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: 110


(a) $e^{x}\geq (1+x+\frac{1}{2}x^{2})$ for $x\geq 0$ (b) $\int_{0}^{1} e^{x^{2}} dx=\frac{43}{30}$

Work Step by Step

(a) As we are given $f(x)=e^{x}\geq 1+x+\frac{1}{2}x^{2}$ if $x\geq 0$ Prove this inequality . Consider $f(x)=e^{x}-[1+x+\frac{1}{2}x^{2}]$ Now, $f'(x)=e^{x}- (1+x)$ $f'(0)=e^{0}- 1=0$ Since, $f(x)$ is increasing on $[0,\infty)$ Thus, $e^{x}\geq (1+x+\frac{1}{2}x^{2})$ if $x\geq 0$ (b) Deduce that $\int_{0}^{1} e^{x^{2}} dx$ Using part (a) which is $e^{x^{2}}\geq (1+x^{2}+\frac{1}{2}x^{4})$ for $x\geq 0$, we have $\int_{0}^{1} e^{x^{2}} dx=\int_{0}^{1}(1+x^{2}+\frac{1}{2}x^{4}) dx$ $= \int_{0}^{1}(x+\frac{x^{3}}{3}+\frac{x^{5}}{10}) dx$ $=1+\frac{1}{3}+\frac{1}{10}$ Hence, $\int_{0}^{1} e^{x^{2}} dx=\frac{43}{30}$
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