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: 26

Answer

$\approx 1.07$

Work Step by Step

$\int^{2}_1 w^{2} \ln (w) dw$ $u = \ln w$ $u' = \frac{1}{w}$ $\frac{du}{dw} = \frac{1}{w}$ $du = \frac{1}{w}dw$ $dv = w^{2}$ $v = \frac{w^{3}}{3}$ $uv - \int vdu$ $= (\ln w)(\frac{w^{3}}{3}) - \int \frac{w^{3}}{3} \frac{1}{w}dw$ $= (\ln w)(\frac{w^{3}}{3}) - \int \frac{1}{3} \frac{w^{3}}{w}dw$ $= (\ln w)(\frac{w^{3}}{3}) - \frac{1}{3} \int\frac{w^{3}}{w}dw$ $= (\ln w)(\frac{w^{3}}{3}) - \frac{1}{3} \int\frac{w^{2}}{1}dw$ $= (\ln w)(\frac{w^{3}}{3}) - \frac{1}{3} \int w^{2}dw$ $= (\ln w)(\frac{w^{3}}{3}) - \frac{1}{3}(\frac{w^{3}}{3}) |^{2}_1$ $= (\ln w)(\frac{w^{3}}{3}) - \frac{w^{3}}{9} |^{2}_1$ $= [(\ln 2)(\frac{2^{3}}{3}) - \frac{2^{3}}{9}] - [(\ln 1)(\frac{1^{3}}{3}) - \frac{1^{3}}{9}] $ $= [(\ln 2)(\frac{8}{3}) - \frac{8}{9}] - [(\ln 1)(\frac{1}{3}) - \frac{1}{9}] $ $= [(\frac{8\ln 2}{3}) - \frac{8}{9}] - [(\frac{\ln 1}{3}) - \frac{1}{9}] $ $= [(\frac{3(8\ln 2)}{9}) - \frac{8}{9}] - [(\frac{3(\ln 1)}{9}) - \frac{1}{9}] $ $= [(\frac{3(8\ln 2)-8}{9}] - [(\frac{3(\ln 1)-1}{9})] $ $= [(\frac{24\ln 2-8}{9})] - [(\frac{3\ln 1-1}{9})] $ $= \frac{24\ln 2-8 -3 \ln 1 + 1}{9}$ $= \frac{24\ln 2-7 -3 \ln 1 }{9}$ $\approx 1.07$
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