## Calculus: Early Transcendentals (2nd Edition)

$= 2\,\left( {e - 1} \right)$
$\begin{gathered} \int_0^1 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}\,dx} \hfill \\ \hfill \\ integration\,\,of\,\,indefinite\,\,integral \hfill \\ \hfill \\ \int_{}^{} {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}} \,dx\,\,\, = \,\,2{e^{\sqrt x }} + C \hfill \\ \hfill \\ use\,\,the\,\,Definition\,\,of\,\,improper\,\,integral\, \hfill \\ \hfill \\ \int_a^b {f\,\left( x \right)} \,dx\,\, = \,\,\mathop {\lim }\limits_{c \to {a^ + }} \int_c^b {f\,\left( x \right)} \,dx \hfill \\ \hfill \\ provided\,\,the\,\,\,limit\,\,exists \hfill \\ \hfill \\ \int_0^1 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}\,dx} = \,\,\mathop {\lim }\limits_{c \to {0^ + }} \int_c^1 {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}\,dx} \hfill \\ \hfill \\ then \hfill \\ \hfill \\ = \,\mathop {\lim }\limits_{c \to {0^ + }} \,\,\,\,\left[ {2{e^{\sqrt x }}} \right]_c^1\,\, \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = \,\mathop {\lim }\limits_{c \to {0^ + }} \,\,\,\,\left[ {2e - {e^0}} \right] \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 2\,\left( {e - 1} \right) \hfill \\ \end{gathered}$