Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.4 The Fundamental Theorem of Calculus - 7.4 Exercises - Page 395: 21

Answer

$$\frac{{91}}{3}$$

Work Step by Step

$$\eqalign{ & \int_{ - 1}^0 {y{{\left( {2{y^2} - 3} \right)}^5}} dy \cr & {\text{use substitution}}{\text{. Let }}u = 2{y^2} - 3,{\text{ so that }}du = 4ydp,\,\,\,\,\frac{1}{4}du = ydy \cr & {\text{the new limits on }}u{\text{ are found as follows}} \cr & \,\,\,\,\,\,{\text{If }}y = 4,{\text{ then }}u = 2{\left( 0 \right)^2} - 3 = - 3 \cr & \,\,\,\,\,\,{\text{If }}y = 1,{\text{ then }}u = 2{\left( { - 1} \right)^2} - 3 = - 1 \cr & {\text{Then}} \cr & \int_{ - 1}^0 {y{{\left( {2{y^2} - 3} \right)}^5}} dy = \int_{ - 1}^{ - 3} {{u^5}} \left( {\frac{1}{4}du} \right) \cr & = \frac{1}{4}\int_{ - 1}^{ - 3} {{u^5}} du \cr & {\text{use }}\int_a^b {{u^n}} du = \left( {\frac{{{u^{n + 1}}}}{{n + 1}}} \right)_a^b \cr & = \frac{1}{4}\left( {\frac{{{u^6}}}{6}} \right)_{ - 1}^{ - 3} \cr & = \frac{1}{{24}}\left( {{u^6}} \right)_{ - 1}^{ - 3} \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = \frac{1}{{24}}\left( {{{\left( { - 3} \right)}^6} - {{\left( { - 1} \right)}^{ - 6}}} \right) \cr & = \frac{1}{{24}}\left( {729 - 1} \right) \cr & = \frac{{728}}{{24}} \cr & = \frac{{91}}{3} \cr} $$
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