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: 76


$2 \sqrt 2 \leq \int_0^{1} \sqrt {x+8} dx \leq 3$

Work Step by Step

When the function $f(x)$ has maximum and minimum values defined on $[a,b]$ then by Maximum-Minimum inequality , we have $min f(b-a) \leq \int_p^q f(x) dx \leq max f(b-a)$ ....(1) Given: $f(x)= \sqrt {x+8} $ defined on $[0,1]$ and $f(0)= \sqrt {0+8}=\sqrt 8=2\sqrt 2 $ and $f(1)= \sqrt {1+8}=\sqrt 9=3$ Equation (1) becomes: $f(0)(1-0) \leq \int_0^{1} f(x) dx \leq f(1) (1-0)$ Hence, $2 \sqrt 2 \leq \int_0^{1} \sqrt {x+8} dx \leq 3$
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