Answer
$ - \frac{4}{3}$
Work Step by Step
$$\eqalign{
& \text{Let }I=\int_{ - 1}^1 {t{{\left( {1 - t} \right)}^2}} dt \cr
& {\text{Expand the binomial}}{\text{, recall that }}{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2} \cr
& I = \int_{ - 1}^1 {t\left( {1 - 2t + {t^2}} \right)} dt \cr
& = \int_{ - 1}^1 {\left( {t - 2{t^2} + {t^3}} \right)} dt \cr
& {\text{Use the power rule for integration}} \cr
& I = \left[ {\frac{{{t^2}}}{2} - \frac{{2{t^3}}}{3} + \frac{{{t^4}}}{4}} \right]_{ - 1}^1 \cr
& {\text{Using the fundamental theorem of calculus}}{\text{, part 2}} \cr
& I= \left[ {\frac{{{{\left( 1 \right)}^2}}}{2} - \frac{{2{{\left( 1 \right)}^3}}}{3} + \frac{{{{\left( 1 \right)}^4}}}{4}} \right] - \left[ {\frac{{{{\left( { - 1} \right)}^2}}}{2} - \frac{{2{{\left( { - 1} \right)}^3}}}{3} + \frac{{{{\left( { - 1} \right)}^4}}}{4}} \right] \cr
& = \frac{1}{{12}} - \frac{{17}}{{12}} \cr
& = - \frac{4}{3} \cr} $$
