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: 75


$\int_0^{1} \sin (x^2) dx$ cannot possibly equal to $2$

Work Step by Step

Since, we have $-1 \leq \sin (x^2) \leq 1$ for all $x \in [0,1]$ Now, we have $ \int_0^{1} f(x) dx \leq \int_0^{1} g(x) dx \leq \int_0^{1} h(x) dx$ But $ \int_0^{1} f(x) dx =-1(1-0)=-1$ and $ \int_0^{1} h(x) dx=1(1-0)=1$ Then, we get $-1 \leq \int_0^{1} \sin (x^2) dx\leq 1$ Therefore, $\int_0^{1} \sin (x^2) dx$ cannot possibly equal $2$
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