Answer
$$ - \frac{3}{2}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^2 {\left( {5t - 3} \right)} dt \cr
& {\text{integrate by using }}\int {{t^n}} dt = \frac{{{t^{n + 1}}}}{{n + 1}} + C{\text{ and }}\int {dt} = t + C \cr
& = \left[ {5\left( {\frac{{{t^{1 + 1}}}}{{1 + 1}}} \right) - 3\left( t \right)} \right]_{ - 1}^2 \cr
& = \left[ {\frac{{5{t^2}}}{2} - 3t} \right]_{ - 1}^2 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( {\frac{{5{{\left( 2 \right)}^2}}}{2} - 3\left( 2 \right)} \right) - \left( {\frac{{5{{\left( { - 1} \right)}^2}}}{2} - 3\left( { - 1} \right)} \right) \cr
& {\text{simplifying}} \cr
& = \left( {10 - 6} \right) - \left( {\frac{5}{2} + 3} \right) \cr
& = 4 - \frac{{11}}{2} \cr
& = - \frac{3}{2} \cr} $$
