Calculus: Early Transcendentals 9th Edition

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

Chapter 5 - Problems Plus - Problems - Page 433: 3

Answer

$\int_0^4xe^{(x-2)^4}dx=2k$

Work Step by Step

Let $y=x-2$. Then, $\int_0^4e^{(x-2)^4}dx=\int_{-2}^2e^{y^4}dy$. $\int_0^4xe^{(x-2)^4}dx=\int_{-2}^2(y+2)e^{y^4}dy$ $=\int_{-2}^2ye^{y^4}+2e^{y^4}dy$ (Apply the properties for integrals) $=\int_{-2}^2ye^{y^4}dy+2\int_{-2}^2e^{y^4}dy$ (See that $f(y)=ye^{4}$ is an odd function) $=0+2\int_{-2}^2e^{y^4}dy$ (Since if $f$ is an odd function, $\int_{-a}^af =0$) $=2\int_{-2}^2e^{y^4}dy$ $=2\int_0^4e^{(x-2)^4}dx$ $=2k$ Thus, $\int_0^4xe^{(x-2)^4}dx=2k$
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