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: 2


$$ M_4=5.333,\ \ \ T_4=5.146 $$

Work Step by Step

Given $$\int_{0}^{4} \sqrt{x} d x $$ Since $\Delta x=\dfrac{b-a}{n}=\dfrac{4}{4}=1$ Therefore, the sub intervals consist of $$[0,1],[1,2],[2,3],[3,4] $$ The midpoints of these sub intervals are $$\{0.5,1.5,2.5,3.5\}$$ 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\\ &=\left[ f(0.5)+ f(1.5)+ f(2.5)+ f(3.5)\right]\\ &\approx 5.33333 \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}&= \frac{1}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+2f(x_{3})+f(x_4)\right]\Delta x\\ &=\dfrac{1}{2}\left[f(0)+2f(1)+2f(2)+2f(3)+f(4)\right] \\ &\approx 5.1462 \end{align*}
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