Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 8 - Techniques of Integration - 8.9 Numerical Integration - Exercises - Page 457: 11


$$T_{5}=0.744,\ \ M_5=0.74$$

Work Step by Step

Given $$\int_{0}^{1} e^{-x^2} d x ,\ \ \ \ N=5$$ Since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{5}=0.2$ Therefore, the sub intervals consist of $$[0,0.2],[0.2,0.4],[0.4,0.6],[0.6,0.8],[0.8,1] $$ The midpoints of these sub intervals are $$\{ 0.1, 0.3,0.5,0.7,0.9\}$$ Hence \begin{align*} M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\ M_{5}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_5)\right]\Delta x\\ &=\frac{1}{5}\left[ f(0.1)+ f(0.3)+ f(0.5)+ f(0.7)+ f(0.9)\right]\\ &\approx 0.74 \end{align*} Now find $T_5$ \begin{align*} T_{n}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+.....+2f(x_{n-1})+f(x_n)\right]\Delta x\\ T_{5}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+2f(x_4)+f(x_5)\right]\Delta x\\ &=\dfrac{1}{10}\left[f(0)+2f(0.2)+2f(0.4)+2f(0.6)+2f(0.8)+f(1)\right] \\ &\approx 0.744 \end{align*}
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