Answer
$$A = 4\sqrt 3 $$
Work Step by Step
$$\eqalign{
& y = \frac{x}{{\sqrt {x + 3} }},{\text{ }}y = 0,{\text{ }}x = 6 \cr
& {\text{The area of the region is given by}} \cr
& A = \int_0^6 {\frac{x}{{\sqrt {x + 3} }}} dx \cr
& {\text{Integrate by tables }}\int {\frac{u}{{\sqrt {bu + a} }}du = \frac{{ - 2\left( {2a - bu} \right)}}{{3{b^2}}}\sqrt {a + bu} + C} \cr
& A = \left[ {\frac{{ - 2\left( {2\left( 3 \right) - x} \right)}}{{3{{\left( 1 \right)}^2}}}\sqrt {x + 3} } \right]_0^6 \cr
& A = \left[ {\frac{{ - 2\left( {6 - x} \right)}}{3}\sqrt {x + 3} } \right]_0^6 \cr
& A = - \frac{2}{3}\left[ {\left( {6 - x} \right)\sqrt {x + 3} } \right]_0^6 \cr
& A = - \frac{2}{3}\left[ {\left( {6 - 6} \right)\sqrt {6 + 3} } \right] + \frac{2}{3}\left[ {\left( {6 - 0} \right)\sqrt {0 + 3} } \right] \cr
& A = \frac{2}{3}\left( 6 \right)\sqrt 3 \cr
& A = 4\sqrt 3 \cr} $$
