Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - Review - Exercises - Page 506: 106

Answer

\[\int_{0}^{1}\sqrt{e^{2x}+1}\,dx\geq e-1\]

Work Step by Step

To show that:- \[\int_{0}^{1}\sqrt{1+e^{2x}}\;dx\geq e-1\] Since on $[0,1]$ \[e^{2x}> 0\] \[\Rightarrow e^{2x}+1> e^{2x}\] \[\Rightarrow \sqrt{e^{2x}+1}> \sqrt{e^{2x}}\] We know that if $f(x)\leq g(x)$ on $[a,b]$ Then \[\int_{a}^{b}f(x)\,dx\leq \int_{a}^{b}g(x)\,dx\] \[\Rightarrow \int_{0}^{1}\sqrt{e^{2x}+1}\,dx>\int_{0}^{1}\sqrt{e^{2x}}\,dx=\int_{0}^{1}e^x \,dx\] \[\Rightarrow \int_{0}^{1}\sqrt{e^{2x}+1}\,dx>\int_{0}^{1}e^x \,dx=\left[e^x\right]_{0}^{1}=e-1\] Hence, \[\int_{0}^{1}\sqrt{e^{2x}+1}\,dx> e-1\]
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