Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Section 7.1 - Integration by Parts - 7.1 Exercises - Page 476: 5


$$\int te^{-3t}dt=-\frac{1}{3}te^{-3t}-\frac{1}{9}e^{-3t}+C$$

Work Step by Step

$$A=\int te^{-3t}dt$$ We would choose $u=t$ and $dv=e^{-3t}dt$ We could easily see that $du=dt$. For $dv=e^{-3t}dy$, we analyze as follows $$\int e^{-3t}dt=\int\frac{1}{-3}e^{-3t}d(-3t)=-\frac{1}{3}\int e^{-3t}d(-3t)=-\frac{1}{3}e^{-3t}+C$$ Therefore, for $dv=e^{-3t}dt$, $v=-\frac{1}{3}e^{-3t}$ Apply Integration by Parts to A, we have $$A=uv-\int vdu$$ $$A=-\frac{1}{3}te^{-3t}-\int(-\frac{1}{3}e^{-3t})dt$$ $$A=-\frac{1}{3}te^{-3t}+\frac{1}{3}\int e^{-3t}dt$$ $$A=-\frac{1}{3}te^{-3t}+\frac{1}{3}(-\frac{1}{3}e^{-3t}+C)$$ $$A=-\frac{1}{3}te^{-3t}-\frac{1}{9}e^{-3t}+C$$
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