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: 73


upper bound=1 and lower bound =$\frac{1}{2}$

Work Step by Step

$f(x)=\frac{1}{1-x^2}$ is decreasing on[0,1] -maximum value that f occurs at 0 =>max f=f(01 -minimum value of f occurs at 1 =>min f=f(1)=$\frac{1}{1+1}=\frac{1}{2}$ -therefore(1-0) min f $\leq\int^{1}_{0}\frac{1}{1-x^2}dx\leq(1-0)$ -max f $\frac{1}{2}\leq\int^{1}_{0}\frac{1}{1+x^2}dx\leq1$ -that is an upper bound=1 and -a lower bound =$\frac{1}{2}$
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