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: 93

Answer

$$4\ln 2 + \frac{3}{2}$$

Work Step by Step

$$\eqalign{ & \int_1^2 {\frac{{{z^2} + 4}}{z}dz} \cr & {\text{split the numerator}} \cr & = \int_1^2 {\left( {\frac{{{z^2}}}{z} + \frac{4}{z}} \right)dz} \cr & = \int_1^2 {\left( {z + \frac{4}{z}} \right)dz} \cr & {\text{integrate using the logarithmic rule and the power rule for integration}} \cr & = \left. {\left( {\frac{{{z^2}}}{2} + 4\ln \left| z \right|} \right)} \right|_1^2 \cr & {\text{using the fundamental theorem}} \cr & = \left( {\frac{{{{\left( 2 \right)}^2}}}{2} + 4\ln \left| 2 \right|} \right) - \left( {\frac{{{{\left( 1 \right)}^2}}}{2} + 4\ln \left| 1 \right|} \right) \cr & {\text{simplify}} \cr & = \left( {\frac{{{{\left( 2 \right)}^2}}}{2} + 4\ln \left| 2 \right|} \right) - \left( {\frac{{{{\left( 1 \right)}^2}}}{2} + 4\ln \left| 1 \right|} \right) \cr & = \left( {2 + 4\ln 2} \right) - \left( {\frac{1}{2} + 0} \right) \cr & = 4\ln 2 + \frac{3}{2} \cr} $$
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