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: 78


$\int_a^b f(x) dx \leq 0$

Work Step by Step

When the function $f(x)$ has minimum and maximum values defined on $[a,b]$ then the maximum-minimum inequality tells that $min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ Here, we have the function $f(x) \leq 0$ on $[a,b]$ then $ min f \leq 0$ and $ max f \leq 0$ on $[a,b]$ This implies that $(b-a) \geq0$ and $(b-a) min f \leq 0$ Also, $(b-a) max f \leq 0$ Here, the both sides of the inequality $min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ are positive. Hence, the result is proved that $\int_a^b f(x) dx \leq 0$
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