Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 5 - Logarithmic, Exponential, and Other Transcendental Functions - 5.4 Exercises: 94

Answer

$-\frac{1}{3}e^{1-3x}+C$

Work Step by Step

$\int e^{1-3x}dx$ let $u=1-3x$ $du=-3dx$ $-\frac{1}{3}du=dx$ $\int e^{1-3x}dx$ $=-\frac{1}{3} \int e^u du$ $=-\frac{1}{3}e^u+C$ $=-\frac{1}{3}e^{1-3x}+C$
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