Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 5 - The Integral - 5.2 The Definite Integral - Exercises - Page 246: 76


$$\frac{1}{3} \leq \int_{4}^{6} \frac{1}{x} d x \leq \frac{1}{2}$$

Work Step by Step

Since $ f(x)=\dfrac{1}{x} $ has a maximum value on $[4,6] $, then $ M=f(4)=\frac{1}{4}$; also, $f(x)$ has a minimum value on $[4,6]$, so $m=f(6)=\frac{1}{6}$: $$m(b-a) \leq \int_{a}^{b} f(x) d x \leq M(b-a)$$ Hence, for $ a=4,\ b=6 $ \begin{aligned} \frac{1}{6}(6-4) & \leq \int_{4}^{6} \frac{1}{x} d x \leq \frac{1}{4}(6-4) \\ \frac{1}{3} & \leq \int_{4}^{6} \frac{1}{x} d x \leq \frac{1}{2} \end{aligned}
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