Answer
$$12\sqrt 2 $$
Work Step by Step
$$\eqalign{
& \int_0^8 {\frac{3}{{\sqrt {8 - x} }}} dx \cr
& \frac{3}{{\sqrt {8 - x} }}{\text{ has an infinite discontinuity at }}x = 8,{\text{ so we can write}} \cr
& \int_0^8 {\frac{3}{{\sqrt {8 - x} }}} dx = \mathop {\lim }\limits_{b \to {8^ - }} \int_0^b {\frac{3}{{\sqrt {8 - x} }}} dx \cr
& {\text{Integrate}} \cr
& = - 3\mathop {\lim }\limits_{b \to {8^ - }} \left[ {2\sqrt {8 - x} } \right]_0^b \cr
& = - 6\mathop {\lim }\limits_{b \to {8^ - }} \left[ {\sqrt {8 - x} } \right]_0^b \cr
& = - 6\mathop {\lim }\limits_{b \to {8^ - }} \left[ {\sqrt {8 - b} - \sqrt {8 - 0} } \right] \cr
& = - 6\mathop {\lim }\limits_{b \to {8^ - }} \left[ {\sqrt {8 - b} - 2\sqrt 2 } \right] \cr
& {\text{Evaluate limits}} \cr
& = - 6\left( {\sqrt {8 - 8} - 2\sqrt 2 } \right) \cr
& = 12\sqrt 2 \cr
& {\text{The following graph confirms the result}} \cr} $$
