## Calculus with Applications (10th Edition)

$$\frac{{35}}{6}$$
\eqalign{ & \int_{ - 2}^3 {\left( { - {x^2} - 3x + 5} \right)} dx \cr & {\text{integrate by using }}\int {{x^n}} dx = \frac{{{x^{n + 1}}}}{{n + 1}} + C{\text{ and }}\int {dx} = x + C \cr & = \left( { - \frac{{{x^{2 + 1}}}}{{2 + 1}} - 3\left( {\frac{{{x^{1 + 1}}}}{{1 + 1}}} \right) + 5\left( x \right)} \right)_{ - 2}^3 \cr & = \left( { - \frac{{{x^3}}}{3} - \frac{{3{x^2}}}{2} + 5x} \right)_{ - 2}^3 \cr & {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr & = \left( { - \frac{{{{\left( 3 \right)}^3}}}{3} - \frac{{3{{\left( 3 \right)}^2}}}{2} + 5\left( 3 \right)} \right) - \left( { - \frac{{{{\left( { - 2} \right)}^3}}}{3} - \frac{{3{{\left( { - 2} \right)}^2}}}{2} + 5\left( { - 2} \right)} \right) \cr & {\text{simplifying}} \cr & = \left( { - 9 - \frac{{27}}{2} + 15} \right) - \left( {\frac{8}{3} - 6 - 10} \right) \cr & = - \frac{{15}}{2} + \frac{{40}}{3} \cr & = \frac{{35}}{6} \cr}