Answer
$$\frac{5}{{81}}$$
Work Step by Step
$$\eqalign{
& \int_2^3 {\left( {3{x^{ - 3}} - 5{x^{ - 4}}} \right)} dx \cr
& {\text{integrate by using }}\int_a^b {{x^n}} dx = \left( {\frac{{{x^{n + 1}}}}{{n + 1}}} \right)_a^b \cr
& = \left( {3\left( {\frac{{{x^{ - 3 + 1}}}}{{ - 3 + 1}}} \right) - 5\left( {\frac{{{x^{ - 4 + 1}}}}{{ - 4 + 1}}} \right)} \right)_2^3 \cr
& {\text{simplifying}} \cr
& = \left( {3\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} \right) - 5\left( {\frac{{{x^{ - 3}}}}{{ - 3}}} \right)} \right)_2^3 \cr
& = \left( { - \frac{3}{2}\left( {{x^{ - 2}}} \right) + \frac{5}{3}\left( {{x^{ - 3}}} \right)} \right)_2^3 \cr
& {\text{use }}\frac{1}{{{x^a}}} = {x^{ - a}} \cr
& = \left( { - \frac{3}{{2{x^2}}} + \frac{5}{{3{x^3}}}} \right)_2^3 \cr
& {\text{use fundamental theorem of calculus }}\int_a^b {f\left( x \right)} dx = F\left( b \right) - F\left( a \right).\,\,\,\,\left( {{\text{see page 388}}} \right) \cr
& = \left( { - \frac{3}{{2{{\left( 3 \right)}^2}}} + \frac{5}{{3{{\left( 3 \right)}^3}}}} \right) - \left( { - \frac{3}{{2{{\left( 2 \right)}^2}}} + \frac{5}{{3{{\left( 2 \right)}^3}}}} \right) \cr
& = \left( { - \frac{1}{6} + \frac{5}{{81}}} \right) - \left( { - \frac{3}{8} + \frac{5}{{24}}} \right) \cr
& = - \frac{{17}}{{162}} + \frac{1}{6} \cr
& = \frac{5}{{81}} \cr} $$
