Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 7 - Integration - 7.4 The Fundamental Theorem of Calculus - 7.4 Exercises - Page 395: 18

Answer

$$2\ln 2 + 10\left( {{e^{ - 0.3}} - {e^{ - 0.6}}} \right)$$

Work Step by Step

$$\eqalign{ & \int_{ - 2}^{ - 1} {\left( {\frac{{ - 2}}{t} + 3{e^{0.3t}}} \right)} dt \cr & {\text{integrate by using }}\int_a^b {{e^{kx}}} dx = \left( {\frac{{{e^{kx}}}}{k}} \right)_a^b{\text{ and }}\int_a^b {\frac{1}{x}} dx = \left( {\ln \left| x \right|} \right)_a^b{\text{ }} \cr & = - 2\left( {\ln \left| t \right|} \right)_{ - 2}^{ - 1} + 3\left( {\frac{{{e^{0.3t}}}}{{0.3}}} \right)_{ - 2}^{ - 1} \cr & = - 2\left( {\ln \left| t \right|} \right)_{ - 2}^{ - 1} + 10\left( {{e^{0.3t}}} \right)_{ - 2}^{ - 1} \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = - 2\left( {\ln \left| { - 1} \right| - \ln \left| { - 2} \right|} \right) + 10\left( {{e^{0.3\left( { - 1} \right)}} - {e^{0.3\left( { - 2} \right)}}} \right) \cr & = - 2\left( { - \ln 2} \right) + 10\left( {{e^{ - 0.3}} - {e^{ - 0.6}}} \right) \cr & = 2\ln 2 + 10\left( {{e^{ - 0.3}} - {e^{ - 0.6}}} \right) \cr} $$
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