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