Calculus: Early Transcendentals (2nd Edition)

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

Chapter 5 - Integration - 5.3 Fundamental Theorem of Calculus - 5.3 Exercises: 46

Answer

$$\frac{{13}}{{162}}$$

Work Step by Step

$$\eqalign{ & \int_4^9 {\frac{{x - \sqrt x }}{{{x^3}}}} dt \cr & {\text{Split the numerator}} \cr & = \int_4^9 {\left( {\frac{x}{{{x^3}}} - \frac{{\sqrt x }}{{{x^3}}}} \right)} dt \cr & {\text{Writting }}\sqrt x {\text{ as }}{x^{1/2}} \cr & = \int_4^9 {\left( {\frac{x}{{{x^3}}} - \frac{{{x^{1/2}}}}{{{x^3}}}} \right)} dx \cr & use\,\,the\,\,rule\,\,\,\frac{{{x^u}}}{{{x^v}}} = {x^{u - v}} \cr & = \int_4^9 {\left( {{x^{ - 2}} - {x^{ - 5/2}}} \right)} dx \cr & {\text{using the power rule for integration}} \cr & = \left. {\left( {\frac{{{x^{ - 1}}}}{{ - 1}} - \frac{{{x^{ - 3/2}}}}{{ - 3/2}}} \right)} \right|_4^9 \cr & = \left. {\left( { - \frac{1}{x} + \frac{2}{{3{x^{3/2}}}}} \right)} \right|_4^9 \cr & {\text{Using The Fundamental Theorem}} \cr & = \left( { - \frac{1}{9} + \frac{2}{{3{{\left( 9 \right)}^{3/2}}}}} \right) - \left( { - \frac{1}{4} + \frac{2}{{3{{\left( 4 \right)}^{3/2}}}}} \right) \cr & {\text{simplify}} \cr & = \left( { - \frac{7}{{81}}} \right) - \left( { - \frac{1}{6}} \right) \cr & = - \frac{7}{{81}} + \frac{1}{6} \cr & = \frac{{13}}{{162}} \cr} $$
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