Answer
\[ = 2\]
Work Step by Step
\[\begin{gathered}
\int\limits_{ - 5}^0 {\frac{{dx}}{{\sqrt {4 - x} }}} \hfill \\
\hfill \\
set\,\,4 - x = u\,\,then - dx = du\,\,\,or\,\,dx = - du \hfill \\
\hfill \\
lower\,\,li\,mit\,\, = \,4 - \,\left( { - 5} \right) = 9 \hfill \\
\hfill \\
upper\,\,\,li\,mit\, = \,4 - 0 = 4 \hfill \\
\hfill \\
integral\,\,becomes \hfill \\
\hfill \\
\int\limits_{ - 5}^0 {\frac{{dx}}{{\sqrt {4 - x} }}} = - \int\limits_9^4 {\frac{{du}}{{\sqrt u }}} \hfill \\
\hfill \\
integrate\,\,\,and\,evalute \hfill \\
\hfill \\
= - 2\,\,\,\left[ {\sqrt u } \right]_9^4 \hfill \\
\hfill \\
= - 2\,\left( {2 - 3} \right) \hfill \\
\hfill \\
{\text{Simplify}} \hfill \\
\hfill \\
= 2 \hfill \\
\end{gathered} \]
