Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 5 - Section 5.3 - The Fundamental Theorem of Calculus - 5.3 Exercises: 34

Answer

$\displaystyle\int\limits_0^3(2\sin x-e^{x})dx=3-2\cos3-e^{3}$

Work Step by Step

$\displaystyle\int\limits_0^3(2\sin x-e^{x})dx$ Integrate each term separately and then apply the second part of the fundamental theorem of calculus to get the answer: $\displaystyle\int\limits_0^3(2\sin x-e^{x})dx=2\int\limits_0^3\sin xdx-\int\limits_0^3e^{x}dx=-2\cos x-e^{x}\Big|_0^3=$ $...=-2\cos3-e^{3}+2\cos0+e^{0}=-2\cos3-e^{3}+2+1=...$ $...=3-2\cos3-e^{3}$
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