Chapter 7 - Integration - 7.6 Numerical Integration - 7.6 Exercises - Page 413: 5

$$ \int_{-1}^{2}\left(2 x^{3}+1\right) d x $$ Here $a=-1, b=2,$ and $n=4,$ with $(b-a) / n=(2+1) / 4=3 / 4 $ as the altitude of each trapezoid. Then $x_{0}=-1, x_{1}=-1 / 4, x_{2}=1/2, x_{3}=5 / 4,$ and $x_{4}=2 .$ Now find the corresponding function values. The work can be organized into a table, as follows. $$ \begin{aligned} &n=4, b=2, a=-1, f(x)=2 x^{3}+\\ &\begin{array}{c|c|c} \hline i & x_{i} & f(x) \\ \hline 0 & -1 & -1 \\ 1 & -\frac{1}{4} & \frac{31}{32} \\ 2 & \frac{1}{2} & \frac{5}{4} \\ 3 & \frac{5}{4} & \frac{157}{32} \\ 4 & 2 & 17 \\ \hline \end{array} \end{aligned} $$ (a) the trapezoidal rule: $$ \begin{array}{l} \int_{-1}^{2}\left(2 x^{3}+1\right) d x \\ \quad \approx \frac{2-(-1)}{4}\left[\frac{1}{2}(-1)+\frac{31}{32}+\frac{5}{4}\right. \\ \left.\quad\quad +\frac{157}{32}+\frac{1}{2}(17)\right] \\ \quad=0.75(15.125) \\ \quad \approx 11.34 \end{array} $$ (b) Simpson’s Rule: $$ \begin{array}{l} \int_{-1}^{2}\left(2 x^{3}+1\right) d x \\ \quad \approx \frac{2-(-1)}{3(4)}\left[-1+4\left(\frac{31}{32}\right)+2\left(\frac{5}{4}\right)+4\left(\frac{157}{32}\right)+17\right] \\ \quad =\frac{1}{4}(42) \\ \quad =10.5 \end{array} $$ (c) Exact value: $$ \begin{array}{l} \int_{-1}^{2}\left(2 x^{3}+1\right) d x \\ \quad=\left.\left(\frac{x^{4}}{2}+x\right)\right|_{-1} ^{2} \\ \quad=(8+2)-\left(\frac{1}{2}-1\right) \\ \quad=\frac{21}{2} \\ \quad=10.5 \end{array} $$

$$ \int_{-1}^{2}\left(2 x^{3}+1\right) d x $$ Here $a=-1, b=2,$ and $n=4,$ with $(b-a) / n=(2+1) / 4=3 / 4 $ as the altitude of each trapezoid. Then $x_{0}=-1, x_{1}=-1 / 4, x_{2}=1/2, x_{3}=5 / 4,$ and $x_{4}=2 .$ Now find the corresponding function values. The work can be organized into a table, as follows. $$ \begin{aligned} &n=4, b=2, a=-1, f(x)=2 x^{3}+1\\ &\begin{array}{c|c|c} \hline i & x_{i} & f(x) \\ \hline 0 & -1 & -1 \\ 1 & -\frac{1}{4} & \frac{31}{32} \\ 2 & \frac{1}{2} & \frac{5}{4} \\ 3 & \frac{5}{4} & \frac{157}{32} \\ 4 & 2 & 17 \\ \hline \end{array} \end{aligned} $$ (a) the trapezoidal rule: $$ \begin{array}{l} \int_{-1}^{2}\left(2 x^{3}+1\right) d x \\ \quad \approx \frac{2-(-1)}{4}\left[\frac{1}{2}(-1)+\frac{31}{32}+\frac{5}{4}\right. \\ \left.\quad\quad +\frac{157}{32}+\frac{1}{2}(17)\right] \\ \quad=0.75(15.125) \\ \quad \approx 11.34 \end{array} $$ (b) Simpson’s Rule: $$ \begin{array}{l} \int_{-1}^{2}\left(2 x^{3}+1\right) d x \\ \quad \approx \frac{2-(-1)}{3(4)}\left[-1+4\left(\frac{31}{32}\right)+2\left(\frac{5}{4}\right)+4\left(\frac{157}{32}\right)+17\right] \\ \quad =\frac{1}{4}(42) \\ \quad =10.5 \end{array} $$ (c) Exact value: $$ \begin{array}{l} \int_{-1}^{2}\left(2 x^{3}+1\right) d x \\ \quad=\left.\left(\frac{x^{4}}{2}+x\right)\right|_{-1} ^{2} \\ \quad=(8+2)-\left(\frac{1}{2}-1\right) \\ \quad=\frac{21}{2} \\ \quad=10.5 \end{array} $$