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: 80


$\int_0^1 \sec x dx \geq \dfrac{7}{6}$

Work Step by Step

As we are given that $f(x)=\sec x \geq 1+\dfrac{x^2}{2}$ for $x \geq 0$ When the function $f(x)$ has minimum and maximum values defined on $[p,q]$ then the maximum-minimum inequality tells that $min f(q-p) \leq \int_p^q f(x) dx \leq max f(q-p)$ Here, we have $ \int_0^1 \sec x dx \geq \int_0^1 (1+\dfrac{x^2}{2}) dx$ $\implies \int_0^1 \sec x dx \geq \int_0^1 dx +(\dfrac{1}{2}) \int_0^1 x^2 dx$ This gives: $\int_0^1 \sec x dx \geq 1+\dfrac{1}{6}$ Hence, $\int_0^1 \sec x dx \geq \dfrac{7}{6}$
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