Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - Chapter Review - Review Exercises - Page 418: 74

Answer

$$ \int_{0}^{2} x e^{-x^{2}} d x $$ Here $a=0, b=2,$ and $n=4,$ with $(b-a) / n=(2-0) / 4=1/2 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1/2 , x_{2}=1 , x_{3}=3/2 ,$ 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=0, f(x)=x e^{-x^{2}}\\ &\begin{array}{l|l|l} \hline i & x_{i} & f\left(x_{i}\right) \\ \hline 0 & 0 & 0 \\ 1 & 0.5 & 0.3894 \\ 2 & 1 & 0.3679 \\ 3 & 1.5 & 0.1581 \\ 4 & 2 & 0.0366 \\ \hline \end{array} \end{aligned} Simpson's Rule: $$ \begin{array}{l} \int_{0}^{2} x e^{-x^{2}} d x \\ \approx \frac{2-0}{3(4)}[0+4(0.3894)+2(0.3679) \\ \quad+4(0.1581)+0.0366] \\ \approx 0.4937 \end{array} $$ The answer is close to the exact value obtained in Exercise 70, which is approximately 0.4908. Where the exact value can be obtained by: \begin{aligned} \int_{0}^{2} x e^{-x^{2}} d x &=-\frac{1}{2} \int_{0}^{2} e^{-x^{2}}(-2 x d x) \\ &=-\left.\frac{1}{2} e^{-x^{2}}\right|_{0} ^{2} \\ &=-\frac{1}{2} e^{-4}+\frac{1}{2} e^{0} \\ &=\frac{1}{2}\left(1-e^{-4}\right) \\ & \approx 0.4908 \end{aligned}

Work Step by Step

$$ \int_{0}^{2} x e^{-x^{2}} d x $$ Here $a=0, b=2,$ and $n=4,$ with $(b-a) / n=(2-0) / 4=1/2 $ as the altitude of each trapezoid. Then $x_{0}=0, x_{1}=1/2 , x_{2}=1 , x_{3}=3/2 ,$ 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=0, f(x)=x e^{-x^{2}}\\ &\begin{array}{l|l|l} \hline i & x_{i} & f\left(x_{i}\right) \\ \hline 0 & 0 & 0 \\ 1 & 0.5 & 0.3894 \\ 2 & 1 & 0.3679 \\ 3 & 1.5 & 0.1581 \\ 4 & 2 & 0.0366 \\ \hline \end{array} \end{aligned} Simpson's Rule: $$ \begin{array}{l} \int_{0}^{2} x e^{-x^{2}} d x \\ \approx \frac{2-0}{3(4)}[0+4(0.3894)+2(0.3679) \\ \quad+4(0.1581)+0.0366] \\ \approx 0.4937 \end{array} $$ The answer is close to the exact value obtained in Exercise 70, which is approximately 0.4908. Where the exact value can be obtained by: \begin{aligned} \int_{0}^{2} x e^{-x^{2}} d x &=-\frac{1}{2} \int_{0}^{2} e^{-x^{2}}(-2 x d x) \\ &=-\left.\frac{1}{2} e^{-x^{2}}\right|_{0} ^{2} \\ &=-\frac{1}{2} e^{-4}+\frac{1}{2} e^{0} \\ &=\frac{1}{2}\left(1-e^{-4}\right) \\ & \approx 0.4908 \end{aligned}
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