Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.3 Derivatives Of Inverse Functions; Derivatives And Integrals Involving Exponential Functions - Exercises Set 6.3 - Page 433: 87

Answer

$$k = \frac{{\ln 7}}{2}$$

Work Step by Step

$$\eqalign{ & y = {e^{2x}} \cr & {\text{The area is given by}} \cr & A = \int_0^k {{e^{2x}}} dx \cr & {\text{Let }}A = 3 \cr & 3 = \int_0^k {{e^{2x}}} dx \cr & {\text{Integrating}} \cr & 3 = \frac{1}{2}\left[ {{e^{2x}}} \right]_0^k \cr & 3 = \frac{1}{2}\left[ {{e^{2k}} - {e^0}} \right] \cr & {\text{Solve for }}k \cr & 3 = \frac{1}{2}{e^{2k}} - \frac{1}{2} \cr & 6 = {e^{2k}} - 1 \cr & {e^{2k}} = 7 \cr & \ln {e^{2k}} = \ln 7 \cr & 2k = \ln 7 \cr & k = \frac{{\ln 7}}{2} \cr} $$
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