Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - 4.5 The Definite Integral - Exercises Set 4.5 - Page 308: 35

Answer

See explanation.

Work Step by Step

By theorem 5.5.6, we then know that: \[ f(x) \leq M \Rightarrow -M+f(x) \leq 0 \] $\int_{a}^{b}[-M+f(x)] d x \leq 0$ $\Rightarrow -\int_{a}^{b} M d x+\int_{a}^{b} f(x) d x \leq 0$ $\Rightarrow \int_{a}^{b} f(x) d x-M(-a+b) \leq 0$ $\Rightarrow \int_{a}^{b} f(x) d x \leq (-a+b)M$ By theorem 5.5.6, we then know that: $f(x) \geq m \Rightarrow -m+f(x) \geq 0$ $\int_{a}^{b}[f(x)-m] d x \geq 0$ $\Rightarrow -\int_{a}^{b} m d x +\int_{a}^{b} f(x) d x\geq 0$ $\Rightarrow -m(-a+b) +\int_{a}^{b} f(x) d x\geq 0$ $\Rightarrow \int_{a}^{b} f(x) d x \geq (-a+b)m$ Combining these two parts, we obtain: $(-a+b)m \leq \int_{a}^{b} f(x) d x \leq (-a+b)M$
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