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: 79

Answer

\[ = 1\]

Work Step by Step

\[\begin{gathered} f\,\left( x \right) = x\sin \,\left( {{x^2}} \right) \hfill \\ f\,\left( x \right) = 0\,\left( {axis - x} \right) \hfill \\ \hfill \\ let\,\,x = 0\,\,\,and\,\,\,x = \sqrt \pi \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ area\,\,\int_0^{\sqrt \pi } {x\sin {x^2}dx} \hfill \\ \hfill \\ or \hfill \\ \hfill \\ = \frac{1}{2}\int_0^{\sqrt \pi } {\sin \,\left( {{x^2}} \right)\,\left( {2x} \right)dx} \hfill \\ \hfill \\ integrate\,\,\,use\,\,\int {\sin u} du = - \cos u + C \hfill \\ \hfill \\ = - \,\,\left[ {\frac{1}{2}\cos \,\left( {{x^2}} \right)} \right]_0^{\sqrt \pi } \hfill \\ \hfill \\ Fundamental\,\,theorem \hfill \\ \hfill \\ = - \,\,\left[ {\frac{1}{2}\cos \,{{\left( {\sqrt \pi } \right)}^2} - \frac{1}{2}\cos \,{0^2}} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = - \,\,\left[ {\frac{1}{2}\,\left( { - 1} \right) - \frac{1}{2}} \right] \hfill \\ \hfill \\ = 1 \hfill \\ \hfill \\ \end{gathered} \]
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