Answer
$$A = \frac{{128}}{{15}}$$
Work Step by Step
$$\eqalign{
& {\text{The area of the region is given by}} \cr
& A = \int_0^4 {x\sqrt {4 - x} } dx \cr
& {\text{Let }}u = 4 - x,{\text{ }}x = 4 - u,{\text{ }}dx = - du \cr
& \int {x\sqrt {4 - x} dx = - \int {\left( {4 - u} \right)\sqrt u du} } \cr
& = \int {\left( {u - 4} \right){u^{1/2}}du} \cr
& = \int {\left( {{u^{3/2}} - 4{u^{1/2}}} \right)du} \cr
& {\text{Integrate}} \cr
& = \frac{2}{5}{u^{5/2}} - \frac{8}{3}{u^{3/2}} + C \cr
& = \frac{2}{5}{\left( {4 - x} \right)^{5/2}} - \frac{8}{3}{\left( {4 - x} \right)^{3/2}} + C \cr
& {\text{Therefore,}} \cr
& A = \left[ {\frac{2}{5}{{\left( {4 - x} \right)}^{5/2}} - \frac{8}{3}{{\left( {4 - x} \right)}^{3/2}}} \right]_0^4 \cr
& {\text{Evaluate}} \cr
& A = \left[ {\frac{2}{5}{{\left( {4 - 4} \right)}^{5/2}} - \frac{8}{3}{{\left( {4 - 4} \right)}^{3/2}}} \right] - \left[ {\frac{2}{5}{{\left( {4 - 0} \right)}^{5/2}} - \frac{8}{3}{{\left( {4 - 0} \right)}^{3/2}}} \right] \cr
& A = - \left[ {\frac{2}{5}\left( {32} \right) - \frac{8}{3}\left( 8 \right)} \right] \cr
& A = - \left( {\frac{{64}}{5} - \frac{{64}}{3}} \right) \cr
& A = \frac{{128}}{{15}} \cr} $$
