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: 6


$$M_5= -0.691,\ \ T_5=-0.695$$

Work Step by Step

Given $$\int_{-2}^{-1}\frac{1}{x}d x,\ \ \ N=5 $$ Since $\Delta x=\dfrac{b-a}{N}=\dfrac{3}{6}=0.2 $ Therefore, the sub intervals consist of $$[-2,-1.8],[-1.8,-1.6],[-1.6,-1.4],[-1.4,-1.2],[-1.2,-1]$$ The midpoints of these sub intervals are $$\{-1.9, -1.7, -1.5, -1.3,-1.1\}$$ 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\\ &=0.2\left[ f(-1.9)+ f(-1.7)+ f(-1.5)+ f(-1.3)+ f(-1.1)\right]\\ &\approx -0.691 \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(-2)+2f(-1.8)+2f(-1.6)+2f(-1.4)+2f(-1.2)+f(-1)\right] \\ &\approx -0.695 \end{align*}
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