Area Solution 7

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SOLUTION 7:

$\ \$ If

$y= \displaystyle{ (1/2)(e^x+e^{-x}) }$ for

$0 \le x \le \ln 2$ , then

$\displaystyle{ { dy \over dx} = (1/2)(e^x+e^{-x}(-1)) = (1/2)(e^x-e^{-x}) }$ so that

$ARC = \displaystyle{ \int_{0}^{\ln 2} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ 1 + ( (1/2)(e^x-e^{-x}) )^2 } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ 1 + (1/2)^2(e^x-e^{-x})^2 } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ 1 + (1/4)((e^x)^2-2 e^xe^{-x}+(e^{-x})^2 ) } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ 1 + (1/4)(e^{2x}-2 e^0+e^{-2x} ) } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ 1 + (1/4)(e^{2x}-2(1)+e^{-2x} ) } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ 1 + (1/4)e^{2x}-(1/2)+(1/4)e^{-2x} } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ (1/4)e^{2x}+(1/2)+(1/4)e^{-2x} } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ { e^{2x} \over 4} + { 1 \over 2 } + { 1 \over 4e^{2x}} } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ { e^{2x} \over 4}{e^{2x} \over e^{2x}} + { 1 \over 2 }{2e^{2x} \over 2e^{2x}} + { 1 \over 4e^{2x}} } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} \sqrt{ { {e^{4x} + 2e^{2x} + 1 } \over 4e^{2x} } } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} { \sqrt{ e^{4x} + 2e^{2x} + 1} \over \sqrt{4e^{2x} } } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} { \sqrt{ (e^{2x} + 1)^2 } \over \sqrt{ (2e^x)^2} } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} { { e^{2x} + 1 } \over 2e^x } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} { \Big( { e^{2x} \over 2e^x } + { 1 \over 2e^x} \Big) } \ dx }$

$= \displaystyle{ \int_{0}^{\ln 2} { \Big( (1/2)e^x + (1/2)e^{-x} \Big) } \ dx }$

$= \displaystyle{ { \Big( (1/2)e^x + (1/2) {e^{-x} \over -1 } } \Big) \ \Big\vert_{0}^{\ln 2} }$

$= \displaystyle{ { \Big( { e^x \over 2 } - { 1 \over 2e^x } } \Big) \ \Big\vert_{0}^{\ln 2} }$

$= \displaystyle{ \Big( { e^{\ln 2} \over 2 } - { 1 \over 2e^{\ln 2} } \Big) - \Big( { e^{0} \over 2 } - { 1 \over 2e^{0} } \Big) }$ (Recall that

$e^{ln z}= z$ .)

$= \displaystyle{ \Big( { 2 \over 2 } - { 1 \over 2(2) } \Big) - \Big( { 1 \over 2 } - { 1 \over 2(1) } \Big) }$

$= \displaystyle{ 1- { 1 \over 4 } }$

$= \displaystyle{ 3 \over 4 }$

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