University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 7 - Section 7.1 - The Logarithm Defined as an Integral - Exercises - Page 401: 9



Work Step by Step

Solve $\int_{\ln (2)}^{\ln (3)} e^x dx$ This implies $\int_{\ln (2)}^{\ln (3)} e^x dx=[e^x]_{\ln (2)}^{\ln (3)}$ Thus, $\int_{\ln (2)}^{\ln (3)} e^x dx=[e^{ln (3)}-e^{ln (2)}]$ or, $\int_{\ln (2)}^{\ln (3)} e^x dx=3-2$ Hence, $\int_{\ln (2)}^{\ln (3)} e^x dx=1$
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