Calculus: Early Transcendentals 9th Edition

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

Chapter 5 - Section 5.4 - Indefinite Integrals and the Net Change Theorem - 5.4 Exercises - Page 416: 44

Answer

$3{e^{\pi /4}} - 4\sqrt 2 + 1$

Work Step by Step

$$\eqalign{ & \text{let }I=\int_0^{\pi /4} {\left( {3{e^x} - 4\sec x\tan x} \right)} dx \cr & {\text{Integrate using basic rules }} \cr & \int {{e^x}} dx = {e^x} + C{\text{ and }}\int {\sec x\tan x} dx = \sec x + C,{\text{ then}} \cr & I = \left[ {3{e^x} - 4\sec x} \right]_0^{\pi /4} \cr & {\text{Using the fundamental theorem of calculus}}{\text{, part 2}} \cr & I = \left[ {3{e^{\pi /4}} - 4\sec \left( {\frac{\pi }{4}} \right)} \right] - \left[ {3{e^0} - 4\sec \left( 0 \right)} \right] \cr & {\text{Simplify}} \cr & I= \left[ {3{e^{\pi /4}} - 4\sqrt 2 } \right] - \left[ {3 - 4\left( 1 \right)} \right] \cr & = 3{e^{\pi /4}} - 4\sqrt 2 + 1 \cr} $$
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