ & \approx 10.20 \end{aligned} $$

$$ \int_{2}^{10} \frac{x d x}{x-1} $$ Here $a=2, b=10,$ and $n=4,$ with $(b-a) / n=(10-2) / 4=2 $ as the altitude of each trapezoid. Then $x_{0}=2, x_{1}=4 , x_{2}=6 , x_{3}=8 ,$ and $x_{4}=10 .$ Now find the corresponding function values. The work can be organized into a table, as follows. $$ \begin{aligned} &n=4, b=10, a=2, f(x)=\frac{x}{x-1}\\ &\begin{array}{c|c|c} \hline i & x_{i} & f\left(x_{i}\right) \\ \hline 0 & 2 & 2 \\ 1 & 4 & \frac{4}{3} \\ 2 & 6 & \frac{6}{5} \\ 3 & 8 & \frac{8}{7} \\ 4 & 10 & \frac{10}{9} \\ \hline \end{array} \end{aligned} $$ The trapezoidal rule: $$ \begin{aligned} \int_{2}^{10} \frac{x}{x-1} d x& \\ &\approx \frac{10-2}{4}\left[\frac{1}{2}(2)+\frac{4}{3}+\frac{6}{5}+\frac{8}{7}+\frac{1}{2}\left(\frac{10}{9}\right)\right]. \\ &\approx 10.46 \end{aligned} $$ Exact value: Let $u=x-1,$ so that $d u=d x$ and $x=u+1$ Then $$ \begin{aligned} \int_{2}^{10} \frac{x}{x-1} d x&=\int_{1}^{9} \frac{u+1}{u} d u\\ &=\int_{1}^{9}\left(1+\frac{1}{u}\right) d u \\ &=\int_{1}^{9} d u+\int_{1}^{9} \frac{1}{u} d u\\ &=\left.u\right|_{1} ^{9}+\left.\ln |u|\right|_{1} ^{9}\\ &=(9-1)+(\ln 9-\ln 1)\\ &=8+\ln 9\\ & \approx 10.20 \end{aligned} $$