Answer
$ - \frac{{13}}{6}$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^0 {\left( {{x^2} + 5x} \right)} dx \cr
& {\text{Integrate}}{\text{, recall that }}\int {{x^n}dx} = \frac{{{x^{n + 1}}}}{{n + 1}} + C \cr
& = \left[ {\frac{{{x^{2 + 1}}}}{{2 + 1}} + 5\left( {\frac{{{x^{1 + 1}}}}{{1 + 1}}} \right)} \right]_{ - 1}^0 \cr
& = \left[ {\frac{{{x^3}}}{3} + 5\left( {\frac{{{x^2}}}{2}} \right)} \right]_{ - 1}^0 \cr
& = \left[ {\frac{{{x^3}}}{3} + \frac{{5{x^2}}}{2}} \right]_{ - 1}^0 \cr
& {\text{Evaluate the limits and simplify}} \cr
& = \left[ {\frac{{{{\left( 0 \right)}^3}}}{3} + \frac{{5{{\left( 0 \right)}^2}}}{2}} \right] - \left[ {\frac{{{{\left( { - 1} \right)}^3}}}{3} + \frac{{5{{\left( { - 1} \right)}^2}}}{2}} \right] \cr
& = \left[ 0 \right] - \left[ { - \frac{1}{3} + \frac{5}{2}} \right] \cr
& = - \frac{{13}}{6} \cr} $$