## Calculus: Early Transcendentals (2nd Edition)

$L = \sqrt 5 - \sqrt 2 + \ln \,\left| {\frac{{2 + 2\sqrt 2 }}{{1 + \sqrt 5 }}} \right|$
$\begin{gathered} find\,\,the\,length\,\,of\,\,the\,\,curve \hfill \\ Let\,\,y = {e^x}\,\,on\,\,the\,\,interval\,\,\,\,\left[ {0,\ln 2} \right] \hfill \\ the\,\,length\,of\,the\,\,curve\,\,y = \,\left( x \right)\,\,is \hfill \\ \hfill \\ \,L = \int_a^b {\sqrt {1 + {{y'}^2}} } \,dx \hfill \\ \hfill \\ y = {e^x}\,\,\,,then\,\,\,differentiating\,\,\,\,\,y' = {e^x} \hfill \\ \hfill \\ L = \int_0^{\ln 2} {\sqrt {1 + {e^{2x}}} dx} = \int_1^2 {\frac{{\sqrt {1 + {u^2}} }}{u}} \,du \hfill \\ \hfill \\ where\,\,u = {e^x} \hfill \\ \hfill \\ Apply\,\,the\,\,formula \hfill \\ \hfill \\ L = \int_{}^{} {\frac{{\sqrt {{a^2} + {x^2}} }}{x}dx} \, = \sqrt {{a^2} + {x^2}} - \ln \left| {\frac{{a + \sqrt {{a^2} + {x^2}} }}{x}} \right| + C \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ L = \int_1^2 {\frac{{\sqrt {1 + {u^2}} }}{u}} \,du = \,\,\left[ {\sqrt {1 + {u^2}} - \ln \left| {\frac{{1 + \sqrt {1 + {u^2}} }}{u}} \right|} \right]_1^2 \hfill \\ \hfill \\ evaluate\,\,the\,\,{\text{limits}}\,\,and\,\,simplify \hfill \\ \hfill \\ L = \sqrt 5 - \ln \,\left| {\frac{{1 + \sqrt 5 }}{2}} \right| - \sqrt 2 + \ln \,\left( {1 + \sqrt 2 } \right) \hfill \\ \hfill \\ L = \sqrt 5 - \sqrt 2 + \ln \,\left| {\frac{{2 + 2\sqrt 2 }}{{1 + \sqrt 5 }}} \right| \hfill \\ \end{gathered}$