Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 7 - Section 7.8 - Improper Integrals - 7.8 Exercises - Page 550: 21

Answer

Diverges

Work Step by Step

$$\eqalign{ & \int_{ - \infty }^\infty {\cos 2t} dt \cr & {\text{Using the definition of improper integrals }} \cr & \int_{ - \infty }^\infty {\cos 2t} dt = \mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\cos 2t} dt + \mathop {\lim }\limits_{b \to + \infty } \int_0^b {\cos 2t} dt \cr & {\text{Multiply and divide each integrand by }}\frac{1}{2} \cr & = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \int_a^0 {\cos 2t\left( 2 \right)} dt + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \int_0^b {\cos 2t\left( 2 \right)} dt \cr & {\text{Integrating}} \cr & = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\sin 2t} \right]_a^0 + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin 2t} \right]_0^b \cr & {\text{Evaluate the integration limits}} \cr & = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ {\sin 2\left( 0 \right) - \sin 2\left( a \right)} \right] + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin 2\left( b \right) - \sin 2\left( 0 \right)} \right] \cr & = \frac{1}{2}\mathop {\lim }\limits_{a \to - \infty } \left[ { - \sin \left( {2a} \right)} \right] + \frac{1}{2}\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin \left( {2b} \right)} \right] \cr & {\text{Evaluate the limits when }}a \to - \infty {\text{ and }}b \to \infty \cr & = \frac{1}{2}\underbrace {\mathop {\lim }\limits_{a \to - \infty } \left[ { - \sin \left( {2a} \right)} \right]}_{{\text{the limit does not exist}}} + \frac{1}{2}\underbrace {\mathop {\lim }\limits_{b \to + \infty } \left[ {\sin \left( {2b} \right)} \right]}_{{\text{the limit does not exist}}} \cr & = {\text{the limit does not exist}} \cr & {\text{Therefore}}{\text{, the integral diverges.}} \cr} $$
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