Calculus 8th Edition

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

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 455: 105


(a) $f(x)=e^{x}\geq 1+x$ if $x\geq 0$ (b) $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$

Work Step by Step

(a) As we are given $f(x)=e^{x}\geq 1+x$ if $x\geq 0$ Prove this inequality . Consider $f(x)=e^{x}- (1+x)$ Now, $f'(x)=e^{x}- 1$ $f'(0)=e^{0}- 1=0$ Since, $f(x)$ is increasing on $[0,\infty)$ Therefore, $f(x)\geq 0$ for $x\geq 0$ Thus, $e^{x}- (1+x)\geq 0$ This implies $e^{x}\geq (1+x)$ if $x\geq 0$ (b) Deduce that $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$ Therefore, $f(x)\geq 0$ for $x\geq 0$ $e^{x^{2}}- (1+x^{2})\geq 0$ Thus, $e^{x^{2}}\geq (1+x^{2})$ For $0\leq x\leq 1$ $(1+x^{2})\leq e^{x^{2}}\leq e^{x}$ Further, $\int_{0}^{1} (1+x^{2}) dx\leq \int_{0}^{1} e^{x^{2}}dx \leq \int_{0}^{1} e^{x} dx$ Hence, $\frac{4}{3}\leq \int_{0}^{1} e^{x^{2}} dx\leq e$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.