University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 5 - Section 5.3 - The Definite Integral - Exercises - Page 310: 73


Upper bound =1 and lower bound : $\dfrac{1}{2}$

Work Step by Step

When the function $f(x)$ has a minimum and maximum on $[a,b]$ then, we have $min f(b-a) \leq \int_a^b f(x) dx \leq max f(b-a)$ Given: $f(x)=\dfrac{1}{1+x^2}$ $x^2(x^2-2) \le 0$ Now, we have $\dfrac{1}{2}(1-0) \leq \int_0^1 \dfrac{1}{1+x^2} dx \leq 1(1-0)=\dfrac{1}{2} \leq \int_0^1 \dfrac{1}{1+x^2} dx \leq 1$ Thus, the limits that maximize the integral are: Upper bound =1 and lower bound : $\dfrac{1}{2}$
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