Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Review - Exercises - Page 430: 58

Answer

$\frac{1}{33}\leq \int_2^4\frac{1}{x^3+2}dx\leq \frac{1}{5}$

Work Step by Step

Property 8 If $m\leq f(x)\leq M$ for $a\leq x\leq b$, then $m(b-a)\leq \int_a^bf(x)dx\leq M(b-a)$ Given $f(x)=\frac{1}{x^3+2}$ Find $m$ and $M$: Notice that $2\leq x\leq 4$ $\Rightarrow 2^3\leq x^3\leq 4^3$ $\Rightarrow 2^3+2\leq x^3+2\leq 4^3+2$ $\Rightarrow \frac{1}{4^3+2}\leq \frac{1}{x^3+2}\leq \frac{1}{2^3+2}$ $\Rightarrow \frac{1}{66}\leq f(x)\leq \frac{1}{10}$ $\therefore m=\frac{1}{66}$ and $M=\frac{1}{10}$ Now, estimate the integral using the Property 8: $m(b-a)\leq \int_a^bf(x)dx\leq M(b-a)$ $\frac{1}{66}(4-2)\leq \int_2^4\frac{1}{x^3+2}dx\leq \frac{1}{10}(4-2)$ $\frac{1}{33}\leq \int_2^4\frac{1}{x^3+2}dx\leq \frac{1}{5}$
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