Answer
$$ - 20\left( {{e^{0.2}} - {e^{0.3}}} \right) + 3\ln \frac{2}{3}$$
Work Step by Step
$$\eqalign{
& \int_{ - 3}^{ - 2} {\left( {2{e^{ - 0.1y}} + \frac{3}{y}} \right)} dy \cr
& {\text{split the integrand}} \cr
& = \int_{ - 3}^{ - 2} {2{e^{ - 0.1y}}} dy + \int_{ - 3}^{ - 2} {\frac{3}{y}} dy \cr
& {\text{integrate by using }}\int_a^b {{e^{kx}}} dx = \left( {\frac{{{e^{kx}}}}{k}} \right)_a^b{\text{ and }}\int_a^b {\frac{1}{x}} dx = \left( {\ln \left| x \right|} \right)_a^b{\text{ }} \cr
& = - 20\left( {{e^{ - 0.1y}}} \right)_{ - 3}^{ - 2} + 3\left( {\ln \left| y \right|} \right)_{ - 3}^{ - 2} \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = - 20\left( {{e^{ - 0.1\left( { - 2} \right)}} - {e^{ - 0.1\left( { - 3} \right)}}} \right) + 3\left( {\ln \left| { - 2} \right| - \ln \left| { - 3} \right|} \right) \cr
& = - 20\left( {{e^{0.2}} - {e^{0.3}}} \right) + 3\left( {\ln 2 - \ln 3} \right) \cr
& = - 20\left( {{e^{0.2}} - {e^{0.3}}} \right) + 3\ln \frac{2}{3} \cr} $$
