Answer
$$\frac{4}{{15}}$$
Work Step by Step
$$\eqalign{
& \int_0^1 {x\sqrt {1 - x} } dx \cr
& {\text{integrate by substitution: set }}u = 1 - x,\,\,\,\,\,x = 1 - u,\,\,\,\,dx = - du \cr
& {\text{write in terms of }}u \cr
& \int {x\sqrt {1 - x} } dx = \int {\left( {1 - u} \right)\sqrt u \left( { - du} \right)} \cr
& = \int {\left( {u - 1} \right){u^{1/2}}} du \cr
& = \int {\left( {{u^{3/2}} - {u^{1/2}}} \right)} du \cr
& {\text{integrating}} \cr
& = \frac{{{u^{5/2}}}}{{5/2}} - \frac{{{u^{3/2}}}}{{3/2}} + C \cr
& = \frac{2}{5}{u^{5/2}} - \frac{2}{3}{u^{3/2}} + C \cr
& {\text{write in terms of }}x \cr
& = \frac{2}{5}{\left( {1 - x} \right)^{5/2}} - \frac{2}{3}{\left( {1 - x} \right)^{3/2}} + C \cr
& {\text{then}}{\text{,}} \cr
& \int_0^1 {x\sqrt {1 - x} } dx = \left( {\frac{2}{5}{{\left( {1 - x} \right)}^{5/2}} - \frac{2}{3}{{\left( {1 - x} \right)}^{3/2}}} \right)_0^1 \cr
& = \left( {\frac{2}{5}{{\left( {1 - 1} \right)}^{5/2}} - \frac{2}{3}{{\left( {1 - 1} \right)}^{3/2}}} \right) - \left( {\frac{2}{5}{{\left( {1 - 0} \right)}^{5/2}} - \frac{2}{3}{{\left( {1 - 0} \right)}^{3/2}}} \right) \cr
& = \left( 0 \right) - \left( {\frac{2}{5} - \frac{2}{3}} \right) \cr
& = \frac{4}{{15}} \cr} $$
