Chapter 7 - Integration - 7.6 Numerical Integration - 7.6 Exercises - Page 413: 10

$$ \int_{0}^{4} x \sqrt{2 x^{2}+1} d x $$ Here $a=0, b=4,$ and $n=4,$ with $(b-a) / n=(4-0) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1 , x_{2}=2 , x_{3}=3 ,$ and $x_{4}=4 .$ Now find the corresponding function values. The work can be organized into a table, as follows. $$ \begin{aligned} &n=4, b=4, a=0, f(x)=x \sqrt{2 x^{2}+1}\\ &\begin{array}{c|c|c} \hline i & x_{i} & f\left(x_{i}\right) \\ \hline 0 & 0 & 0 \\ 1 & 1 & \sqrt{3} \\ 2 & 2 & 6 \\ 3 & 3 & 3 \sqrt{19} \\ 4 & 4 & 4 \sqrt{33} \\ \hline \end{array} \end{aligned} $$ (a) the trapezoidal rule: $$ \begin{array}{l} \int_{0}^{4} x \sqrt{2 x^{2}+1} d x \\ \quad \approx \frac{4-0}{4}\left[\frac{1}{2}(0)+\sqrt{3}+6+3 \sqrt{19}+\frac{1}{2}(4 \sqrt{33})\right] \\ \quad =1(\sqrt{3}+6+3 \sqrt{19}+2 \sqrt{33}) \\ \quad \approx 32.30 \end{array} $$ (b) Simpson’s Rule: $$ \begin{array}{l} \int_{0}^{4} x \sqrt{2 x^{2}+1} d x \\ \quad \approx \frac{4-0}{3(4)}[0+4(\sqrt{3})+2(6) \\ \quad\quad+4(3 \sqrt{19})+4 \sqrt{33}] \\ \quad =\frac{1}{3}(4 \sqrt{3}+12+12 \sqrt{19}+4 \sqrt{33}) \\ \quad \approx 31.40 \end{array} $$ (c) Exact value: $$ \begin{aligned} \int_{0}^{4} x \sqrt{2 x^{2}+1} d x& =\int_{0}^{4} \frac{1}{4} \sqrt{2 x^{2}+1} (4xd x)\\ &=\int_{0}^{4} \frac{1}{4} \sqrt{2 x^{2}+1} d (2 x^{2})\\ &=\left.\frac{\left(2 x^{2}+1\right)^{3 / 2}}{6}\right|_{0} ^{4} \\ &=\frac{33^{3 / 2}-1}{6}\\ & \approx 31.43 \end{aligned} $$

$$ \int_{0}^{4} x \sqrt{2 x^{2}+1} d x $$ Here $a=0, b=4,$ and $n=4,$ with $(b-a) / n=(4-0) / 4=1 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1 , x_{2}=2 , x_{3}=3 ,$ and $x_{4}=4 .$ Now find the corresponding function values. The work can be organized into a table, as follows. $$ \begin{aligned} &n=4, b=4, a=0, f(x)=x \sqrt{2 x^{2}+1}\\ &\begin{array}{c|c|c} \hline i & x_{i} & f\left(x_{i}\right) \\ \hline 0 & 0 & 0 \\ 1 & 1 & \sqrt{3} \\ 2 & 2 & 6 \\ 3 & 3 & 3 \sqrt{19} \\ 4 & 4 & 4 \sqrt{33} \\ \hline \end{array} \end{aligned} $$ (a) the trapezoidal rule: $$ \begin{array}{l} \int_{0}^{4} x \sqrt{2 x^{2}+1} d x \\ \quad \approx \frac{4-0}{4}\left[\frac{1}{2}(0)+\sqrt{3}+6+3 \sqrt{19}+\frac{1}{2}(4 \sqrt{33})\right] \\ \quad =1(\sqrt{3}+6+3 \sqrt{19}+2 \sqrt{33}) \\ \quad \approx 32.30 \end{array} $$ (b) Simpson’s Rule: $$ \begin{array}{l} \int_{0}^{4} x \sqrt{2 x^{2}+1} d x \\ \quad \approx \frac{4-0}{3(4)}[0+4(\sqrt{3})+2(6) \\ \quad\quad+4(3 \sqrt{19})+4 \sqrt{33}] \\ \quad =\frac{1}{3}(4 \sqrt{3}+12+12 \sqrt{19}+4 \sqrt{33}) \\ \quad \approx 31.40 \end{array} $$ (c) Exact value: $$ \begin{aligned} \int_{0}^{4} x \sqrt{2 x^{2}+1} d x& =\int_{0}^{4} \frac{1}{4} \sqrt{2 x^{2}+1} (4xd x)\\ &=\int_{0}^{4} \frac{1}{4} \sqrt{2 x^{2}+1} d (2 x^{2})\\ &=\left.\frac{\left(2 x^{2}+1\right)^{3 / 2}}{6}\right|_{0} ^{4} \\ &=\frac{33^{3 / 2}-1}{6}\\ & \approx 31.43 \end{aligned} $$