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: 1


$$T_4=2.75,\ \ M_4= 2.625 $$

Work Step by Step

Given $$\int_{0}^{2} x^{2} d x $$ We know that $\Delta x=\dfrac{b-a}{n}=\dfrac{2}{4}=\frac{1}{2}$ Therefore, the sub intervals consist of $$[0,0.5],[0.5,1],[1,1.5],[1.5,2] $$ The midpoints of these sub intervals are $$\{0.25,0.75,1.25,1.75\}$$ Hence \begin{align*} M_{n}&= \sum_{i=1}^{n}f(m_i)\Delta x\\ M_{4}&= \left[ f(m_1)+ f(m_2)+ ..............+f(m_4)\right]\Delta x\\ &=\frac{1}{2}\left[ f(0.25)+ f(0.75)+ f(1.25)+ f(1.75)+ f(9)\right]\\ &\approx 2.625 \end{align*} To find $T_4$ \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_{4}&=\dfrac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+f(x_4)\right]\Delta x\\ &=\dfrac{1}{4}\left[f(0)+2f(0.5)+2f(1)+2f(1.5)+f(2)\right] \\ &\approx 2.75 \end{align*}
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