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: 13



Work Step by Step

Solve $\int_{\ln 4}^{\ln 9} e^{x/2} dx$ Let us $p= \dfrac{x}{2}$ and $dp=\dfrac{1}{2}dx$ This implies $\int_{\ln 4}^{\ln 9} e^{x/2} dx= 2\int_{\ln 4}^{\ln 9} e^p dp$ $= 2[e^p]_{\ln 4}^{\ln 9}+c$ $= 2[e^{ \frac{x}{2}}]_{\ln 4}^{\ln 9} +c$ $= 2[e^{ \frac{\ln 9}{2}}-e^{ \frac{\ln 4}{2}}]+c$ Also, $=2[3-2]$ Hence, $\int_{\ln 4}^{\ln 9} e^{x/2} dx=2$
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