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: 12


$$T_6=22.216,\ \ \ \ M_6= 15.89 $$

Work Step by Step

Given $$\int_{-2}^{1} e^{x^2} d x ,\ \ \ \ N=6$$ Since $\Delta x=\dfrac{b-a}{n}=\dfrac{3}{6}=0.5$ Therefore, the sub intervals consist of $$[-2,-1.5],[-1.5,-1],[-1,-0.5],[-0.5,0],[0,0.5],[0.5,1], $$ The midpoints of these sub intervals are $$\{ -1.75, -1.25,-0.75,-0.25,0.25,0.75\}$$ 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}{5}\left[ f(-1.75)+ f(-1.25)+ f(-0.75)+ f(-0.25)+ f(0.25)+f(0.75)\right]\\ &\approx \end{align*} Now 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)+2f(x_5)+f(x_6)\right]\Delta x\\ &=\dfrac{1}{4}\left[f(-2)+2f(-1.5)+2f(-1)+2f(-0.5)+2f(0)+2f(0.5)+f(1)\right] \\ &\approx 22.216 \end{align*}
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