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: 73

Answer

\[ = - \frac{{32}}{3}\]

Work Step by Step

\[\begin{gathered} \int_0^2 {{x^3}\sqrt {16 - {x^4}} \,dx} \hfill \\ \hfill \\ \,set\,\,u = 16 - {x^4}\,\,\,\,\,then\,\,\,\, - \frac{{du}}{4} = {x^3}dx \hfill \\ \hfill \\ {\text{Changing limits of integration}} \hfill \\ \hfill \\ x = 0\,\,\,\,\,implies\,\,\,\,u = 16 - {0^4} = 16 \hfill \\ x = 2\,\,\,\,\,implies\,\,\,u = 16 - {2^4} = 0 \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ = - \frac{1}{4}\int_{16}^0 {{u^{\frac{1}{2}}}} du \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = - \frac{1}{6}\,\,\left[ {{u^{\frac{3}{2}}}} \right]_{16}^0 \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = - \frac{1}{6}\,\,\left[ {{0^{\frac{3}{2}}} - {{16}^{\frac{3}{2}}}} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = - \frac{1}{6}\,\left( {64} \right) \hfill \\ \hfill \\ = - \frac{{32}}{3} \hfill \\ \end{gathered} \]
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