Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.4* General Logarithmic and Exponential Functions - 6.4* Exercise - Page 464: 51

Answer

$$\frac{16}{5 \ln 5}-\frac{1}{2 \ln 2}$$

Work Step by Step

Given $$ y=2^x,\ \ y=5^x ,\ \ \ x=-1,\ \ x=1 $$ first, we find the intersection points \begin{aligned} 2^x&= 5^x\\ x\ln(2)&= x\ln (5)\\ x(\ln (5)-\ln (2))&=0\\ x&=0 \end{aligned} Since \begin{aligned} 2^x&\geq 5^x,\ \ \ \ \ -1\leq x\leq 0\\ 2^x&\leq 5^x,\ \ \ \ \ 0\leq x\leq 1 \end{aligned} Then area given \begin{aligned} A &=\int_{-1}^0\left(2^x-5^x\right) d x+\int_0^1\left(5^x-2^x\right) d x\\ &=\left[\frac{2^x}{\ln 2}-\frac{5^x}{\ln 5}\right]_{-1}^0+\left[\frac{5^x}{\ln 5}-\frac{2^x}{\ln 2}\right]_0^1 \\ &=\left(\frac{1}{\ln 2}-\frac{1}{\ln 5}\right)-\left(\frac{1 / 2}{\ln 2}-\frac{1 / 5}{\ln 5}\right)+\left(\frac{5}{\ln 5}-\frac{2}{\ln 2}\right)-\left(\frac{1}{\ln 5}-\frac{1}{\ln 2}\right) \\ &=\frac{16}{5 \ln 5}-\frac{1}{2 \ln 2} \end{aligned}
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