Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 7 - Integration Techniques - 7.8 Improper Integrals - 7.8 Exercises: 74

Answer

\[ = \frac{1}{{1 - p}}\,\,\,for\,\,p < 1\]

Work Step by Step

\[\begin{gathered} \int_0^1 {f\,\left( x \right)} \,dx \hfill \\ \hfill \\ \,find\,\,value\,\,of\,\,p\,\,for\,\,which \hfill \\ \hfill \\ \int_0^1 {f\,\left( x \right)} \,dx{\text{ }}exists \hfill \\ \hfill \\ then \hfill \\ \hfill \\ \int_0^1 {f\,\left( x \right)} \,dx\,\, = \mathop {\lim }\limits_{a \to 0} \,\,\left[ {\frac{{{x^{ - p + 1}}}}{{1 - p}}} \right]_a^1 \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \mathop {\lim }\limits_{a \to 0} \,\,\left[ {\frac{1}{{1 - p}} - \frac{{{a^{ - p + 1}}}}{{1 - p}}} \right] \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \frac{1}{{1 - p}}\,\,\,for\,\,p < 1 \hfill \\ \end{gathered} \]
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