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: 74

Answer

$$ - 410$$

Work Step by Step

$$\eqalign{ & \int_{ - 1}^1 {\left( {x - 1} \right){{\left( {{x^2} - 2x} \right)}^7}dx} \cr & {\text{substitute }}u = {x^2} - 2x,{\text{ }}du = 2\left( {x - 1} \right)dx \cr & {\text{express the limits in terms of }}u \cr & x = - 1{\text{ implies }}u = {\left( { - 1} \right)^2} - 2\left( { - 1} \right) = 3 \cr & x = 1{\text{ implies }}u = {\left( { - 1} \right)^2} - 2\left( { - 1} \right) = - 1 \cr & {\text{the entire integration is carried out as follows}} \cr & \int_{ - 1}^1 {\left( {x - 1} \right){{\left( {{x^2} - 2x} \right)}^7}dx} = \frac{1}{2}\int_3^{ - 1} {{u^7}du} \cr & {\text{find the antiderivative}} \cr & = \frac{1}{2}\left. {\left( {\frac{{{u^8}}}{8}} \right)} \right|_3^{ - 1} \cr & {\text{use the fundamental theorem}} \cr & = \frac{1}{{16}}\left( {{{\left( { - 1} \right)}^8} - {{\left( 3 \right)}^8}} \right) \cr & {\text{simplify}} \cr & = \frac{1}{{16}}\left( {1 - 6561} \right) \cr & = - 410 \cr} $$
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