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: 50

Answer

$$\frac{1}{{16}}$$

Work Step by Step

$$\eqalign{ & \int_0^{1/4} {\frac{x}{{\sqrt {1 - 16{x^2}} }}dx} \cr & {\text{substitute }}u = 1 - 16{x^2},{\text{ }}du = - 32xdx \cr & {\text{express the limits in terms of }}u \cr & x = 1/4{\text{ implies }}u = 1 - 16{\left( {1/4} \right)^2} = 0 \cr & x = 0{\text{ implies }}u = 1 - 16{\left( 0 \right)^2} = 1 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_0^{1/4} {\frac{x}{{\sqrt {1 - 16{x^2}} }}dx} = \int_1^0 {\frac{{ - 1/32}}{{\sqrt u }}du} \cr & = - \frac{1}{{32}}\int_1^0 {{u^{ - 1/2}}du} \cr & {\text{find the antiderivative}} \cr & = - \frac{1}{{32}}\left. {\left( {\frac{{{u^{1/2}}}}{{1/2}}} \right)} \right|_1^0 \cr & {\text{use the fundamental theorem}} \cr & = - \frac{1}{{16}}\left( {{0^{1/2}} - {1^{1/2}}} \right) \cr & {\text{simplify}} \cr & = \frac{1}{{16}} \cr} $$
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