Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 5: Integrals - Section 5.3 - The Definite Integral - Exercises 5.3 - Page 276: 74

Answer

$\frac{1}{4}+\frac{2}{5}\leq\int{0.5}{0}\frac{1}{1+(x)^2}dx+\int^{1}_{0.5}\frac{1}{1+(x)^2}dx\leq\frac{1}{2}+\frac{2}{5}=\frac{13}{20}\leq\int_{0}^{1}\frac{1}{1+(x)^2}dx\leq\frac{9}{10}$

Work Step by Step

--On[0,0,5], -max f=$\frac{1}{1+(0)^2}=1$, -min f=$\frac{1}{1+(0.5)^2}=0.8$, therefore (0.5-0) min $f\leq\int^{0.5}_{0}f(x)dx\leq(0.5-0) $ -max f=$\frac{2}{5}\leq\int^{0.5}_{0}$ $\frac{1}{1+(x)^2}dx\leq\frac{1}{2}$ --on [0,5,1] -max f=$\frac{1}{1+(0.5)^2}=0.8$ and - min f=$\frac{1}{1+(1)^2}=0.5$ --therefore (1-0.5) - min $\leq\int^{1}_{0.5}$ $\frac{1}{1+(x)^2}dx\leq(1-0.5)$ -max=$\frac{1}{4}\leq\int^{1}_{0.5}$ $\frac{1}{1+(x)^2}\leq\frac{2}{5}$ then $\frac{1}{4}+\frac{2}{5}\leq\int{0.5}{0}\frac{1}{1+(x)^2}dx+\int^{1}_{0.5}\frac{1}{1+(x)^2}dx\leq\frac{1}{2}+\frac{2}{5}=\frac{13}{20}\leq\int_{0}^{1}\frac{1}{1+(x)^2}dx\leq\frac{9}{10}$
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