University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 8 - Section 8.7 - Improper Integrals - Exercises - Page 471: 25



Work Step by Step

Re-arrange the given integral as follows: $\int_{0}^{1} x \ln x dx=[(1/2) (\ln x-(1/2)]_0^1$ ....(1) Then, we have $[(1/2) (\ln x-(1/2)]_0^1=(\dfrac{1}{2}) (1)^2 (\ln 1-\ln (1/2))-(\dfrac{1}{2}) (0)^2 (-\infty-\dfrac{1}{2})$ Thus, equation (1) becomes $\dfrac{1}{2}(0-\dfrac{1}{2})=-\dfrac{1}{4}$
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