Calculus: Early Transcendentals (2nd Edition)

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

Chapter 7 - Integration Techniques - 7.1 Basic Approaches - 7.1 Exercises - Page 514: 24

Answer

$$\frac{{65}}{2} + \ln \left( {\frac{4}{9}} \right)$$

Work Step by Step

$$\eqalign{ & \int_4^9 {\frac{{{x^{5/2}} - {x^{1/2}}}}{{{x^{3/2}}}}} dx \cr & {\text{split numerator}} \cr & = \int_4^9 {\left( {\frac{{{x^{5/2}}}}{{{x^{3/2}}}} - \frac{{{x^{1/2}}}}{{{x^{3/2}}}}} \right)} dx \cr & {\text{use }}\frac{{{x^m}}}{{{x^n}}} = {x^{m - n}} \cr & = \int_4^9 {\left( {{x^{5/2 - 3/2}} - {x^{1/2 - 3/2}}} \right)} dx \cr & = \int_4^9 {\left( {x - {x^{ - 1}}} \right)} dx \cr & {\text{find the antiderivative}} \cr & = \left. {\left( {\frac{{{x^2}}}{2} - \ln \left| x \right|} \right)} \right|_4^9 \cr & {\text{evaluate limits}} \cr & = \left( {\frac{{{9^2}}}{2} - \ln \left| 9 \right|} \right) - \left( {\frac{{{4^2}}}{2} - \ln \left| 4 \right|} \right) \cr & {\text{simplify}} \cr & = \frac{{81}}{2} - \ln \left| 9 \right| - 8 + \ln \left| 4 \right| \cr & = \frac{{65}}{2} + \ln \left( {\frac{4}{9}} \right) \cr} $$
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