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: 9


$$M_{5}= 0.387,\ \ \ T_5= 0.384 $$

Work Step by Step

Given $$\int_{1}^{2} \ln x d x ,\ \ \ \ N=5$$ Since $\Delta x=\dfrac{b-a}{n}=\dfrac{1}{5}=0.2$ Therefore, the sub intervals consist of $$[1,1.2],[1.2,1.4],[1.4,1.6],[1.6,1.8],[1.8,2] $$ The midpoints of these sub intervals are $$\{ 1.1, 1.3,1.5,1.7,1.9\}$$ 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_4)\right]\Delta x\\ &=\frac{1}{2}\left[ f(1.1)+ f(1.3)+ f(1.5)+ f(1.7)+ f(1.9)\right]\\ &\approx 0.387 \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(1)+2f(1.2)+2f(1.4)+2f(1.6)+2f(1.8)+f(2)\right] \\ &\approx 0.384 \end{align*}
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