Answer
$$ - \frac{{116}}{{15}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 3}^0 {x\sqrt {1 - x} } dx \cr
& u = 1 - x,{\text{ then }}du = - dx \cr
& {\text{With this substitution}}{\text{,}} \cr
& {\text{if }}x = 0,{\text{ }}u = 1 - 0 = 1 \cr
& {\text{if }}x = - 3,{\text{ }}u = 1 - \left( { - 3} \right) = 4 \cr
& {\text{so}} \cr
& \int_{ - 3}^0 {x\sqrt {1 - x} } dx = \int_4^1 {\left( {1 - u} \right)} \sqrt u \left( { - du} \right) \cr
& = - \int_4^1 {\left( {{u^{1/2}} - {u^{3/2}}} \right)} du \cr
& {\text{find the antiderivative by the power rule}} \cr
& = - \left( {\frac{{{u^{3/2}}}}{{3/2}} - \frac{{{u^{5/2}}}}{{5/2}}} \right)_4^1 \cr
& = - \left( {\frac{2}{3}{u^{3/2}} - \frac{2}{5}{u^{5/2}}} \right)_4^1 \cr
& {\text{part 1 of fundamental theorem of calculus}} \cr
& = - \left( {\frac{2}{3}{{\left( 1 \right)}^{3/2}} - \frac{2}{5}{{\left( 1 \right)}^{5/2}}} \right) - \left( {\frac{2}{3}{{\left( 4 \right)}^{3/2}} - \frac{2}{5}{{\left( 4 \right)}^{5/2}}} \right) \cr
& = - \left( {\frac{2}{3} - \frac{2}{5}} \right) + \left( {\frac{{16}}{3} - \frac{{64}}{5}} \right) \cr
& = - \frac{4}{{15}} - \frac{{112}}{{15}} \cr
& = - \frac{{116}}{{15}} \cr} $$
