Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 7 - Techniques of Integration - 7.7 Approximate Integration - 7.7 Exercises - Page 564: 15

Answer

$$ \int_{0}^{1}\frac{x^{2}}{1+x^{4}} dx ,\quad n=10 $$ Here $$ f(x)=\frac{x^{2}}{1+x^{4}}, \quad \Delta x=\frac{1-0}{10}=\frac{1}{10} $$ (a) the Trapezoidal Rule $$ \begin{aligned} T_{10} &=\frac{1}{10 \cdot 2}\{f(0)+2[f(0.1+f(0.2)+\cdots+f(0.9)]+f(1)\} \\ & \approx 0.243747 \end{aligned} $$ (b) The midpoints Rule is given by: $$ \begin{aligned} M_{10}&=\frac{1}{10}[f(0.05)+f(0.15)+\cdots+f(0.85)+f(0.95)] \\ &\approx 0.243748 \end{aligned} $$ (c) Simpson’s Rule: $$ \begin{aligned} S_{10}&=\frac{1}{10 \cdot 3}[f(0)+4 f(0.1)+2 f(0.2)+4 f(0.3)+2 f(0.4)+4 f(0.5) \\ & \quad\quad +2 f(0.6) +4 f(0.7)+2 f(0.8)+4 f(0.9)+f(1)] \\& \approx 0.243751 \end{aligned} $$

Work Step by Step

$$ \int_{0}^{1}\frac{x^{2}}{1+x^{4}} dx ,\quad n=10 $$ Here $$ f(x)=\frac{x^{2}}{1+x^{4}}, \quad \Delta x=\frac{1-0}{10}=\frac{1}{10} $$ and so the Trapezoidal Rule can be given by: $$ \int_{a}^{b} f(x) d x \approx T_{n}=\frac{\Delta x}{2}\left[f\left(x_{0}\right)+2 f\left(x_{1}\right)+2 f\left(x_{2}\right)+\cdots+2 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] $$ Therefore, (a) the Trapezoidal Rule $$ \begin{aligned} T_{10} &=\frac{1}{10 \cdot 2}\{f(0)+2[f(0.1+f(0.2)+\cdots+f(0.9)]+f(1)\} \\ & \approx 0.243747 \end{aligned} $$ (b) The midpoints Rule is given by: $$ \int_{a}^{b} f(x) d x \approx M_{n}=\Delta x\left[f\left(\bar{x}_{1}\right)+f\left(\bar{x}_{2}\right)+\cdots+f\left(\bar{x}_{n}\right)\right] $$ where $$ \bar{x}_{i}=\frac{1}{2}\left(x_{i-1}+x_{i}\right)=\text { midpoint of }\left[x_{i-1}, x_{i}\right] $$ So, we have: $$ \begin{aligned} M_{10}&=\frac{1}{10}[f(0.05)+f(0.15)+\cdots+f(0.85)+f(0.95)] \\ &\approx 0.243748 \end{aligned} $$ (c) Simpson’s Rule: $$ \begin{aligned} \int_{a}^{b} f(x) d x & \approx S_{n} \\ &=\frac{\Delta x}{3}\left[f\left(x_{0}\right)+4 f\left(x_{1}\right)+2 f\left(x_{2}\right)+4 f\left(x_{3}\right)\\ +\cdots\right. \left.+2 f\left(x_{n-2}\right)+4 f\left(x_{n-1}\right)+f\left(x_{n}\right)\right] \end{aligned} $$ So, we obtain: $$ \begin{aligned} S_{10}&=\frac{1}{10 \cdot 3}[f(0)+4 f(0.1)+2 f(0.2)+4 f(0.3)+2 f(0.4)+4 f(0.5) \\ & \quad\quad +2 f(0.6) +4 f(0.7)+2 f(0.8)+4 f(0.9)+f(1)] \\& \approx 0.243751 \end{aligned} $$
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