University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 310: 76


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

Work Step by Step

When the function $f(x)$ has minimum and maximum on $[a,b]$ then, we have $min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ We are given that $f(x)= \sqrt {x+8} $ and $f(0)= \sqrt {0+8}=\sqrt 8=2\sqrt 2 $; $f(1)= \sqrt {1+8}=\sqrt 9=3$ Now, we have $f(0)(1-0) \leq \int_0^{1} f(x) dx \leq f(1) (1-0)$ Thus, $2 \sqrt 2 \leq \int_0^{1} \sqrt {x+8} dx \leq 3$
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