Calculus: Early Transcendentals 9th Edition

Published by Cengage Learning
ISBN 10: 1337613924
ISBN 13: 978-1-33761-392-7

Chapter 5 - Section 5.5 - The Substitution Rule - 5.5 Exercises - Page 425: 10

Answer

$- \frac{{{{\left( {5 - 3x} \right)}^{11}}}}{{33}} + C$

Work Step by Step

$$\eqalign{ & \int {{{\left( {5 - 3x} \right)}^{10}}} dx \cr & {\text{Let }}u = 5 - 3x,{\text{ then }}du = - 3dx,{\text{ }}dx = - \frac{1}{3}du \cr & {\text{Applying the substitution}}{\text{, we obtain}} \cr & \int {{{\left( {5 - 3x} \right)}^{10}}} dx = \int {{u^{10}}} \left( { - \frac{1}{3}} \right)du \cr & = - \frac{1}{3}\int {{u^{10}}} du \cr & {\text{Integrate apply the power rule }}\int {{u^n}du} = \frac{{{u^{n + 1}}}}{{n + 1}}{\text{ + C }} \cr & = - \frac{1}{3}\left( {\frac{{{u^{11}}}}{{11}}} \right) + C \cr & = - \frac{{{u^{11}}}}{{33}} + C \cr & {\text{Write in terms of }}x,{\text{ substitute }}u = 5 - 3x \cr & = - \frac{{{{\left( {5 - 3x} \right)}^{11}}}}{{33}} + C \cr} $$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.