Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 5 - Integration - 5.5 Substitution Rule - 5.5 Exercises: 69

Answer

$$\frac{1}{3}$$

Work Step by Step

$$\eqalign{ & \int_0^1 {x\sqrt {1 - {x^2}} } dx \cr & {\text{substitute }}u = 1 - {x^2},{\text{ }}du = - 2xdx \cr & {\text{express the limits in terms of }}u \cr & x = 0{\text{ implies }}u = 1 - {\left( 0 \right)^2} = 1 \cr & x = 1{\text{ implies }}u = 1 - {\left( 1 \right)^2} = 0 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^1 {x\sqrt {1 - {x^2}} } dx = - \frac{1}{2}\int_1^0 {{u^{1/2}}du} \cr & {\text{find the antiderivative}} \cr & = - \frac{1}{2}\left. {\left( {\frac{{{u^{3/2}}}}{{3/2}}} \right)} \right|_1^0 \cr & {\text{use the fundamental theorem}} \cr & = - \frac{1}{3}\left( {{0^{1/2}} - {1^{1/2}}} \right) \cr & {\text{simplify}} \cr & = \frac{1}{3} \cr} $$
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