Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.2 - The Definite Integral - 5.2 Exercises - Page 390: 57


$2 \leq \int_{-1}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$

Work Step by Step

Note the following: $\int_{-1}^{1} \sqrt{1+x^2}~dx = \int_{-1}^{0} \sqrt{1+x^2}~dx+\int_{0}^{1} \sqrt{1+x^2}~dx = 2\int_{0}^{1} \sqrt{1+x^2}~dx$ On the interval $0 \leq x \leq 1$: $0 \leq x^2 \leq 1$ $1 \leq 1+x^2 \leq 2$ $\sqrt{1} \leq \sqrt{1+x^2} \leq \sqrt{2}$ $1 \leq \sqrt{1+x^2} \leq \sqrt{2}$ Therefore, by the Comparison Property 8: $1 \leq \int_{0}^{1} \sqrt{1+x^2}~dx \leq \sqrt{2}$ $2 \leq 2\int_{0}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$ $2 \leq \int_{-1}^{1} \sqrt{1+x^2}~dx \leq 2\sqrt{2}$
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