## Calculus: Early Transcendentals (2nd Edition)

$= 2\sqrt {{e^a}}$
$\begin{gathered} \int_{ - \infty }^a {\sqrt {{e^x}} dx} \hfill \\ \hfill \\ integrating \hfill \\ \hfill \\ \int_{}^{} {\sqrt {{e^x}} dx} = \int_{}^{} {{e^{\frac{x}{2}}}dx} \hfill \\ \hfill \\ = 2{e^{\frac{x}{2}}} + C\,\, = 2\sqrt {{e^x}} + C \hfill \\ \hfill \\ If\,\,f\,\left( x \right)\,\,is\,\,continuous\,\,on\,\,\left( { - \infty ,b} \right]\,\,,\,\,then \hfill \\ \hfill \\ \int_{ - \infty }^b {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{a \to - \infty } \int_a^b {f\,\left( x \right)dx} \hfill \\ \hfill \\ use\,\,Definition\,\,of\,\,improper\,\,integral\,\, \hfill \\ \hfill \\ \int_a^\infty {f\,\left( x \right)} \,dx = \,\,\,\,\,\mathop {\,\lim }\limits_{b \to \infty } \int_0^b {f\,\left( x \right)dx} \hfill \\ \hfill \\ \int_{ - \infty }^a {\sqrt {{e^x}} \,dx} \, = \,\mathop {\lim }\limits_{t \to - \infty } \int_t^a {\sqrt {{e^x}} \,dx} \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ = \mathop {\lim }\limits_{t \to - \infty } \,\,\left[ {2\sqrt {{e^x}} } \right]_t^a \hfill \\ \hfill \\ use\,\,the\,\,ftc \hfill \\ \hfill \\ = 2\sqrt {{e^a}} - 2\sqrt 0 \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 2\sqrt {{e^a}} \hfill \\ \hfill \\ \end{gathered}$