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: 79

Answer

$\dfrac{1}{2}$

Work Step by Step

When the function $f(x)$ has minimum and maximum values defined on $[p,q]$ then the maximum-minimum inequality tells that $min f(q-p) \leq \int_p^q f(x) dx \leq max f(q-p)$ As we are given that $f(x)=\sin x \leq x$ for $x \geq 0$ $\implies \sin x-x \leq 0$ for $x \geq 0$ or, $ \int_0^1 (\sin x-x) dx \leq 0$ This implies that $\int_0^1 \sin x dx \leq \int_0^1 x dx$ Thus, $\int_0^1 \sin x dx \leq \dfrac{1}{2}$ So, we get the upper bound: $\dfrac{1}{2}$
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