Calculus: Early Transcendentals 8th Edition

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

Chapter 7 - Review - Exercises - Page 537: 6


$\displaystyle \frac{32\ln 2}{3}-\frac{7}{4}$

Work Step by Step

$\left[\begin{array}{lll} u=\ln x & e^{u}=x & x^{6}=e^{6u}\\ du=\frac{1}{x}dx && \\ x\in[1,2] & \Rightarrow u\in[0,\ln 2] & \end{array}\right]$ $\displaystyle \int_{1}^{2}x^{5}\ln xdx=\int_{1}^{2}x^{6}\ln x\cdot\frac{1}{x}dx=\int_{0}^{\ln 2}ue^{6u}du$ by parts, $\displaystyle \left[\begin{array}{ll} r=u & ds=e^{6u}du\\ dr=du & s=\frac{1}{6}e^{6u} \end{array}\right],\quad\int_{0}^{\ln 2}rds=rs|_{0}^{\ln 2}-\int_{0}^{\ln 2}sdr$ $=\left[\dfrac{ue^{6u}}{6}\right]_{0}^{\ln 2}-\left[\dfrac{e^{6u}}{36}\right]_{0}^{\ln 2}$ $=\displaystyle \frac{2^{6}\cdot\ln 2}{6}-(\frac{2^{6}}{36}-\frac{1}{36})$ $=\displaystyle \frac{32\ln 2}{3}-\frac{63}{36}$ $=\displaystyle \frac{32\ln 2}{3}-\frac{7}{4}$
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