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: 5

Answer

$$ M_6= 1.37,\ \ \ T_6=1.40 $$

Work Step by Step

Given $$\int_{1}^{4}\frac{1}{x}d x,\ \ \ N=6 $$ Since $\Delta x=\dfrac{b-a}{N}=\dfrac{3}{6}=0.5 $ Therefore, the sub intervals consist of $$[1,1.5],[1.5,2],[2,2.5],[2.5,3],[3,3.5],[3.5,4] $$ The midpoints of these sub intervals are $$\{ 1.25, 1.75, 2.25, 2.75, 3.25, 3.75\}$$ Hence \begin{align*} M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\ M_{6}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_6)\right]\Delta x\\ &=\frac{1}{2}\left[ f(1.25)+ f(1.75)+ f(2.25)+ f(2.75)+ f(3.25)+f(3.75)\right]\\ &\approx 1.37 \end{align*} Now, we find $T_6$ \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_{6}&=\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}{4}\left[f(1)+2f(1.5)+2f(2)+2f(2.5)+2f(3)+2f(3.5)+f(4)\right] \\ &\approx 1.40 \end{align*}
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