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: 39

Answer

\[ = \frac{7}{2}\]

Work Step by Step

\[\begin{gathered} \int_0^1 {2x\,\left( {4 - {x^2}} \right)dx} \hfill \\ \hfill \\ let\,\,u = 4 - {x^2}\,\,\,\,{\text{which implies that}}\,\,du = - 2xdx \hfill \\ and\,\, - du = 2xdx \hfill \\ \hfill \\ apply\,\,the\,\,\,substitution \hfill \\ \hfill \\ = \int_{}^{} {u\,\left( { - du} \right)} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{{ - {u^2}}}{2} + C \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \frac{{ - \,{{\left( {4 - {x^2}} \right)}^2}}}{2} + C \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \,\,\left[ {\frac{{\,{{\left( {4x - {x^2}} \right)}^2}}}{2}} \right]_1^0 \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = \frac{{\,{{\left( {4 - {0^2}} \right)}^2}}}{2} - \frac{{\,{{\left( {4 - {1^2}} \right)}^2}}}{2} \hfill \\ \hfill \\ = \frac{7}{2} \hfill \\ \end{gathered} \]
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